SOLU Introducción a La Probabilidad y Estadística, 12ma Edición - W. Mendenhall, R. Beaver

8/17/2019 SOLU Introducción a La Probabilidad y Estadística, 12ma Edición - W. Mendenhall, R. Beaver

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S T U D E N T S O L U T I O N S M A N U A LF O R M E N D E N H A L L , B E A V E R , A N D B E A V E R ’ S

I N T R O D U C T I O N T O

P R O B A B I L I T Y

S T A T I S T I C S

http://www.solucionarios.net/








Duxbury

Student Solutions Manual

for

Mendenhall, Beaver, and Beaver's

tro d u ctio n to P ro b a b ility and Statistics

Twelfth Edition

Barbara M. BeaverUn iversity o f California, Riverside

T H O M S O N

* — -------

B R O O K S / C O L E

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Table of Contents

1: Describing Da ta with Graphs

2: De scribing Data with Nu merical M easures

3: Describing Biva riate Data

4: Prob ability and Probability Distributions

5: Several Usefu l Discrete Distributions

6: The Norm al Probability Distribution

7: Sam pling Distributions

8: Large-S am ple Estimation

9: Large-S am ple Tests of Hypotheses10: Inference from Small Sam ples

11: Th e Analysis o f Variance

12: Linear Regression and Correlation

13: M últiple Regression Analysis

14: Analysis o f Categ orical Data

15: N on parametric Statistics

1

9

23

31

43

53

65

71

81

91

107

119

135

141

151



1: Describing Data with Graphs

a The experime ntal unit , the individual or object on which a v ariable is measured,

is the student.

b The ex perimental u nit on w hich the nu m ber of errors is me asured is the exam.

c The experimental unit is the patient.

d The experimental unit is the azalea plant.

e The experime ntal unit is the car.

The popu lat ion of interest consists o f voter opinions (for or against the candidate) at

the time of the election for all perso ns voting in the election. No te that when a sample

is taken (at some t im e p rior or the elect ion), we are not actual ly sam pling from the popu la tion o f in te re st. A s tim e passes, vote r opin io ns change. H enee, the popula tion

o f voter opinions chan ges w ith t ime, and the sample may not be representative of the

popu la tion o f in te re st.

a The variable “reading score ” is a quanti tat ive variable, which is probably

integer-valued and henee discrete.

b Th e individual on wh ich the variable is m easured is the student.

c Th e pop ulation is hypo thetical - i t does not exist in fact - but consists of the

reading scores for all students who could possibly be taught by this method.

a The percentages given in the exercise only add to 94% . W e should add another

category called “Other”, which will account for the other 6 % o f the responses.

b Eithe r type of cha rt is appropriate. Since the data is already p resented as

perc enta ges o f th e w hole g ro up. we choose to use a p ie chart, sh ow n in the fi gure

belo w .

c Answers will vary.

1



1.15 a Th e total percentage of responses given in the table is only

(40 + 34 + 19)% = 93% . Henee there are 7% o f the opinions not record ed. w hichshould go into a category cal led “O ther” or “More than a few days” .

b Yes. The bars are very cióse to the correct proportions.

c Similar to previou s exercises. The pie chart is shown below. The ba r cha rt is

1.21 a Since the variable of interest can only take the valúes 0, 1, or 2, the classe s can

be cho sen as th e in te ger valú es 0 , 1, and 2. The ta ble belo w show s th e c la sses, th eir

corresp ond ing frequencies and their relative frequencies. Th e relative frequeney

histogram is shown below. ___________ _______________________

V alué Frequeney R elative Frequeney

0 5 .25

1 9 .452 6 .30

0.5

b Using the table in part a, the proport ion of measurements greater then 1 is the

same as the p roport ion o f “2”s, or 0.30.

2

c Th e propo rt ion o f m easurem ents less than 2 is the same as the proport ion of “0”s

and “ l ”s , or 0 .25 + 0 .45 = .70 .

d The proba bil ity of select ing a “2” in a random select ion from these twenty

m easurem ents is 6 / 2 0 = 30 .

e Th ere are no outl iers in this relat ively sym m etric, m oun d-shaped distr ibution.

a Th e test scores are graphed u sing a stem and leaf plot generated by Mi ni t ab .

Stem-and-Leaf Display: ScoresStem-and-leaf of Scores N = 2 0Leaf Unit = 1.0

2 5 575 6 1238 6 5789 7 2(2) 7 56

9 8 247 8 66793 9 134

b-c The d istribution is not m ound -shaped , but is rather bimoda l w ith two peaks

centere d around the scores 65 and 85. Th is might indícate that the students are

divided into two group s - those w ho unde rstand the ma terial and do well on exams,

and those who do not have a thorough com m and o f the m ater ial.

a The data ranges from .2 to 5.2, or 5.0 units. Since the num ber of class intervals

should be betw een five and tw enty, we cho ose to use eleven class intervals, with each

class interval having length 0.50 (5.0/11 = .45 , w hich, round ed to the nearest

con venien t fract ion, is .50). W e m ust now select interval bou ndaries such that nom easurem ent can fal l on a bou nda ry point . Th e subintervals .1 to < .6 , .6 to < 1.1,

and so on, are convenient and a tally i s constructed. ^ ^ _______________________

Class

i

Class

Boundaries

Tally í Relative frequency,

f j n

1 0.1 to < 0.6 11111 11111 10 .167

2 0.6 t o < 1.1 11111 11111

11111

15 .250

3 1.1 to < 1.6 11111 11111

11111

15 .250

4 1.6 to < 2.1 11111 11111 10 .1675 2.1 to < 2.6 1111 4 .067

6 2.6 to < 3.1 1 1 .017

7 3.1 to < 3.6 11 2 .033

8 3.6 to < 4.1 1 1 .017

9 4.1 to < 4 .6 1 1 .017

10 4.6 to < 5.1 0 .0 0 0

11 5.1 to < 5.6 1 1 .017

The relat ive frequency histogram is shown below.

3



15/60

10/60

IV

I 5/60

0

a The distribution is skewed to the right, with several unusually large

observations.

b Fo r som e reason, one person had to wait 5.2 minutes. Perhaps the superm arket

was un derstaffed that day, or there may have been an unu sually large num ber of

custom ers in the store.

c The two graphs convey the same information. The stem and leaf plot allows us

to actually recreate the actual data set, while the histogram does not.

1.35 a H istograms will vary from student to student. A typical histogram, generated by

Mi n i í ab is show n on the next page. ^

b Since 2 o f the 20 players have averages abov e 0.400, the chance is 2 out of 20 or

2 / 2 0 = 0 . 1 .

1.39 To d etermine w hether a distr ibution is likely to be skewed, look for the likelihood of

observing extremely large or extrem ely small v alúes o f the variable of interest.

4



a The distr ibution of non-secured loan sizes might be skewed (a few extremely

large loans are possible) .

b The distr ibution o f secured loan sizes is not l ikely to contain unu sually large or

small valúes.

c N ot likely to be sk ew ed.

d N ot likely to be sk ew ed.e If a packag e is drop ped , it is likely that allthe shells will be bro ken. H enee, a

few large n um ber of broken sh ells is po ssible.The distribution will be sk ew ed.

f If an animal has one t ick, he is likely to have m ore than one. There will be some

“0”s w ith uninfected rabbits, and then a larger num ber of large valúes. The

distribution will not be symmetric.

1.43 a Stem and leaf displays may va ry from student to student. Th e most obvious

choice is to use the tens digit as the stem and the ones digit as the leaf.

7 | 8 9

8 | 0 1 79 | 0 1 2 4 4 5 6 6 6 8 8

1 0 1 1 7 9

11 | 2

Th e display is fair ly mo und-sha ped, w ith a large pe ak in the m iddle.

1.47 A nsw ers wil l vary from student to student . The students should notice that the

distr ibution is skewed to the r ight w ith a few presiden ts (Truman , Cleveland, and F.D.

Roosevelt) east ing an unusually large number of vetoes. _________________

Vetoes

1.51 a The popu lar vote w ithin each state should vary depen ding on the size of the

state. Since there are several v ery large States (in po pulation ) in the U nited States, the

distr ibution should be skew ed to the r ight .

b-c Histogram s w ill vary from stud ent to student , bul should resem ble the histogram

generated by Mi n i t ab in the f igure below . Th e distribution is indeed skewed to the

right , with two o utl iers - Ca lifornia and New York.

5



O 1000 2000 3000 4000 5000Popubr Vote

1.55 a-b An swers w ill vary from student to student . The l ine chart should loo k sim ilar to

the one show n below.

c Th e percentage of people who were not w orried was r ising at a slow rate until

Sep tem ber 11, 2001 , at wh ich time the percentages reversed them selves dram atical ly.

d The horizontal axis on the www.g a l lup . co m chart is not an actual time line, so

tha t the t ime f rame in which these changes occur m ay be dis tor ted.

1.59 a-b An sw ers wil l vary. A typical histogram is show n below. N otice the gaps and the

b im odal natu re o f th e h is to gram , p robab ly due to th e fa ct th at th e sam ple s were

collected at different locat ions.

6

http://www.gallup.com/

http://www.gallup.com/



1.63 a-b The M ini t ab s tem and le af plot is show n below. The d istr ibution is sl ightly

skew ed to the left .

Stem-and-Leaf Display: PercentStem-and-leaf of Percent N = 5 1

Leaf Unit = 1.0

1 0 72 0 83 1 04 1 36 1 4512 1 66677720 1 88888999

(11) 2 0000000111120 2 2222233312 2 444445554 2 6771 2 9

7



c G eorgia (7.5) and Arkan sas (8.0) have gasoline taxes that are som ew hat sm aller

than m ost, but they m ay no t be “outliers” in the sense that they lie far away from therest o f the m easurem ents in the data set.

1.67 a-b The distr ibution is approximately mound-shaped. with one unusual

m easurem ent, in the class w ith midpoint at 100.8°. Perhaps the person w hose

temperature w as 100 .8 has som e sort of i l lness coming on ?

c Th e valué 98.6° is slightly to the r ight of center.

1.69

a The distr ibution is somewhat mound-shaped (as much as a small set can be);

there are no outliers.

b 2/10 = 0. 2

1.73 a Th ere are a few extreme ly large num bers, indicating that the distr ibution is

p ro bably skew ed to th e right.

b-c The d istribution is indeed skewe d r ight with three possible outliers - Y ahoo!,

Time Warner and MSN-Microsof t .



2: Describing Data with Numerical Measures

2.1 a The dotplot shown below plots the f ive measurements along the horizontal axis.

Since there are two “ 1 ”s, the correspon ding d ots are placed one abo ve the other. The

approxim ate center of the data appears to be around 1 .

b The mean i s the sum o f the m easurements d ivided by the num ber of

measurements , or

2.5

_ = ^ = 0 + 5 + 1+ U 3 = m = 2 x ~ n 5 ~ 5 ~

To ca lcúlate the m edian, the observations are f i rst ranked from sm allest to largest : 0,

1, 1, 3, 5. Then since

n = 5 ,the positiono f the m edian is 0.5(n + l) = 3, and the m edian is the 3rd ranked

measurement , orm = 1 .The mode is the measurement occurring most frequently, or

mode = 1 .

c The three measures in part b are located on the dotplot . Since the m edian and

m ode are to the left of the mean, w e conclude that the me asurem ents are skewed to

the right.

a Although there may be a few households who own more than one DVD player,

the m ajori ty should own e i ther 0 or 1. Th e distribution should be sl ightly skew ed to

the right.b Since most households wil l have only one DVD player, we guess that the mode

is 1.

c The m ean is

Z jL = 1 + 0 + - + , a 27

n 25 25

To calcúlate the m edian, the ob servations are f i rst ranked from sm allest to largest:

Th ere are six Os, thír teen ls, four 2s, and two 3s. The n since n = 25 , the position o f

9



the me dian is 0.5(n +1) = 13, w hich is the 13th ranked m easurem ent, or m = 1. The

mode is the measurement occurring most frequently, ormo d e = 1 .

d The relat ive frequency histogram is shown below, with the three measures

superim pose d. N otice that the m ean fal ls sl ightly to the right of the median an d

m ode, indicat ing that the m easurem ents are slightly sk ewe d to the r ight .

0 1 2 3 VCRsmedian meanmode.

2.9 The distr ibution o f sports salaries wil l be skew ed to the r ight , becau se o f the very high

salaries of som e sports figures. Henee , the me dian salary w ould be a bet ter measureo f cen ter than the mean.

n 5

b Create a table of differences, ( x ¡ - x ) and their squares, (*. - T ) 2 .

X¡ x¡ - x ( * / - * )

2 - 0 .4 0.161 - 1 .4 1.96

1 -1 .4 1.96

3 0 .6 0.36

5 2 .6 6.76

Total 0 1 1 .2 0

Then^

j3 _ ! ( * , - ? ) • _ ( 2 - 2 A ) 2 -i------H 5 - 2 .4 ): . 11.20 . „

n - 1 4 4

c The sam ple standard deviat ion is the posi t ive square root of the variance or

s = V 7 = V l 8 =1 .673

Calcúlate ' Z x f = 2 2 + 1 2 +- - - + 52 = 4 0 . T he n

10



2.17

2.19

2.21

n _ 1 - = — = 2.8 a nd s = y f ? = - J l .8 = 1.673 .4n - 1 4

Th e results o f parts a and b are identical.

a The range is R = 2 .39 -1 .28 = 1 .11 .

b Calcúlate I *,2 = 1,282 + 2 .3 9 2 + • •• + 1 .5 l2 = 15.415 . Th en

^ , ( I x ) 2 (8 .56 )2 Z * ? - - 15.451 —-— —1— 7 . n ? 8 — ---------" — = -------------------5 — = ^ 6 0 2 8 _ 1 9 0 0 7

n —1 4 4

and 5 = 7 7 = V. 1 90 07 = .4 36

c The range, R = 1.11, is 1.1 1/.436 = 2.5 standard d eviations.

The range of the data is R = 6 - 1 = 5 and the range approxima t ion wi th n = 10 is

s ~ — = 1.673

The s tandard d eviation o f the sample is

= V 7 =U -

& * , f n _

n - 1

130 — (32)

= 7 3 . 0 6 6 7 = 1.7 51

which is very cióse to the estimate for part a.

c-e From the dotplot on the next page, you can see that the data set is not mound-

shaped. Henee you can use T ch eb ysh eff s Theo rem, but not the Em pirical Rule to

describe the data.

a The interval from 40 to 60 represents f i ± g = 50 ± 10. S ince the distribu tion is

relatively m ound-shaped , the proportion o f m easurem ents between 40 and 60 is 6 8 %

according to the Empirical Rule and is shown on the next page

11



b A gain, using the Em pirical Rule, the interval / / ± 2 o = 50 ± 2(10) or between 30

and 70 co nta ins approximately 95% o f the measurements .

c Re fer to the figure below .

Since approximately 6 8 % o f the measurements are between 40 and 60, the sym metry

o f the dis t ribut ion im plies that 34% o f the measurem ents are b etween 50 and 60.

Similarly , s ince 95% o f the measurements are betw een 3 0 and 70, approximately

47.5% are between 30 and 50. Thus , the propor t ion o f m easurements between 30 and60 is

0.34 + 0.475 = 0.815

d From the f igure in part a, the propor t ion o f the m easuremen ts between 50 and 60

is 0.34 and the proportion o f the mea surem ents w hich are greater than 50 is 0.50.

Th erefore, the propo rtion that are greater than 60 must be0 . 5 - 0 . 3 4 = 0 .1 6

2.25 A ccording to the Em pirical Rule, if a distr ibution o f m easurem ents is appro xim ately

mound- shaped .

12

a approximately 6 8 % or 0 . 6 8 o f the m easu rem ents fall in the interval

H ± o = 12± 2.3 or 9.7 to 14.3

b app roxim ately 95% or 0.95 of the me asurem ents fall in the interval

/ / ± 2 a = 12 ± 4.6 o r 7.4 to 16.6

c app roxim ately 99.7% or 0.997 o f the measu reme nts fall in the interval

/y ±3<r = 12 ± 6.9 or5 .1 to 18.9

Th erefore, app roxim ately 0.3% or 0.003 wil l fal l outside this interval.

2.31 a W e choo se to use 12 classes of length 1.0. The tal ly and the relat ive frequency

-------------

C la ss i C la ss Bo un da rie s Tally fi Relat ive frequency , / / / !

1 2 to < 3 1 1 1/70

2 3 to < 4 1 1 1/70

3 4 to < 5 111 3 3/70

4 5 to < 6 11111 5 5/70

5 6 to < 7 11111 5 5/70

6 7 to < 8 1 1 11 1 11111 11 12 12/70

7 8 to < 9 11111 11111 11111 111 18 18/70

8 9 to < 10 11111 11111 11111 15 15/70

9 10 to < 11 11111 1 6 6/70

10 11 to < 12 111 3 3/70

11 12 to < 13 0 0

12 13 to < 14 1 1 1/70

20/70

« 10/70■E

L1

5 10TRH3

15

V* y 1b Calcúla te n = 70, X x = 5 4 1and X x f = 4453 . T hen x = —— = ------ = 7.729 is

n 70

an es t ím ate of / / .

c The sample standard deviat ion is

13



U - í 541):J = V — = \ -------------- ^ — = V 3 .9 39 8 = 1.985

\ n - 1 \ 69

Th e th ree in te rv a ls , T ± ^ f o r k = 1,2,3 are calculated below. Th e table shows the

actual percentage of me asurem ents fal ling in a part icular interval as well as the

percenta ge pred ic te d by T cheby sh eff’s T heorem and th e E m pir ic a l R ule . N ote that

the Em pirical R ule should be fair ly accurate, as indicated by the m ound-sha pe of thehistogram in part a. ______________________________________________________________

k x ± k s Interval F raction in

Interval

Tchebyshef f Empir ical

Rule

1 7 .729 ±1 .985 5.744 to 9.714 50/70 = 0.71 at least 0 - 0 . 6 8

2 7 . 7 2 9 ± 3 . 9 7 0 3.759 to

11.699

67/70 = 0 .96 at least 0 .75 - 0 . 9 5

3 7 .7 2 9 ± 5 .9 5 5 1.774 to

13.6847 0 /7 0 = 1.00 at least 0 .89 - 0 .9 9 7

2.35 a Calcúlate R = 2 .3 9 -1 .28 = 1 .11 so that s - R / 2 .5 = 1.11/2.5 = .444 .

b In Exercise 2 .17, we calculated X*. =8.56 and

I *,2 = 1.2 82 + 2 .3 9 2 +--- + 1.512 = 15.41 5. Th en

,2■> ( Z * (8. 56)"

I xf-± 15.451 - - L.76028

= .19007n - 1 4 4

and 5 = V 7 = V-19007 = .436, which i s very cióse to our es tímate in par t a.

2.39 a T he data in this exercise have been arranged in a frequen cy table.

Xi 0 1 2 3 4 5 6 7 8 9 10

f i 10 5 3 2 1 1 1 0 0 1 1

Using the frequency table and the grouped formu las, calcúlate

' Lx i f = 0(10) +1(5) H------1-10(1) = 51

X * ,2/; = 0 2(10) + 12(5) h------1-102 (1) = 293

Then

I = I i A = 5 1 = 2 0 4

25n

U f -( Z x i í ) 2 2 9 3 -

s~ =

(51)1

^ - = 7.873 a ndn - 1 24

s = y ¡ l . 8 7 3 = 2 .8 0 6 .

b-c The three intervals x ± k s for k = 1,2,3 are calculated in the table a long with the

actual proport ion of m easurem ents fal ling in the intervals. T ch eb ys he ff s Th eorem is

14



satisf icd and the approximation given by the Empirical Rule are fair ly cióse for

k = 2 and k = 3 .

k x ± k s Interval F raction in

Interval

T ch eb y sh eff E m pirical

Rule

1 2.04 ± 2 .8 06 -0 .7 6 6 to 4.846 21/25 = 0 .8 4 at least 0 = 0 . 6 8

2 2 .0 4 ± 5 .6 1 2 -3 .5 7 2 to 7.652 23/25 = 0 .92 at least 0.75 = 0.95

3 2.04 + 8.418 -6 .3 7 8 to

10.458

25/25 = 1.00 at least 0.89 - 0 .9 9 7

2.41 The data have already been sorted. Find the positions of the quartiles. and the

measurements that are just above and below those positions. Then f ind the quartiles

by in te rp o la ti on .

Sorted D ata Set P osition o f

Q.

Above

and below

Qi

1, 1.5, 2 ,2 , 2.2 .2 5 (6 )= 1.5 1 and 1.5 1.25

0, 1.7, 1.8, 3.1, 3.2,

7, 8 , 8 .8 , 8.9, 9, 10

.25(12) = 3 None 1.8

.23, .30, .35, .41,

.56, .58, .76, .80

.25(9) = 2.25 .30 and .35 .30 + .25(.05) = .3125

Posit ion o f Q 3 Above and below q3

.75(6) = 4.5 2 and 2 .2 2.1

.75(12) = 9 None 8.9

.75(9) = 6.75 .58 and .76 .58 + .75(.18) = .7150

2.45 The ordered data are:

2 , 3 , 4 , 5 , 6 , 6 , 6 ,7 ,8 , 9 , 9, 10, 22

Fo r « = 13, the p os i tion o f the median is 0 .5(« +1) = 0.5(13 +1) = 7 and m = 6 . The

p osit io ns o f th e q uartil es are 0.2 5(n + l) = 3.5 and 0.7 5(n + l) = 10 .5 , so th atQ¡ = 4.5, Qy = 9, and IQR = 9 - 4.5 = 4.5 .

The l ower and upper f ences are:

Qx - 1 . 5 I Q R = 4 .5 - 6 .75 = -2 .25

Q, + 1.5/0/? = 9 + 6.75 = 15.75

The valué x = 22 lies outside the upp er fence and is an outlier . The b ox plot is shown

belo w . T he low er w h is ker connects th e box to th e sm allest valué th at is not an

outlier , which happens to be the mínimum valué, x = 2. The upper whisker connects

the box to the largest valué that is not an outlier or x = 1 0 .

15



2.49 a For n = 18 , the po sition of the m edian is 0.5(n +1) = 9.5 an d the p ositions o f the

qua rtiles are 0.25(n + 1) = 4.75 and 0.75(n + l) = 14.25 . Th e low er quartile is V a the

way betw een the 4 lh and 5111 measurements and the upper quartile is V a the way

betw een th e 14Ih an d 15,h m easure m ents . T he sorte d m easurem en ts are show n belo w .Favre: 10, 12, 13, 14, 15, 15, 18, 19, 21, 22, 22, 23, 23, 23, 23, 25, 25, 26

McNabb: 9, 10, 11, 15, 15, 16. 16, 17, 18, 18, 18, 18, 19, 21, 21, 23, 24, 27

For Brett Favre, m = ( 21 + 2 2 ) / 2 = 2 1 . 5 , 0 , = 1 4 + 0 . 7 5 ( 1 5 - 1 4 ) = 1 4.7 5 a nd

0 3 = 2 3 + 0 . 2 5 ( 2 3 - 2 3 ) = 2 3 .

For Donovan M cNabb, m = ( 18 + 1 8 ) / 2 = 1 8 , 0 , = 15 + 0.75 (15 -1 5) = 15 and

0 3 = 2 1 + 0 .2 5 ( 2 1 - 2 1 ) = 2 1 .

Then the f ive-num ber summ ar ies are

Min Q. Median 03 M ax

Favre 10 14.75 21.5 23 26M cN abb 9 15 18 21 27

b For B rett Favre, calcúlate IQR = 0 , - 0 , = 2 3 - 1 4 . 7 5 = 8 .2 5 . T h e n th e lower

a n d u p p er f en ees are:

0 , - 1 .5 IQR = 14.75 -1 2.3 75 = 2.375

0 , + 1 . 5 / 0 / ? = 2 3 + 1 2 . 3 7 5 = 3 5.3 75

For Donovan McNabb, ca lcúla te IQR = 0 , - 0 , = 2 1 -1 5 = 6 . Th en the l ower and

u p p er f en ces are:

0 , - 1 . 5 / 0 / ? = 1 5 - 9 = 6

0 , + 1 .5 IQ R = 21 + 9 = 30

Th ere are no outliers, and the box plots are shown on the next page.

16

3

5

c Answers will vary. The Favre distribution is skewed left, while the Donovan

distribution is roughly symmetric , probably mound-shaped. The McNabb dis tribution

is s l ightly m ore variable; Favre has a higher m edian num ber of com pleted passes .

Answers will vary. The student should notice the outliers in the female group, that the

med ian female tem perature is higher than the median m ale temperature.

The ordered sets are shown below:

G eneric Sunm aid

24 25 25 25 26 2 2 24 24 24 24

26 26 26 26 27 25 25 27 28 2827 28 28 28 28 28 29 30

For n = 14, the position o f the m edian is0.5(/i + l) = 0.5(14 + 1) = 7.5 and the p ositions

o f the quarti les are 0.25 (n +1) = 3.75 and 0.75(n +1 ) = 11.25 , so that

Generic: m = 26, Qx= 25, Qy = 27.25,and IQR = 27.25 - 25 = 2.25

Sunmaid: m = 26, (2, = 2 4, <2, = 2 8, an d IQR = 28 - 24 = 4

Generic: L ow e r a nd u pp e r f e n c e s are:

<2, -1 5 I Q R = 25 - 3 .375 = 21.625

Q, +1 .5 IQ R = 27.25 + 3.375 = 30.625

Sunmaid: L ow e r an d up pe r f e n c e s are:

Q , - 1 . 5 I Q R = 2 4 - 6 = 18

Q3 + X.5 IQR = 28 + 6 = 34

T he box plots are shown on the next page. Th ere are no outliers.

17

Sunmaid -

Generic

n ----------1----------1----------1----------1--------- 1---------- 1----------1----------1----------1-21 22 23 24 25 26 27 28 29 30

Raisins

d If the b oxes are not being und erfil led, the av erage size of the raisins is roughly

the same for the two brands. How ever, since the num ber of raisins is more va riable

for the Sun m aid brand, i t wou ld appear that some o f the Sun m aid raisins are large

wh ile others are small. Th e individual sizes of the gene ric raisins are not as variable.

2.59 The ordered data are shown below.0.2 2.0 4.3 8.2 14.7

0.2 2.1 4.4 8.3 16.7

0.3 2.4 5.6 8.7 18.00.4 2.4 5.8 9.0 18.0

1.0 2.7 6.1 9.6 18.4

1.2 3.3 6.6 9.9 19.2

1.3 3.5 6.9 11.4 23.1

1.4 3.7 7.4 12.6 24.01.6 3.9 7.4 13.5 26.7

1.6 4.1 8.2 14.1 32.3

Since n = 5 0 , the posi tion of the median is 0.5 (n +1) = 25.5 and the po si t ions of the

lower and upper qu art iles are 0.25(n +1) = 12.75 and 0.15(n +1) = 38.25 .

Then m = (6.1 + 6.6 ) /2 = 6.35, 0 , = 2 .1 + 0.7 5 (2.4 -2 .1) = 2.325 and<2, = 12 .6 + 0.25(13 .5-12 .6) = 12.825. Then 1QR = 12 .82 5-2 .325 = 10 .5 .

The l ower and upper f enees are:

0 , - 1 . 5 / 0 / ? = 2 . 3 2 5 - 1 5 . 7 5 = - 1 3 .4 2 5

- 0 , + 1 . 5 / 0 / ? = 1 2.8 25 + 1 5.7 5 = 2 8 .5 7 5

and the box plot is shown on the next page. There is one outlier, x = 32.3 . The

distribution is skewed to the right.

18



2.63 The following inform ation is available:

n = 400, x = 600. s 2 = 4900

The s tandard dev ia t ion o f these scores is then 70, and the resul ts o f T chc by she f fsTheo rem fol low:

k x ± k s Interval T cheb ysheff

1 6 0 0 ± 7 0 530 to 670 at least 0

2 600 ± 140 4 6 0 to 740 at least 0.75

3 6 00 ± 2 1 0 390 to 810 at least 0.89

If the distr ibution o f scores is mou nd-shap ed, w e use the Em pirical Rule, and

conclude that approximately 6 8 % o f the scores w ould lie in the interval 530 to 670

(which is x ± 5 ). A pprox imately 95 % of the scores would lie in the interval 460 to

740.

2.69 If the distribution is m oun d-shap ed, then alm ost all of the measurem ents will fall in

the interval j u ± 3(7 , which is an interval 6(7 in length. Th at is, the range o f the

measurem ents should be approximately 6(7 . In this case , the range is8 0 0 - 2 0 0 = 6 0 0 , s o th at (7 = 6 0 0 / 6 = 1 0 0 .

2.73 Th e diameters o f the trees are approxim ately moun d-shaped with mean 14 and

standard deviation 2 .8 .

a The valué x = 8.4 lies two standard de viations below the m ean, while the valué x

= 22.4 is three standard deviations above the mean. Use the Empirical Rule. The

fraction o f trees with diam eters betw een 8.4 and 14 is half of 0.95 o r 0.475, while the

fraction o f trees with diameters betw een 14 and 22.4 is half of 0.997 or 0.4985. The

total fraction o f trees w ith diam eters betwee n 8.4 and 22 .4 is

0.475 + 0.4985 = .9735

19



b The valué x = 16.8 l ies one standard deviat ion above the mean. Using the

Em pirical Rule, the fraction o f t rees with diam eters between 14 and 16.8 is half of0.68 or 0.34, and the fraction o f t rees with d iameters g reater than 16.8 is

0 .5 -0 .34 = .16

2.77

y x 4182.81 a Calcúlate n = 50, Z x = 418 , so that x = — — = ------ = 8 . 3 6 .

n 50

b Th e posi t ion o f the med ian is ,5(n +1) = 25.5 and m = (4 + 4)/2 = 4.

c Since the m ean is larger than the median, the distr ibution is skew ed to the r ight.d Since n = 5 0 , the posi tions of Q¡ and 0 3 are .25(51) = 12.75 and .75(51) =

38 .25 , r e spec tively Then 0 , = 0 + 0 .7 5(1 -0) = 12 .75 ,

0 3 =1 7 + .25 (19-1 7) = 17 .5 and I Q R = 17 .5- . 75 = 16.75 .

The l ow e r and uppe r f e ne e s are:

_ 0 , - 1 .5 /0 / ? = .7 5 -2 5 .1 2 5 = -2 4 .3 7 5

0 , + 1 .5 IQR = 17.5 + 25.125 = 42.625

and the box plot is show n on the next page. There are no outl iers and the data is

skew ed to the r ight.

a Th e percentage o f col leges that have betw een 145 and 205 teach ers correspo nds

to the fract ion of me asurem ents expected to l ie within two standard deviat ions o f the

m ean. T ch eb ys he ff s Theo rem States that this fract ion will be at least 3á or 75% .

b If the popu lat ion is normally distr ibuted, the Em pirical Rule is approp riate and

the desired fraction is calculated. Referring to the normal distr ibution sh ow n below,

the fract ion of area lying between 175 and 190 is 0.34, so that the fract ion o f col leges

m ore than 190 teachers is 0.5 - 0.34 = 0.16 .

20

87 a-b As the valué of x gets smaller , so does the mean.

c The med ian does not chan ge until the green dot is smaller than x = 10, at which

poin t th e green do t becom es the m edia n.

d The largest and sm allest possible valúes for the median are 5 < m < 10.

91 Th e box plot show s a distribution that is skew ed to the left, but with one outlier to the

right of the other observations ( x = 5 2 0 ) .

21



3: Describing Bivariate Data

3.1 a The side-by-side pie charts are constructed as in Chapter 1 for each o f the two

groups (men and women) and are displayed below using the percentages shown in thetable below.

Group 1 G roup 2 G roup 3 T otal

M en 23% 31% 46% 100%

W o m e n 8 % 57% 35% 100%

b-c The side-by-side and stacked bar charts in the next two f igures m easure the

frequency o f occu rrence for each o f the three groups. A sepárate bar (or portion o f a

bar) is used fo r m en and w om en. — - ■*>-., ----- ~~.y--'.------ . ... 7---------- i

'' “— —' ' >*

60

Gender Male Female Male Female Male FemaleGroup 1 Group 2 Group 3

23



d The d ifferences in the proport ions of men and wom en in the three groups is mostgraphical ly port rayed by the pie char ts , s ince the unequal num ber of m en and wom en

tend to confuse the interpretation o f the bar charts. H ow ever, the bar charts are useful

in retaining the actual frequen cies of occurrence in each group , w hich is lost in the pie

chart.

3.5 a The popu lation o f interest is the populat ion of responses to the que st ion about

free time for all parents and children in the U nited States. Th e sam ple is the set of

responses generated for the 198 parents and 2 0 0 children in the survey.

b The data can be considered bivariate if , for each person interviewed, we record

the person’s relationship (Parent or Child) and their response to the question (just ther ight amount , not enough , too m uch, do n’t know). Since the mea surem ents are not

num erical in nature, the variables are qualitative.

c The entry in a cell represents the number of people who fell into that

relationship-opinion category.

d A pie chart is created for both the “parent” and the “childre n” categories. The

size of each sector angle is propo rtional to the fraction o f m easu rem ents falling into



e Either stacked or com parative bar charts could be used, but since the height of

the bar rcpresents the frequen ey o f occu rrence (and henee is t ied to the sample size),

this type o f chart w ould be misleading . The com parative pie charts are the best

choice.

3.9 Follow the instructions in the M y Personal Tra iner section. T he correct answ ers are

shown in the table.

X y xy C alcúlate: C ovaria nce

1 6 6 n = 32 0 - ( 6 ) ( , 2 )

s *y 2 3 23 2 6

Sx = 1

2 4 8 sy = 2 Corre lat ion Coeff íc ient

I * - 6 Z y = 12 2 > = 2 0 r = _ 2 = —1

1 ( 2 )

3.13 a The scatterplot is shown below.

b There appears to be a negative relationship between x and y; that is, as x

increase, y d ecreases.

c Use your scientif ic calculator to calcúlate the sums, sums of squares and sum of

cross prod ucís for the pairs {xt, y , ) .

I * , = 2 1 ; I y . = 2 4 .3 ; I * 2 = 9 1 ; I y ,2 = 1 0 3 .9 9 ; I * , y . = 7 5 .3

Then the covar iance is

1 53J m ±

s „ = * ^---------

= -----------------

£-------

= - 1 . 9 5* n - 1 5

25

3.17

and the samp le standard deviations are

s . = \ \ --------------^— = — = 1.8708 andn - 1 5

i r a no <24-3)26 _ s v = \ f — = ------------- =1.0559

l / i - l 5

The correlation coeff icient is

s —1 95r = — = ------------ '-------------= - 0 . 9 8 7

5I5> (1.8708X1.0559)

This valué of r indicates a strong negative relationship betw een x and y.

a- b The scatterplot is shown below. The re is a slight positive trend betw een pre-

and posttest scores, but the trend is not too pronoun ced.

c Calcúlate

n = 7; I x¡ = 677; I y. = 719; Z x f = 65 ,9 93 ; I y? = 74,585; I x i y i = 70,00 6 . Then

the covariance is

5 n = — = 7 8 .0 7 1 42 9* n - 1

The sample standard deviations are sx = 9.286447 and s y = 11.056134 so that

r = 0.7 60 . Th is is a relatively strong positive correlation, confirm ing the

interpretation of the scatterplot.

26



23 a For each person interviewed in the survey, the following variables are recorded:

the service branch (qua litative), the age (quantitative con tinuo us) and the rank

(enlisted vs. officer, qualitative).

b The population of interest is the population of ages for all people in the military.

A lthough the source o f this data is not given, i t is probably b ased on censusinformation, in which case the data represents the entire population.

c A com parative (s ide-by-side) bar chart has been used. A n alternative

p resenta ti on can be ob ta in ed by usi ng com parative p ie chart s, w ith th e ages d iv id ed

into eight age groups, and com pared for the Army versus the M arine Corps.

d The enlisted men tend to be younger.

e The M arine Corps tends to have a much h igher percentage o f young er enlis ted

personnel.

29 a-c N o. T here se em s to be a la rge clu ste r o f poin ts in th e lo w er le ft hand córner

show ing no apparent relationship between the variables , while 7-10 d ata points fromtop left to bottom right show a negative linear relationship.b Th e pattern described in parts a and c would indícate a weak correlation:

r = ̂ -8 3 .5 3 0 = ^ Q32

sxs y ^ (712.603X9346.603)

d N um ber o f w aste si te s is only sl ig htly affecte d by th e si ze o f th e st at e. Som e

other possible ex planatory v ariables m ight be local environmental regulations,

p opu la ti on per square m ile, o r geographic al re gió n in th e U nit ed Sta te s.

.33 a The 1) num ber of home networks (quanti tat ive discrete) have been measured,along with the year (q uanti tative continuous) and the type of netwo rk (quali tat ive).

b-c Answers will vary. We choose to use a line chart for the two types of networks.

As the num ber of wireless netw orks increase, the num ber of wired netw orks

decreases .

.37 a-b T he scatterplot is show n on the next page. Th ere is a strong positive linear

relationship between x and y.

27



3.38

3.43

Test 1 (x)

a Calcúlate

n = 8 ; I x, = 6 34 ; I y, = 386; X x2 = 52270;

I >’(2 = 1 9 8 7 6 ; I * , ? , = 3 2 1 3 6

Then the covariance is

sn = — ^ = 2 20 .7 85 71* n —1

The sample s tandard devia t ions are sx = 17 .010501 and $v = 13.37 107 75 so that

r = 0 . 9 7 1 .

b Since the co rrelation co eff icient is so cióse to 1, the strong correlation indicates

that the secon d and q uicker test could be used in place o f the long er test- interview.

a Calcúla te

,2 = 8 ; I * . = 4 5 1 ; L y , = 5 55 ; X x 2 = 2 9 , 6 1 9 ; X y f = 4 3 ,2 0 5 ; I x , ^ = 3 5 , 0 8 2 . T hen

the covariance is

= = 5 41 .9 82 1a; - 1

The sample standard deviations are sx = 24.4770 and s y = 25.9171 so that

r = 0 . 8 5 4 4 .

b-c The scatterplot should look like the one shown on the next page. The

correlation coe ff icient should be cióse to r = 0.85 . Th ere is a strong positive trend.

28



29



4: Probability and Probability Distributions

4.1 a Th is experim ent involves tossing a s ingle die and observing the outcom e. The

sample space for this expe rimen t consis ts o f thefollowing s im ple eyents: E\. O bserve a 1 £ 4: O bservea 4

E 2: O bserve a 2 £ 5: Ob servea 5

£ 3: O bserve a 3 £ 6: O bservea 6

b E vents A th rough F are com pound even ts and ar e com posed in th e fo llow in g

manner:

A: (E2) D: (E2)

B: ( E 2, E 4 , E b) E: ( E 2, E ^ E 6)

C: ( £ 3, £4, £5, £ 6) F: contains no s imple events

c Since the s imple events £ ,- , /= 1, 2, 3, 6 are equally l ikely, £ (£ ,) = 1/6 .

d To find the probab il ity o f an event, we sum the proba bil i t ies assigned to the

simple events in that event. F or example,

/>( A) = /> (£ , ) = j6

Similarly, P(D) = \ /6: P (B ) = P ( E ) = £ ( £ 2) + £ ( £ 4) + £ ( £ 6) = - = - ; and6 2

4 2 P (C ) = — = — . S ince event £ conta ins no s imple events , P (£ ) = 0 .

4.5 a The exp er iment cons is ts of choos ing three coins a t random from four. The order

in which the coins are drawn is unimpo rtant. Henee, each s imp le event consis ts of a

triplet, indicating the three coins draw n. U sing the letters N, D, Q, and H to represen!

the nickel, dime, quarter, and half-dollar, respectively, the four possible simple events

are listed below.

£ , : (NDQ ) £ 2: (ND H) £ 3: (NQH ) £ 4: (DQH )

b T he event th at a half -doll ar is chosen is associa te d w ith th e sim ple even ts £ 2, £3,and £4. Henee ,

P lchoos e a ha lf -do ll ar ] = £ (£ , ) + P ( E , ) + P ( £ 4) = - + - + - = -4 4 4 4

since each simple event is equally likely.

c Th e s imple events along with their m onetary valúes follow:

E^ N D Q $0.40

e 2 N D H 0.6 5

E i N QH 0.80

£4 D Q H 0.85

31



4.9

4.13

4.17

4.21

4.25

4.29

Hen ee, P[total am ount is $0.60 or more] = P ( E 2) + P ( E y) + P (E 4) = 3 /4 .

The four possible outeome s o f the experime nt, or simple events, are represented as

the c el ls o f a 2 x 2 table, and have proba bil i ties as given in the table.

a R[adult judg ed to need glasses] = .44 + .14 = .58

b P[adult needs glasses but does not use them] = . 14

c F[a du lt uses glasses] = .44 + .02 = .46

a E xper iment : A taster tastes and ranks three variet ies o f tea A, B, and C,

according to preference.

b Simple events in S are in triplet form. E l : (1,2,3) E 4 : (2,3,1)

E 2 : (1,3,2) £ 5 : (3,2,1)

E y : (2,1,3) £ 6 : (3,1,2)

Here 1 is assigned to the mos t desirab le, 2 to the next m ost desirable, and 3 to the

least desirable.

c De fine the even ts D: variety A is ranke d first

F: variety A is ranke d third

Then

P (D ) = />(£,) + P ( E 2) = 1/6 + 1/6 = 1/3

The probability that A is least desirable is

£ ( £ ) = P(E 5) + P ÍE , ) = 1 /6 +1 /6 = 1/3

Use the mn Rule. Th ere are 10(8) = 80 possible pairs.

8 'Since order is important, you use permutat ions and p \ = —j = 8(7)(6)(5)(4 ) = 672 0 .

Since order is unim portant, you use combinat ions a nd C í° = ^ = 120.3! 17! 3(2)(1)

a Each student has a choice of 52 cards, since the cards are replaced between

select ions. The mn R ule al low s you to f ind the total num ber o f configurat ions for

three students as 52(52)(52) = 140,60 8.

b N o w e ac h student m ust p ic k a d if fe re nt card . T hat is, the fir st stu dent has 52

choices, but the second and third students have only 51 and 5 0 choices, respectively.

The total num ber of configurat ions is found using the mn Rule or the rule for

perm uta tions:

mnt = 52(51)(50) = 132,600 or £ 5,2= — = 132,600 .49!

c Let A be the event of interest. Since there are 52 differen t card s in the deck,

there are 52 configurations in which all three students pick the same card (one for

32



each card). Th at is , there are nA = 52 ways for the event A to occur, out of a total of

N =1 40 ,608 poss ible conf igurations f rom par t a. The probahil i ty of interest is

P (A ) = ^ - = — — — = .00037 N 140,608

d Ag ain, let A be the event o f interest . T here are nA =1 32 ,600 ways (f rom par t b)for the eve nt A to occu r, out of a total of N = 140,608 possible configurations froTn

part a, and the prob ahil i ty o f interest is

= ̂ = 1 3 2 , 6 0 0 ^943

N 140,608

33 N otice that a sam ple of 10 nurses will be the same no matter in w hich orde r they were

selected. Hen ee, order is unim portant and com binations are used. The num ber of

sam ples of 10 selected from a total of 90 is

9 0 i 2.07 590 76 Í1019) .Cm = - = -------------- = 5.720645 (10 )10 !80! 3.628 8(10 ) V ’

.37 Th e si tuation presented here is analog ous to draw ing 5 i tems from a ja r ( the f ive

m em bers v oting in favor of the plaint iff). If the jar con tains 5 red and 3 white i tems

(5 wo m en and 3 m en), wh at is the proba hility that all five items are red ? T ha t is, if

there is no sex bias, f ive o f the eight m em bers are random ly chosen to be those voting

for the plaint iff . W hat is the probah il ity that all five are wo m en? There are

8 ' N = C \ = — — = 56

5!3!

simple events in the experiment, only one of which results in choosing 5 women.

Henee,

P (fiv e w om en) = ̂ .

.41 Follow the instruct ions given in the My Personal Trainer section. The an swe rs are

given in the table.

P(A) P (B ) C o n ditio ns fo r e ve nts A a n d B P (A n B) P(A u B) P(A|B)

.3 .4 Mutually exclusive 0 .3 + .4 = .7 0

.3 .4 Independent .3(4 ) = .12 .3 + .4- ( .3) ( .4 ) = .58 .3

.1 .5 Independent .1 (5 ) = .05 , l + . 5 - ( . l ) ( .5 ) = .55 .1

.2 .5 Mutually exclusive 0 .2 + .5 = .7 0

.47 Refer to the solut ion to Exe rcise 4.1 where the six simple events in the exp erime nt are

given, w ith P{Ej ) = 1 /6 .

a 5 = { E l , E 2 , E i , E A , E 5 , E 6 } and P (S ) = 6 / 6 = 1

33



4.48

4.49

4.55

4.59

b P ( A \ B ) = « A n B ) = y i = x

v ' P (B ) 1/3c B = { £ „ £ , } and P (B ) = 2 /6 = 1/3

d A n f i n C c o nta in s n o s im p le e ve n ts , a nd P ( A n B n C ) = 0

e P ( A n B ) = P (A \ B ) P ( B ) = 1(1/3) = 1/3

f A n C c on ta in s no sim p le ev en ts , an d P ( A n C ) = 0

g B n C c o n t a in s no sim p le e ve nts , a nd P ( f i n C ) = 0

h A k j C = S and P ( A u C ) = l

¡ 8 u C = { £ l, £ , , £ 4, £ s , £ j a nd P ( P u C ) = 5 /6

a From Exerc ise 4 .47 , P(A n B ) = 1/3 , P (A | B ) = 1, P(A ) = 1/2 ,

P ( A | B ) * P ( A ) , so t ha t A and B are no t i ndependen t. P (A n B )^ 0 , so t hat A and

B are not m utually exclusive.

b P ( A |C ) = P ( A n C ) /P ( C ) = 0 , P (A ) = 1 /2 , P ( A n C ) = 0 . Since

P(A | C ) * P (A ) , A and C are dependent . S ince P ( A r \ C ) = 0 , A and C are

m utually exclusive.

a Since A and B are independen t , P(A r \ B ) = P { A ) P { B ) = .4(.2) = .08 .

b P ( A u B ) = P ( A ) + P ( P ) - P ( A n f i ) = . 4 + .2 - (.4 )( .2 ) = . 52

Define the fol lowing events:

A: project is approved for funding

D: project is disappro ved for funding

For the first group, P { \ ) = .2 and P(D ,) = .8 . Fo r the second group,

P [same decisión as f i rst group] = .7 and P [reversal] = .3. Th at is,

P( A 2 \ A t ) = P (D 1 \ D t ) = . l and P ( \ \ D, ) = P (D 7 \ A ) = .3 .

a P ( A n 4 ) = P ( A ) í>(A2 |A ) = -2 (.7 ) = .14

b P(D¡ n D 2 ) = P ( D ¡)Pi ,D2 \ £),) = . 8(.7) = . 56

c

P(D, n A 2) + P ( A i n D 2) = P ( D , ) P ( A , | D , ) + P ( A , )P ( D 2 1A,) = ,8(.3) + ,2(.3) = .30

Fix the birth date o f the f irst person e ntering the room . Th en define the fol lowing

events:

A2: seco nd perso n’s birthda y differs from the first

A 3: third perso n’s birthda y differs from the first and sec ond

A 4: fourth pers on ’s birthday differs from al l preceding

An: nlh pe rso n’s birthda y differs from all prece ding

34



' 364̂1̂363 p 65- / i + nv 365J ,365 J l 365 )

Then

P (A ) = P (A 2) P ( A ) - P ( A ii) =

since at each step, one less birth date is available for selection. Since event B is the

complem en t o f even t A ,

P (B ) = \ - P ( A )

a F o r n = 3 , P(A ) = (364)(363) = 99l8 and P (B ) = 1 - .99 18 = .0082(365)

b F or n = 4 , P (A ) = (364X363X362) = 9836 and P (B)= 1 - .9 8 3 6 = .0164(365)

63 D efine A: smo ke is detected by device A

B: smok e is detected by device BIf i t is given that P (A ) = .95, P (B ) = .98, and P ( A n B ) = .94 .

a P{A k j B ) = P{A) + P ( B ) - P( A n B ) = .95 + .98 - .94 = .99

b P ( A C n B c ) = l - P ( A k j B ) = 1- .99 = .01

69 a Use the Law of Total Probabil i ty, writing

P (A ) = P(5 , )P (A 15 ,) + P ( S 2 ) P ( A | S2) = ,7 (.2) + .3(.3) = .23

b U se th e resu lt s o f part a in th e fo rm o f B ayes’ Rule :

P ( S |M ) = -------------W , ) P ( » | 5 . )

P ^ P i A l S ^ + P i S ^ P i A l S , )

F o r i = l , P(S, | A) = ----------------- = — = .60871 ,7(.2) + .3(.3) .23

For í = 2 , P(5 , |A) = -------------- = ^ = .39132 ,7(.2 ) + .3(.3) .23

73 De fine A: m achine produces a defective i tem

B: wo rker follows instructions

Then P ( A \ B ) = . 0\ , P (B ) = . 90, P (A \ B c ) = .03, P ( B C) = .10 . The probabi l ity of

interest is

P (A ) = P ( A r ^ B ) + P ( A n B c )

= P{A | B ) P ( B ) + P ( A | B C) P ( B C)

= ,01(.90) + .03(.10) = .012

77 Th e pro bab ility o f interest is P (A \ H ) wh ich can be calculated using B ayes’ Rule

and the proba bil i t ies given in the exercise.

35

P ( A \ H ) = ________________ P ( A ) P ( H \ A ) ________________

P ( A ) P ( H | A ) + P ( B ) P ( H | B ) + P ( C ) P ( H \ C)

__________ .01(.90) ____________ .009

.01(.90) + .005(.95) + .02( .75) ” .02875= .3130

4.83 a Since one of the requiremen ts of a probabi l ity dis t ribut ion is that ^ p ( x ) = 1 X

we need

p (3 ) = 1- (. 1+ .3 + .3 + . 1) = 1- .8 = .2

b The probability histogram is shown below.

c For the random var iable x given here,

H = E ( x ) = Z xp ( x ) = 0( . 1) + 1(.3) H— + 4(. 1) = 1.9

The var iance of x is defined as

( j 1 ^ E ^ x - p f Y ^ x - p ) 1 p (x ) = (0 - 1.9)2(.1) + (1 -1 .9 )2(.3) + --- + ( 4 - 1.9)2(.l) = 1.29

a n d í t = VT 2 9 = 1 . 1 3 6 .d Using the table form o f the probab ility distr ibution given in the exercise,

P ( x > 2 ) = .2 + . l = . 3 .

e P ( x < 3) = 1 - P ( x = 4) = 1- . 1 = .9 .

4.87 a-b On the f irst try, the probab ility o f selecting the p rope r key is 1/4. I f the key is

not found oír the f irst try, the proba bility chan ges on the se cond try. Let F denote a

failure to f ind the key and S deno te a success. Th e random variable is x, the number

o f keys tr ied before the correct key is found. Th e four associated sim ple events are

shown below.

E , : S ( x = \) E3: FFS (x = 3)

E 2:F S (^ = 2) E4: FFFS (* = 4)

36



c-d Then

P ( l ) = P ( x = 1) = P ( S ) = 1/4

p{ 2 ) = P ( jc = 2) = P (F 5) = P ( F ) P ( S ) = (3 /4 ) (1/3) = 1/4

p ( 3 ) = P (* = 3) = P (F FS ) = P ( F ) P ( F ) P ( S ) = (3 /4) (2/3) (1/2) = 1/4

p (4 ) = F(jc = 4 ) = P ( F F F S ) = P ( F ) P ( F ) P ( F ) P ( S ) = ( 3 / 4 ) ( 2 / 3 ) ( l / 2 ) ( l ) = 1 /4

The probability distr ibution and

JC 1 2 3 4

p( x) 1/4 1/4 1/4 1/4

pro babil ity h is to gra m fo llow .

1 Let x be the nu m ber o f drilling s until the first success (oil is struck). It is given that

the prob ability o f striking oil is P (O) = .1, so that the proba bility o f no oil is

P ( N ) = .9

a p{ 1) = F [o il struck on first drilling] = P ( 0 ) = .1

p ( 2) = P [o il s truck on second d rill ing ] . Th is is the probab ility that oil is not

found on the first drilling, but is found on the second drilling. Using the

M ult ipl ica tion Law ,

p (2) = P ( N O ) = (.9)(. 1) = .0 9.

Finally, p ( 3) = P ( N N O ) = (.9)(.9 )(. 1) = .081 .

b -c For th e fir st success to occur on tr ia l x, (jc - 1) failures must oc cur b efore the

first success. Thu s,

p ( x ) = P ( N N N ... N N O ) = (.9)* "' (. 1)

since there are (jc - 1) N ’s in the sequence. The pro bability histogram is show n on the

next page.

37



1 3 5 7 9 11 13 15 17

x

4.95 The random variable G, total gain to the insurance company, will be D if there is no

theft, but D - 50,000 if there is a theft during a given year . The se two even ts will

occu r with probab ility .99 and .01, respectively. H enee, the prob ability distr ibution

for G is given below.

G _____________ p(G) The expe cted gain is

D .99 E{G ) = l G p ( G ) = . 99 D + . 0 1(D - 5 0 ,0 0 0 )

= D -5 0 ,0 0 0

D - 5 0,0 00 .01

In order that E (G ) = 1000 , it is necessary to h ave 1000 = D - 5 0 0 o r D = $ 1 5 0 0 .

4.97 a Similar to E xercise 4 .9 1. Fo r the fírst no n-b eliev er to be fo und on cali jc, (x - 1)

people w ho do believe in heaven mu st be called before the f irst non-believ er is found.

Thus , p ( x ) = P ( N N N . . . N N Y ) = ( .8 1)*"1(. 19)

b A s w ith o th er phone surv eys, th ere is a lw ay s a p rob lem o f no n-r espon se -

p eople who do not answ er th e tele phone or dec li n e to parti c ipa te in th e surv ey. A lso,

there is a problem of truthfulness of the response for a question such as this w hich

may be a sensitive subject for som e people.

4.101 Define the following events:

A: wo rker fails to repo rt fraud

B: w orke r suffers reprisal

It is given that P ( B | A c ) = .23 and P (A) = .69 . The p robability of interest is

P ( A C r \ B ) = P (B | Ac )P(A r ) = .23( .31) = .0713

4.105 Tw o systcms are selected from seven, three of w hich are defective. D enote the seven

system s as G 1( G 2, G 3, G 4, D(, D 2, D3 acco rding to w heth er they are go od o r defective.

38



Each simple ev ent w ill represen t a particular pair of systems cho sen for testing, and

the sam ple space, consisting o f 21 pairs , is shown below.

G1G2 G jD | G2D3 G4D2 G1G3 G1G2 G3D1 G4D3

G,G4 G,D3 g ] d 2 D ,D2 G2G3 G 2D, G3D3 D ,D3

G2G4 G2D2 G4D1 D2D3 G3G4

N ote th at th e tw o syste m s are dra w n sim ulta neously and th at o rder is unim port ant in

identifying a sim ple event. H enee, the pairs G |G 2 and G 2G! are not considered to

represen t two d ifferent simple events. The eve nt A, “no defectives are selected”,

consis ts of the s imple events G |G 2, G1G3, G1G4, G 2G 3, G2G4, G3G4. Since the

systems are selected at rando m , any pa ir has an equal proba bility o f being selected.

He nee, the probab ility assigned to each simple event is 1/21 and P (A) = 6 /21 = 2 /7 .

4.109 a P ( c o l d ) =

Define:

49 + 43 + 34 126= .4565

276 276

F: person has four or f ive relationshipsS: person has six or more relationships

Then for the two people chosen from the total 276,

P ( o n e F and one S ) = P ( F n S ) + P ( S n F )

100

276

( 96+

r 96 ^ í k x T

275y ^ 2 7 6 , 1,275,= .2530

. P í t h r e e o r f e w e r n c o l d ) 4 9 / 2 7 6 4 9P Í T h r e e o r fe w e r c o id ) = = — = ----- = .

v 1 ’ P ( co ld) 126/276 1263889

4.113 a Define the following events:Bj: client buy s on f irst contact

B2: client buys on second contact

Since the client m ay b uy on e ither the f irst o f the second con tact, the desired

p rob abili ty is

P[cl ien t w ill buy] = P[cl ient b uys on f i rs t contact]

+ P [client doesn' t buy on f irst , but buys on second ]

= P ( P , ) + ( l - P ( f í 1) ) P ( f i 2) = .4 + ( l- .4 ) ( .5 5 )

= .73

b Th e pro bability that the client w ill not buy is one m inus the proba bility that the

client will buy, or 1- .73 = .27 .

4.117 Ea ch hall can be chosen from the set (4, 6) and there are three such balls. Henee,

there are a total of 2(2)(2 ) = 8 potential w inning numb ers.

4.121 a Co nside r a single tr ial which consists of tossing two coins. A m atch occurs

w hen either HH or TT is observed. Henee, the probability of a match on a single tr ial

39



is P ( H H ) + P(T T ) = 1/4+ 1/4 = 1 /2. Let MM M denote the event “match on tr iá is 1,

2, and 3”. Then

P ( M M M ) = P ( M ) P ( M ) P ( M ) = (1 /2)3 = 1/8.b On a single tr ial the event A, “two trails are ob served ” has probability

P (A ) = P(T T ) = 1/4 . H enee , in thre e triáis

P ( A A A ) = P ( A ) P ( A ) P ( A ) = (1/4) ' = 1/64

c This low probability would not suggest collusion, since the probability of three

matches is low only if we assume that each student is merely guessing at each answer.

If the students have stud ied togethe r or if they b oth know the correct answ er, the

pro babil it y o f a m atc h on a si ng le tr ia l is no longer 1/2 , but is substa nti a ll y hig her.

Hen ee, the occurrence of three m atches is not unusual.

4.125 Define the events: A: the man waits f ive m inutes or longer

B: the wom an waits f ive minu tes or longer

The two events are independent, and P (A ) = P (B ) = .2 .

4.129 Since the f irst pooled test is positive, we a re interested in the p robab ility o f requir ing

five single tests to detec t the disease in the single affected pe rson. Th ere are

(5)(4)(3)(2)(1) w ays of ordering the f ive tests , and there are 4(3 )(2)(1) w ays of

ordering the tests so that the diseased person is given the f inal test . H enee, the desired

4! 1 pro babili ty is — = - .

I f two p eop le are diseased, six tests are need ed if the last two tests are given to the

diseased p eople. There are 3(2)(1) ways o f ordering the tests of the other three people

and 2(1) ways of ordering the tests of the two diseased peo ple. H enee, the probability

2!3! 1that six tests will be needed is ------ = — .

5! 10

4.133 a Define P: shopp er prefers Pepsi and C: shop per prefers Coke . Then if there is

actually no difference in the taste, P(P) = P(C) = 1/2 and

a P ( A r ) = l - / > M ) = .8

b P ( A r f lr ) = / > ( / l c ) /> ( £ c ) = (.8 )(.8 ) = .64

c P [at least one waits f ive minu tes or longer]

= 1- P [ne i the r wa i ts f ive minutes o r longer ] = 1- P ( A ( B c ) = 1 - . 6 4 = .3 6

b

P(exactly one prefers Pepsi) = P ( P C C C ) + P ( C P C C ) + P ( C C P C ) + P(C CC P)

40



4.137 Refer to the Toss ing D ice applet, in which the simple events for this experiment are

displayed. Each sim ple event has a part icular valué of T associated with it , and by

sum m ing the p robabil i t ies o f all simple eve nts producing a part icular valué of T, the

fol lowing probah il ity distribution is obtained. The d istribution is mound -shaped.

a-b

T p( T) T p(T)2 1/36 8 5/36

3 2/36 9 4/36

4 3/36 10 3/36

5 4/36 11 2/36

6 5/36 12 1/36

7 6/36

2 4 6 8 10 12X

- ______________

41

5: Several Useful Discrete Distributions

5.1 Follow the instructions in the Personal Trainer section. Th e answ ers are show n in thetables below.

k 0 1 2 3 4 5 6 7 8

P(x < k) .000 .001 .011 .058 .194 .448 .745 .942 1.000

T h e P r o b l e m Lis t the

V a lú e s o f x

W r i te t h e

p ro b a b ili ty

Rewr i te the

p ro b a b il it y

F ind the

p ro b a b il it y

Three or less 0 , 1 , 2 , 3 P (x < 3 ) .058Three or more 3, 4. 5, 6. 7, 8 P (x > 3) 1 - P(x < 2) 1—.011 = .989

M ore than three 4 , 5, 6, 7, 8 P (x > 3) 1 - P(x <, 3) 1 - . 058 = .942

Few er than three 0 , 1 ,2 P (x < 3) P (x < 2 ) .011

Between 3 and 5 (inclusive) 3 , 4 , 5 P<3 < jc < 5) P (x < 5) - P (x < 2) . 448 - .011 = .437

Exactly three 3 P fx = 3) P (x < 3 ) - P (x < 2) r -~

s I I ©

r o o

O

5.5 a

b

c

d

5.11 a

b

P ( x > 4 ) = 1- P ( x < 4) = 1- P ( x < 3).

These probabilit ies can be foun d individually using the binom ial formula, or

alternatively using the cumulative binomial tables in Appendix I .

P ( x = 0) = Co°( .4)° ( .6) '° = .006 P ( x = 1) = C !°( .4 )‘ (.6 )9 = .040

P ( x = 2 ) = C!.0(.4 )2 (.6 )8 = .121 P ( x = 3) = C 3°(-4)3 (.6 )? = .215

The sum of these probabilit ies gives P ( x < 3) = .382 an d P ( x > 4) = 1 - .382 = .618.

c Use the results o f parts a and b.

P ( x > 4 ) = 1- P ( x < 4) = 1 - ( .3 8 2 + .25 1) = .367

C ¡ ( -3) ; ( ,7 )6 = |Z 2 ( .0 9 ) ( . l 17649) = .2965

C S ( .0 5 ) °( .9 5 ) 4 = ( . 9 5 ) ‘ = . 8 1 4 5

C'J»(.5)S(.5)7= ^ ^ (.5 r= .1 17 2

Ci7(-2)‘ (.8)6 = 7 (.2 )(.8 )6 = .3670

For n = 10 and p = .4, P ( x = 4 ) = C 4°(-4)4 (.6)6 = .251 .

To calcúlate P ( x > 4 ) = p ( 4 ) + p ( 5) + ---+ p (10) it is easiest to write

43

5.15

5.19

5.23

d From par t c , P { x < 4 ) = P ( x < 3) + P ( x = 4) = .382 + .251 = .63 3.

e n = np = 10( .4) = 4f o = yj ñpq = 7 l 0 ( . 4 ) (. 6 ) = y ¡2A = 1.549

a P f j t < 12] = P [ x < 1 1] = .74 8

b P [ x < 6] = .610

c />[jc > 4] = 1- ^ [ jc < 4 ] = 1- .6 3 3 = .367

d P [ x > 6 ] = 1 - P [ x ^ 5] = 1 —.034 = .966

e P [ 3 < J t < 7 ] = P [ x < 6 ] - P [ ; t < 3 ] = .8 2 8 - .1 7 2 = .656

a p (0 ) = C o°(-l )° (- 9 )20 = .1 215767 p (3 ) = C f ( - l ) ' (- 9)17 = .1 901199

M D = C?0(.1)1(.9)19 = .2701703 p(4 ) = C 4 ,( . l )4 ( .9 ) '6 = .0897788

^( 2 ) = C l°(-1)2 (-9) '8 = .2851798

so that P [ x < 4] = p ( 0) + p ( 1) + p { 2) + p ( 3) + p ( 4) = .9568255

b Using Table 1, Appendix I , P [ x < 4] is read directly as .957.

c A dding the entr ies for x = 0 ,1 ,2 ,3 ,4 , we have P [x < 4] = .956826 .

d p = np = 20(. 1) = 2 an d O = yjnpq = VE8 = 1.3416

e For ¿ = 1, / /± c r = 2 ± l .342 or .658 to 3 .342 so tha t

P [ .658 < jc < 3.342] = P [ 1 < x < 3] = .2702 + .2852 +. 1901 = .7455

For k = 2 , / j ± 2 ct = 2 ± 2 .683 or - .68 3 to 4.683 so that

P [ - . 6 8 3 < x < 4 .6 8 3 ] = P [ 0 < * < 4 ] = .9 56 9

For k =3 , p ± 2 o = 2 ± 4.025 or -2.02 5 to 6.025 so that

P [-2 .02 5 < jc < 6.025] = P [0 < x < 6] = .9977

f The resul ts are consistent with T ch eb ysh ef fs Theo rem and the Emp irical Rule .

Define x to be the num ber of alarm system s that are tr iggered. Then

p = P [alarm is tr iggered ] = .99 and n = 9 . Since there is a table availab le in

Appendix I for n = 9 and p = .99 , you sho uld use it rather than the binom ial formula

to calcúlate the necessary probabilit ies.

a P[at least one alarm is tr iggered] = P ( x > 1) = 1 - P ( x = 0 ) = 1- . 0 0 0 = 1 . 00 0 .

b P[more than seven] = P ( x > 7 ) = \ - P ( x < 7 ) = 1- .0 0 3 = .997

c P[e ight or fewer] = P ( x < 8) = .086

5.25 Define x to be the num ber of cars that are black. Th en p = P [bla ck ] = .1 and ti = 25.

Use Table 1 in App endix I.

44

a />( jc> 5 ) = 1 - / >( jc< 4 ) = 1- .9 02 = . 098

b P ( x < 6 ) = .991

c P ( x > 4 ) = l - P ( x < 4 ) = 1 -.9 0 2 = .098

d P ( x = 4 ) = P ( x < 4 ) - P ( x < 3 ) = .9 0 2 -.7 6 4 = .138

e P ( 3 < jc < 5 ) = P ( x < 5 ) - P ( x < 2 ) = .9 6 7 - .5 3 7 = .4 30

f P ( m o r e th an 2 0 not b lack) = P ( less than 5 b lack) = P ( x < 4) = .902

5.29 De fine x to be the num ber of f ields infested with whitefly.

Then p = P [infec ted field] = .1 and n = 100.

a p = n p = 10 0(.l) = 10

b S in ce n is large, this binom ial distr ibution should be fair ly m ound -shaped , even

though p = .1 . Henee you would expect approximately 95% o f the measurem ents to

l ie w ithin two standa rd deviat ion of the m ean w ith o = y j npq = ^10 0(.1)( .9) = 3. The

lim its are calculated as

/i ± 2 ít => 10 ± 6 or from 4 to 16

c From par t b, a valué of x = 25 would be very u nlikely, assum ing that the

characterist ics of the binom ial experime nt are met and that p = . 1 . If this valué were

actually observed, it might be possible that the triáis (fields) are not independent.

This could e asi ly be the case, since an infestation in one field m ight quickly sprea d to

a neighboring field. This is evidenc e of contagión.

5.33 Def ine x to be the numb er of Am ericans who are “tas ters”. T hen, n = 20 and p = .7 .

Using the binom ial tables in Appe ndix I ,

a P ( x > 17 ) = 1 - P ( x < 16)= 1 - . 8 9 3 = .1 07

b P ( x < 15) = .762

5.35 Follow the instruct ions in the My Personal T rainer sect ion. The answ ersare shown in

the table below.

P robability F orm ula C alcula ted va lu é

P(x = 0) 2 . 5 V 250!

.0821

P ( * = l ) 2.51e~25

1!

.2052

P(* = 2) 2 . 5 V 25

2!

.2565

P(2 or fewer

successes)

P (x = 0 ) + P (x = 1) + P (x = 2) .5 438

45

5.39

5.43

5.47

5.51

Using p ( x ) = p x e~^

x \2°e~2

X -22* e

X !

a p[ jc = 0l = ---------- = .1353351 J O!

2 V 2 b /> [* = l] = —— = .2 7067

c P [ x > 1] = 1- P [ x < 1] = 1 - .13 5 33 5 - .27 0 67 = .593994

9-V2d P [ * = 5] = --------- = .036089

1 J 5!

Let x be the num ber of misses during a given m onth. The n x has a Poisson

distribution with p = 5.

p ( 0) = e '5 = .00675 V 5

p( 5) = ---------= .17555!

c P [ x > 5] = 1 - />[* < 4] = 1- .4 4 0 = .560 f rom Table 2 .

The random var iable x, n um ber of bacteria, has a Poisson distr ibution w ith p = 2 .

The probab il ity of interest is

P [ x exceeds máx imum coun t ] = P [ x > 5]

Using the fact that p = 2 and < j = 1.414 from E xercise 5.47, m ost of the obse rvations

should fall within p ± 2cr or 0 to 4. H enee, it is unlikely that x will exc eed 5. In fact,

the exact Poisson probab il ity is P [ x > 5] = .017.

C4C"The formula for p {x ) is p ( x ) = x_ ¿ x for x = 0,1,2,3

p ( 0) =c 4c " 165

C j5 45 5= .36

^ = £ ^ = « - = .15Cj5 455

C 4r “ 2 2 0 p ( \ ) = = — — = .48

Cj5 455

C 4 C “ 4 P( 3) = - b r - = — - .0 1

c 15 455

b T h e p robabil it y h is to gram is sho wn belo w .

5.55

5.61

c U sing the form ulas given in Section 5.4.

( í 4 ^

b i i

’ j j

, 1 5 ]

( N - M > Í N - n > o í 4 "l f 15 —4 ̂i * U - l y i , 5 ) I . 5 - 1 )

= .50286.

d Ca lcúlate the intervals

p ± 2 a = .8 ±2>/ .50286 = .8 + 1 .418 or - .6 1 8 to 2 .218

/ / ± 3 cr = . 8 ± 3-V-50286 = .8 ± 1 .418 or -1 .3 2 7 to 2 .927

Then,

P [- .6 18 < jc < 2.218] = p ( 0) + p ( 1) + p ( 2) = .99

P [ -1 .327 < x < 2 .927] = p ( 0) + p ( 1) + p {2) = .99

These results agree with Tchebysheff’s Theorem.

a Th e random variable x has a hype rgeom etric distr ibution with

N = 8,M = 5 and rc = 3. Th en

-.5 -̂3

p (x ) = * for x = 0,1,2 ,3C

b P ( x = 3) =C¡C¡ _ 10

C*= — = .1 78 6

56= — = . 0 17 86

56

r 5r ? r 5r 3 1 + 15

d P ( x < 1) = + - 4 - ^ - = I j l i l = .2857V ' C¡ C3* 56

Refer to E xercise 5 .60 and assume that p = . 1 instead o f p = .5 .

a P [ . t = 0 ] = p ( 0 ) = C o(- l )° ( -9) ' = .729

P [ j : = 1]= />(1) = C¡'(-1)‘(-9 )2 = .24 3

/>[.v = 2] = p(2 ) = C 5 (.l )2 (.9) ' = .027

P [ . i = 3] = p ( 3 ) = C ! ( . l ) ' ( . 9 ) “ = .0 0 1

b N o te th at th e p robab il it y d is tr ib ution is no lo nger sym m etr ic ; th at is , sin ce th e

p robabil ity o f observ in g a head is so sm all, th e p robabil it y o f observ in g a sm all

num ber of heads on three f lips is increased (see the f igure on the nex t page).

47

0 1 2 3

c f j = n p = 3(. 1) =.3 and (T = yjn pq = ^ 3 ( . l ) ( .9 ) = . 52 0

d Th e desired intervals are

/ / + ít = . 3 ± . 5 2 0 o r - . 2 2 0 t o .8 20

/y ±2(7 = .3±1 .04 or - .7 4 0 to 1 .34

The only valué of x which falls in this first interval is jc = 0, and the fraction of

me asurem ents in this interval will be .729. Th e valúes of * = 0 and jc = 1 are enclosed

by th e second in te rv al, so th at .7 2 9 + .243 = .9 72 o f th e m easure m en ts fa ll w ithin tw o

standard devia t ions of the mean, cons is tent with both Tch eby shefPs Theorem and the

Empirical Rule.

5.65 Refer to Ex ercise 5.64. Red efine x to be the numb er of people who cho ose an inter ior

num ber in the samp le of n = 20. Then x has a binomial distr ibution with p = .3 .

a /> [jc > 8] = 1 - / > [ . * < 7 ] = 1 - 7 7 2 = .2 2 8

b Observing eight or more people choosing an inter ior number is not an unlikely

event, assum ing that the integers are all equally likely. Therefore, there is no

evidence to indícate that people are more likely to choose the inter ior numbers than

any others.

5.69 It is given that jc = nu m ber of pat ients with a psychosomatic problem, n = 25 , and

p = f ’ fpatient has psych osom atic pro ble m ]. A psyc hiatr ist w ishes to determine

whether or not p = .8 .

a A ssum ing that the psyc hiatr ist is correct ( that is, p = .8 ), the expected valué of

jc is E ( x ) = n p = 25(.8) = 20.

b ( f L = n pq = 25( .8) (.2) = 4

c G iven that p = .8 , P [ x < 14] = .006 from Tab le 1 in Appen dix I.

d Assum ing that the psych iatr ist is correct,the probab ility of observing jc = 14 or

the more u nlikely valúes, x - 0, 1, 2, . . . , 13 is very unlikely. H enee, one of twocon clusions can be drawn . Either we have observed a very unlikely even t, or the

p sychiatr is t is in corr ect and p is actually less than .8. W e would probab ly conc lude

that the psychiatr ist is incorrect. Th e prob ability that we have m ade an incorrect

decisión is

48

P [ x < 14 given p = .8] = .006

which is quite small.

5.71

5.75

5.79

Define x to be the n um ber of students 30 years or older ,

with ti = 200 and p = ^ [stud en t is 30+ years] = .25.

a S ince x has a binom ial distribution, p = n p = 200(.25) = 50 and

G - y fñ p q - yj 2 0 0 (. 2 5 ) (. 1 5 ) = 6.124.

b The observed valué, x = 35 , lies

^ « = - 2 . 4 56.124

standard deviations below the mean. I t is unlikely that p = .25.

a The random variable x, the num ber of plants with red petáis , has a binomialdistr ibution with n = 10 and p = P[re d petáis] = .75.

b Since the valué p = .75 is not given in Table 1, you m ust use the binom ial

formula to calcúlate

P ( x > 9 ) = C¿° (.75 f ( .25)' + C¡¡¡ ( .75) '° ( .25)° = . 1877 + .0563 = .2440

c P ( x < 1) = C (.75)° (.25 )'° + C,10 (.75)' ( ,2 5)9 = .0000 296 .

d Refer to part c. The proba bility o f observing x = 1 or som ething even more

unlikely ( x = 0) is very small - .0000296. This is a highly unlikely event if in

fact p = .75 . Perhaps there has been a nonrandom choice o f seeds, or the 75% figureis not co rre d for this particular genetic cross.

a The distr ibution of x is actually hyp ergeom etr ic, w ith N = 1200, n = 20 and M =

num ber of defectives in the lot. How ever, s ince N is so large in com parison to n , the

distr ibution of * can be closely approxim ated by the binom ial distribution w ith

// = 20and p = P [defective].

b If p is small, with np < 7, the Poisson approxim ation can be used.

c If there are 10 defe ctive s in the lot, then p = 10/1200 = .008333 and p = .1667.

The probability that the lot is shipped is, ( .1 6 6 7 )% -1667

p ( x = 0 W ---------= .850 !

If there are 20 d efectives, p = 20/1200 and p = .3333. Then

x ( .3 3 3 3 Í V 3333/> (jc = 0 ) * - = .72V ’ 0!

If there are 30 d efectives, p = 30/1200 and p = . 5. The n

( 5 )° éf 5/> (* = 0 ) = ^ = .61

’ 0 !

49

5.83 a The random variable x, the num ber o f tasters w ho pick the correct sam ple, has a

b in om ia l d is tr ib ution w ith n =5 and, if there is no difference in the taste of the thrce

samples, p = P{ taster picks the correct sam ple) = ^

b T he pro babil ity th at exactl y one o f th e five ta s te rs chooses th e la te st batc h as

different from the others is

5.87 The random variable x has a Poisson distribution with p = 2. Use Ta ble 2 in

Appendix I or the Poisson formula to find the following probabilities.

5.91 The random variable x , the num ber of California hom eow ners w ith earthquake

insurance, has a binom ial distr ibution with n = 15 and p = . 1.

a P ( x > 1) = 1- P ( x = 0 ) = 1 - .206 = .794

b P ( x > 4) = 1 - P(x < 3) = 1- .944 = .056

c Calcúlate p = np = 15(. 1) = 1.5 an d a = yfñpq = >/ l5( .l ) (.9) =1.16 19. Then

approximately 95% of the valúes of x should lie in the interval

/ /± 2 c r = > 1 .5 ± 2 ( 1 .1 6 1 9 ) = > - .8 2 to 3 .8 2.

or between 0 and 3.

5 .95 Use the C alcu lat ing Bino m ial P rob ab i l i ties applet . T he correct answers are g iven

below .

a P ( x < 6 ) = 6 .0 ( 1 0) "5 = 0 .0 0 0 0 6 d P ( 2 < x < 6) = .5948

b P ( x = 8 ) = .0 42 e P ( * > 6 ) = 1

c P ( x > 14) = .0207

5.99 Define x to be the num ber of young adu lts who prefer M cD ona ld’s. Then x has a

b in om ia l d is tr ib ution w ith n = 100 and p = .5 . Use the C alcu lat ing Binom ial

P rob ab i l i ties applet.

a P ( 6 l < x < 100) = .0176

a P(jc = 0) = ----------= e~2 = .1353350 !

= . 135335 + .270671 + .270671 = .676676

50

b P ( 4 0 < x < 60) = .9648

c If 40 p refer Burger King, then 60 prefer M cDo nalds, and vice versa. The

pro bab il ity is th e sam e as th at calc u la te d in part b. sin ce p = .5.

51



6: The Normal Probability Distribution

6.3 The first few exercises are designed to provide pract ice for the student in evaluat ing

areas under the normal curve. The fol low ing notes m ay be of som e assistance.1 Ta ble 3, Ap pendix I tabulates the cum ulat ive area under a standard norma l curve

to the left of a specified valué o f z.

2 Since the total area under the curve is one, the total area lying to the r ight o f a

specified valué of z and the total area to its left m ust add to 1. Th us, in order to

calcúlate a “tail area” , such as the one sho wn in Figure 6.1, the valué o f z = z0 will be

indexe d in T ab le 3, and the area that is obta ined will be subtrac ted from 1. De note

the area obtained by indexing z = z0 in Tab le 3 by / \ (z0) and the desired area by A.

Then, in the above e xam ple, A = 1- A (z 0) .

3 T o find the area under the standard normal curve betw een tw o valúes, z¡ and z2,

calcúlate the difference in their cum ulat ive areas, A = A { z 2 ) - A ( z x).

4 N ote that z, sim ilar to x, is actual ly a random variable which may take on an

infinite num ber of valúes, both posi t ive and negative. N egative valúes of z lie to

the left o f the m ean, z = 0 , and posi t ive valúes l ie to the r ight .

a I t is necessa ry to find the area to the left o f z = 1.6. Th at is, A = A (1.6) = .9452.

b T he area to th e le ft o f z = 1.83 i s A = /ftl .83) = .9664.

c A = A (.90) = .8159

d A = A(4.58) = 1. N otice that the valúes in Tab le 3 app roach 1 as the valué of z

increases. W hen the valué o f z is larger than z = 3.49 (the largest valué in the table) ,

we can assume that the area to its left is approximately 1.

6.7 N ow we are asked to fi nd the z-valu e co rr espond in g to a part ic u la r area.

53

a W e need to f ind a zo such that P ( z > z0 ) = .025. T his is eq uiva lent to finding an

indexed area o f 1- .025 = .975 . Search the interior of Ta ble 3 unti l you find the four-

digi t numbe r .9750. Th e correspond ing z-value is 1.96; that is, A ( \ .96) = .975 0.

Therefore, z0 = 1.96 is the de sired z-value (see the f igure below).

b W e need to fi nd a zo s uch th a t P { z < zQ) = .9251 (see below ). U sing Ta ble 3, we

find a valué such that the indexed area is .9251. Th e correspon ding z-value is

* 0 = 1 . 4 4 .

The pth percenti le ,of the standa rd no rmal d istribution is a valué of z wh ich has area

p / 100 to its left. Since all fou r perc entiles in this exerc ise are gre ater than the 50 ,h

percenti le , th e valu é o f z will all lie to the right of z = 0 , as show n for the 90 th

perc enti le in th e fi gure on th e next page.

54



a Fro m the figure, the area to the left o f the 90 lh perc entile is .9000. Fro m Ta ble 3,

the appro priate valué o f z is closest to z = 1.28 with area .8997. He nee the 9 0lh

perc enti le is appro x im ate ly z = 1 .28.

b A s in part a, the area to the left of the 95th perce ntile is .9500. From Ta ble 3, the

appropriate valué o f z is found using l inear interpolat ion (see E xercise 6.9b) as

z = 1.64 5. H enee the 95lh percen tile is z = 1 .645.

c Th e area to the left of the 98th percen tile is .9800. From T able 3, the appropriate

valué o f z is closest to z = 2.05 with area .9798. H enee the 98 lh perc entile is

approximately z = 2 .05.

d Th e are a to the left of the 99lh percen tile is .9900. From Ta ble 3, the approp riate

valué of z is close st to z = 2.33 with area .9901. H enee the 9 9,h perc entile is

approximately z = 2 .33.

6.15 Th e 99,h perc entile of the standa rd normal distribution was found in Ex ercise 6.1 Id to

be z = 2 .3 3 . S in ce the re la tionship betw een the genera l norm al random vari able *

x — uand the standard normal z is z = — , the corresponding percenti le for this general

o

normal random variable is found by solving for x = f j + z (J \

10

jc- 35 = 23.3 or x = 58.3

6.19 The random variable jc, the height o f a male hum an, has a normal distr ibution

with fJ = 6 9 and o = 3.5 .

a A height of 6 ’0” represents 6(12) = 72 inches, so that

P ( x > 7 2) = p Í z > — = p { z > -86) = 1 - .805 1 = .1949

b H eights o f 5’8” and 6 ’1” represen t 5(12) -t- 8 = 68 and 6(12) +1 = 73 inches,

respectively. Then

55

P (68 < jc < 73) = p \ —— — < z < — j = ^ ( - - 2 9 < z < 1.1 4)

= . 8 7 2 9 - . 3 8 5 9 = . 48 70

c A height o f 6 ’0 ” represents 6(12) = 72 inches, which h as a z-value of

6 8 - 6 9 7 3 -6 9 ^1

3.5

This w ould not be considered an unusu ally large valué, since it is less than two

standard deviations from the mean.

d The probab ility that a man is 6 ’0” o r taller was found in part a to be .194 9, w hich

is not an unusual occurrence. Ho we ver, if you define y to be the number of men in a

random samp le of size « = 36 wh o are 6 ’0” o r taller, then y has a binom ial

distr ibution with mean f i = np = 36(. 1949) = 7.02 and standard deviation

a = y¡ ñp q = y l 3 6 ( . \ 9 4 9 ) ( M 5 \ ) = 2 .3 8 . The valué y = 17 l ies

standard deviations from the mean, and would be considered an unusual occurrence

for the general population of male humans. Perhaps our presidents do not represent a

random sample from this population.

6.23 The random variable jc, total we ight of 8 people, has a m ean o f/r = 1200 and a

variance o 1 = 980 0. I t is necessary to f ind P ( x > 1300) and P ( x > 1500) if the

distr ibution o f jc is approxim ately norm al. Re fer to the next f igurc-

o 2.38

1200 1300 1300

- p 1 3 0 0 - 1 2 0 0 _ 100- 4 - = — . = ------------ = 1.01

o V9800 98.995

P ( x > 1300) = P ( z > 1.01) = 1—A(1.01) = 1—.8438 = .1 5 6 2 .

56

6.27

6.31

6.37

Similar ly, the z-value corresponding to x 2 =1 50 0 is

_ *2 - / / _ 1 5 0 0 -1 2 0 0 ^ Zi — — .------- j .V j j .

(7 >/9800

and P ( x > 1500) = P ( z > 3 .03) = 1 - ¿ (3 .03) = 1- .99 8 8 = .0012 .

a It is given that the prime interest rate forecasts, x, are approximately normal with

mean /y = 4.5 and standard d eviation o - 0.1 . I t is necessary to determ ine the

p ro b ahili ty th at x exceed s 4 .7 5 . C alc úla te

x - n 4 . 7 5 - 4 . 5 z = — = --------------- = 2 . 5 . T h e n

o 0.1

P ( x > 4 .75) = P ( z > 2 .5 ) = 1- .9 9 38 = .0062 .

x — u 4 3 7 5 - 4 5b Calcúlate z = ----- -- = -1 -25 . Th en

o 0.1

P ( x < 4 .375) = P ( z < - 1 . 2 5 ) = .1 0 5 6 .

Let w be the num ber of words specif ied in the contract. Then jc, the num ber o f words

in the man uscript, is normally distr ibuted with /y = w + 20 ,000 and cr = 10 ,000 . The

pu b li sh er w ould like to specify w so that

P ( x < 100,000) = .95.

As in Ex ercise 6.30, calcúlate

1 00 ,0 00 0- ( w + 2 0,0 00 ) 8 0 , 0 0 0 - w Z - -

10,000 10,000

8 0 , 0 0 0 - w } —_ T ,= .95 . It is nec essary thatThen P ( x < 100,000 ) = P z <

K ’ 10,000 )

z 0 = ( 8 0 , 0 0 0 - w ) / l 0 , 0 0 0 b e s u ch t ha t

/ , ( z < z 0 ) = -9 5 => A ( z 0 ) = .9500 or z0 = 1.6 45 .

Henee,

8 0 , 0 0 0 - w

10,000= 1.645 or w = 6 3 , 5 5 0 .

a The normal app roxima tion w ill be approp riate if both n p and nq are greater than 5.For this binomial experiment,

np = 2 5(3 ) = 1.5 and nq = 25( .7) = 17.5

and the normal approximation is appropriate.

b For the binomial random variable,

p = n p = 7.5 and o = y jnpq = -v/25 (.3)(.7 ) = 2.2 91 .

c The probahility of interest is the area under the binomial p robah ility histogram

corresponding to the rectangles x = 6,7,8 and 9in the f igure on the next page.

57

6.41

6.45

To approximate this area. use the “correction for continuity” and f ind the area under a

normal curve with mean fu = 7.5 and o = 2.291 between x { = 5.5 and x 2 = 9.5 . The

Z-values corresponding to the two valúes of x are

z, = ^ = -. 87 9 . 5 - 7 . 5 ^and z, = ------------- = .87

2 2.2912.291

The approximating probabi l i ty is

P ( 5 .5 < * < 9 .5 ) = P ( - . 8 7 < z < .87) = .80 78 - .19 22 = .6156 .

d From Table 1, Appendix I ,/>(6 < jc < 9 ) = / >( jc < 9 ) —/>( jc < 5 ) = .811 —. 193 = .618

which is not too far from the approximate probability calculated in part c.

Using the binomial tables for n = 20 and p = .3 , you can ver ify that

a P ( .r = 5) = P ( . x < 5 ) - P ( j t < 4 ) = .4 1 6 - .2 3 8 = .178

b P ( x > 7 ) = 1 - P ( a : < 6 ) = 1- .608 = .392

a The approximating probability will be P ( x > 2 0.5) wh ere x has a normal

distribution with ¡a = 50( .32) = 16 and o = >/50( .32 )( .68) = 3.298 . Then

2 0 . 5 - 1 6 ' P ( x > 20.5) = P z > ■

\

b The approximating probability is

3.298

P ( x < 14.5) = P z <1 4 . 5 - 1 6

3.298

= P ( Z > 1.36 ) = 1—.9131 = .0869

= P ( z < - .4 5 ) = .3264

c If few er than 28 students do not prefer cherry, then more than 50 - 28 = 22 do

p re fe r cherr y. T he appro xim atin g p ro babil it y is

P ( x > 22 .5) = p ( z > 22-5 ~ .16.v ’ 1, 3 .298

d As long as your class can be a ssum ed to be a representative sam ple o f all

Americans, the probabilit ies in parts a-c will be accurate.

= P ( z > 1 .97) = 1- .9756 = .0244

58



6.49 D efine x to be the num ber of elect ions in wh ich the tal ler cand idate won. If

Am ericans are not biased by height , then the random variable x has a binom ial

distr ibution with n = 31 and p = .5 . Calcúlate

p = np = 31(.5) = 15.5and G = y¡ 3\(.5 )(.5 ) = V7/75 = 2.784

a Using the normal approximation with correction for continuity, we find the area tothe r ight o f x = 16.5 :

P ( x > 16.5) = P\^z > 1625 ~g45 '5 ) = P ( z > -36) = 1 - 6406 = .3594

b S in ce the occurr ence o f 17 ou t o f 31 ta ll er choic es is not unusual, based on th e

results of part a, i t appe ars that Am ericans do not consider height w hen cast ing a vote

for a can didate.

6.53 Re fer to Exercise 6.52, and let x be the num ber of working wom en w ho put in more

than 40 hours per w eek on the job. Then x has a binomial distr ibution with n = 50

and p = .62 .

a The average valué of * is ( i = np = 50(.62) = 31.

b The sta ndard devia tion o f x is G = yjnpq = >/50(.62 )(.38) = 3.432.

c The z-score for x = 25 is z = ——— = — — — = -1 .7 5 w hich is within twoG 3.432

standard deviat ions o f the m ean. This is not considered an u nusual occu rrence.

6 .55 a Th e de sired are A| , as shown in the f igure on the next page, is found by

subtract ing the cum ulat ive areas corresponding to z = 1.56 and z = 0 .3 , respectively.

A, = A ( 1 .56) - A( .3) = .9406 - .6179 = .3227 .

b T he desir ed area is show n on the next page :

A, + A, = A(. 2) - A { - .2 ) = .5793 - .4207 = .1586

59

6.59

6.63

desired valué, zo, will be between z, = -67 and z2 = .68 w ith asso ciated p robabilit ies

Px = . 2 5 1 4 a n d P2 = .2483 . Sinc e the de sired tail area, .25 00, is clos er to P[ = .25 14 ,

we approximate zo as z0 = .67 . Th e valú es z = -.6 7 and z = .67 repre sen t the 25 1*1

and 75th percentiles of the standard normal distribution.

It is given that x is normally distr ibuted with fu = 10 and <7 = 3. L et t be the

guaran tee time for the car . I t is necessary that only 5% of the cars fail before time t (see below). That is,

P ( x < í ) = . 05 o r p f z < L - J 5 j = .05

From Table 3, we know that the valué of z that satisf ies the abov e prob ability

statement is z = -1 .645 . Henee,

60

6.67

6.71

6.75

-— - = -1 .6 4 5 or t = 5.065 months

W e mu st ha ve

P ( x < X o ) = p [ z < ^ y .

' v - 7 0

l 12

90

= .90

xn —70Consider z0 = — . W ithout interpolat ing, the approxim ate valué for zo is

12

x , - 7 0 _ o

For this exercise ¡u =10 and o = 12. The o bject is to determ ine a part icular valué, x 0,

for the random variable x so that P ( x < x 0 ) = .9 0 (that is, 90% of the students will

f inish the exam ination before the set t im e l imit) . Re fer to the f igure below.

-------------------

I t is given that the ran dom variable * (ounce s of f i ll ) is norm ally distr ibuted with

mean /y and standard deviat ion o = .3 . It is nec ess ary to find a valu é o f /y so that

P { x > 8) = .01 . Th at is, an 8-ounce cup w ill overflow w hen x > 8, and this should

happen only 1% of the t ime. Then

7 - / / ' P ( x > 8) = P ^ z > = .0 1 .

•3

From T able 3 , the valué of z correspond ing to an area ( in the upp er tail o f the

distribution) o f .01 is z0 = 2.3 3. Flence, the valué of /y can be ob tained by solving

for/y in the fol lowing equation:

2 .33 = - ^ ^ or / i = 7.301.3

Def ine x = num ber o f incom ing cal is that are long distance

p = P[inco m ing cal i is long distance] = .3

61

n = 200

The desired probab ility is P ( x > 5 0 ) , wh e r e x is a binomial random variable with

H = n p = 200( .3) = 60 and ^ = V2Ó 0(3)C7) = >/42 = 6.481

A correction for continuity is mad e to include the entire area un der the rectangle

cor responding to x = 50 and henee the approximation wil l be

6.79 The random var iable jc,the gestation time for a human baby is normally distr ibutedwith p = 278 and <7 = 12 .

a From Exercise 6.59, the valúes (rounded to two decimal places) z = - .6 7 and

Z = .67 represe nt the 25 lh and VS01 pe rcentiles o f the sta nd ard norm al distribu tion.

Converting these valúes to their equivalents for the general random variable x using

the relationship jc = p + z o , you have:

b If you consider a month to be approximately 30 days, the valué x = 6(30) = 180 is

unusual, since it lies

standard deviations below the mean gestation time.

6.83 In order to implem ent the traditional interpretation of “curving the g rades” , the

p ro port io ns show n in th e ta ble need to be applied to th e norm al curv e, as show n in th e

figure below. _________________________________________________ ______

The lower quartile: jc = - . 6 7 ( 1 2 ) + 2 78 = 26 9 .9 6 a nd

The upper quartile: jc = .67(12) + 2 78 = 286.04

_ x - f j _ 1 8 0 - 2 7 8

o 12= - 8 . 1 6 7

A

0.3-

S 0.2-

0 . 1 -

0.0 -

-3 -2 -1 0 2 3x

62

a The C grades constitute the middle 40%, that is , 20% on either side of the mean.

Th e low er bound ary has a rea .3000 to its lef t. From T able 3, we need to Find a valué

o f z such that A(z) = .3000 . The close st valué in the table is .3015 w ith z = - .52 . The

upper boundary is then z = +.52 .

b The cutoff for the lowest D and highest B grades constitute the lower and upper bou ndari es o f th e m id dle 80% , th at is, 40% on e it her sid e o f th e m ean . T he lo w er

boundary has arc a .1 000 to its le ft , so we need to Find a valu é o f z such that A(z) =

.1000. The closest valué in the table is .1003 with z = -1 .28 . The upper boundary is

then z = 1.28.

6.87 Use eithe r the Normal Distribution Probabilit ies or the Normal Probabil i t ies and

z-scores applets.

a P ( - 2 . 0 < z < 2 .0) = .97 72 - .022 8 = .9544

b P ( - 2 3 < z < - 1 . 5 ) = . 0 6 6 8 - . 0 1 0 7 = .0561

6.91 a Use the Norm al D istr ibut ion Probabil i ties applet. En ter 5 as the mean and 2 as

the standa rd deviation, and the ap propriate lower and upp er boun daries for the

pro bab il iti es you need to calc úla te . T he pro babil ity is re ad fr om th e apple t as Prob =

0.9651.

b Use the Norm al Proba bil it ies and z-scores applet. En ter 5 as the mean and 2 as

the standard deviation and x = 7.5 . Ch oose One-tail from the dropdown list and

read the prob ability as Prob = 0.1056.c Use the No rm al P robabil i ties and z-scores applet. En ter 5 as the mean and 2 as

the stand ard deviation and x = 0 . Choose Cumulat ive fromthe dropd ow n list andread the prob ability as Prob = 0.0062.

6.95 a I t is given that the scores on a national achievem ent test were approxim ately

norm ally distr ibuted with a mean of 540 and standard deviation o f 110. I t is

necessary to determine how far. in standard deviations, a score of 680 d eparts from

the mean of 540. Calcúlate

x - / J 6 8 0 - 5 4 0 ,Z = --------- = -------------------= 1 .2 / .

o 110

b To Find the percentage of people who scored h igher than 680 , we f ind the area

und er the standardized no rmal curve greater than 1.27. Using T able 3, this area isequal to

P ( x > 680) = P ( z > 1 .27) = 1- .89 8 0 = .1020

Th us, approx imately 10.2% of the people w ho took the test scored h igher than 680.

(The applet uses three decimal place accuracy and shows z = 1.273 with Prob =

0.1016.)

63

7: Sampling Distributions

7.1 You can select a s imple rando m sample o f s ize n = 20u sing T able 10 in Ap pendix I .

First choo se a starting point and consid er the first three digits in each nu m ber. Since

the experimen tal units have already been n um bered from 0 00 to 999, the first 20 can be

used. The three digits OR the (three digits - 500) will identify the prop er experimen tal

unit . For exam ple, if the three digits are 742, you should select the experimen tal unit

num bered 7 4 2 -5 0 0 = 242 . T he probahil ity that any three digit num ber is selected is

2/10 00 = 1/500 . One p ossible selection for the sample size n = 20 is

242 134 173 128 399

056 412 188 255 388

469244

332 439 101399 156 028 238 231

7.5 If a l l o f the town cit izenry is l ikely to pass this com er, a samp le obtained by selecting

every tenth person is probably a fairly random sample.

7.9 Use a random ization schem e similar to that used in Exercise 7 .1. N um ber each of the

50 rats from 01 to 50. To ch oose the 25 rats who will receive the dose of MX , select

25 two -digit rando m nu m bers from Table 10. Each tw o-digit num ber OR the (two

digits - 50) will identify the prop er experimental unit .

7.13 a Th e firs t question is more unbiased. b N oti ce th at th e percenta ge favorin g th e new sp ace pro gram drops dram ati call y

wh en the phrase “spending b il lions o f dollars” is added to the question.

7.19 Re gardless o f the shape of the population from wh ich we are sampling, the sam pling

distribution o f the sample m ean will have a mean / / equ al to the mean o f the

popu la ti on fr om w hic h we are sam pling, and a st andard devia tion equal to o ¡ \ f ñ .

a /y = 10; ( j / y fñ =3/>/36 =.5

b ju = 5; ( j / y f ñ = 2 / -n/ToO =.2

c n = 120; cr/Vñ = l/>/8 =.3536

7.22-23 For a popula t ion wi th o = 1 , the standard error of the mean is

<7 / y fñ = l / y f ñ

The va lúes of <7 / y fñ for various valúes of n are tabulated below and plotted on the next

page. N oti ce th at th e sta ndard e rro r decreases as the sample size increases. _ n ______________________ 1 __________ 2 __________ 4 _________ 9 _________ 16 __________ 25 ___________ 100

SE(x ) = a /4 ñ L(X) _7Q7 _5(X) J33 _25Q _20Q 100

65

7.25

7.29

7.33

a Ag e o f equipm ent, technician error, technician fat igue, equipm ent fai lure,

difference in Chemical puri ty, contam ination from outside sources, a nd so on.

b T he vari abil it y in th e avera ge m easure m ent is m easured by th e sta ndard err or,

o / y f ñ . In order to decrease this variabil ity you should increase the sam ple size n.

a The po pulat ion f rom w hich we are random ly sampl ing n = 35 measurem ents i s not

necessari ly normally distr ibuted. Ho wev er, the sam pling distr ibution of x does have

an approximate normal distr ibution, with mean ¡a and standard deviat ion <r/ y fñ . T he p robab il it y o f in te re st is

P ( \ x - j u \ < l ) = P ( - 1 < ( * - / / ) < l ) .

Since z = X , has a stand ard norm al distribution, we need only find o / y f ñ to( j /y jn

approx ima te the above probabili ty. Th ough o is unknown, i t can be approximated by

s = 12 and o / y f ñ = 12 /^35 = 2 .028 . T hen

P ( \ x - f i \ < 1) = P ( - 1/2.028 < z < 1/2.028)

= P ( - A 9 < z < .49) = .6879 - .31 21 = .3758

b N o. T here are m any possib le valú es fo r* , th e actu al percen t tax savin gs, as giv en

by th e probabil it y d is tr ib u tion fo r * .

a Since the-eriginal populat ion is norm ally distr ibuted, the sam ple m ean * is also

norm ally distr ibuted (for any sample size) with mean /y and standard d eviat ion

(7/y fñ ~ 0 .8 /V l30 = .07016

The z-value corresponding to * = 98.25 is

9 8 . 2 5 - 9 8 .6 _

• Cj/yfñ 0.8A/I3Ó

66

7.37

7.41

7.45

an d

P ( x < 98.25) = P ( z < - 4 . 9 9 ) = O

b Sin ce th e pro babil ity is ex tr em ely sm all , th e avera ge te m pera ture o f 98.2 5 degre es

is very unlikely.

a p = .3; S E ( p ) = = .0458V ' V n V 100

b p = .l ; S£ ( p ) = ̂ = J | | = .015

c p = . * 0310

The valúes S E = p q / n for n = 100 and v arious valúes o f p are tabulated and graphed

belo w . N oti ce th at S E is máximum for p = .5 and becom es very small for p nea r zero

and one. p .01 .10 .30 .50 .70 .90 .99

SE (P ) .0099 .03 .0458 .05 .0458 .03 .0099

a The random variable p , the sample propor t ion o f brown M &M s in a package of

n = 55 , has a binomial distr ibution with n = 55 and p = . 13. Since n p = 7.15 and

nq = 47.85 are both greater than 5, this binomial distr ibution can be approx imated by a

I 13( .87 )normal distr ibution with mean p = . 13 and S E = . : 1 = .04535.

F V 55

b / - ( p < . 2 ) = / > ^ < | ^ | j = P ( c < 1 .5 4 ) = .9 382

67



c P (p > . 3 5 ) = ̂ > ^ ^ j = /> (j > 4 .8 5 ) = 1 -1 = 0

d From the Empirical Rule (and the general properties of the normal distr ibution) ,approximately 95% o f the m easurem ents will l ie w ithin 2 (or 1.96) standard d eviations

of the mean:

p ± 2 S E => . 1 3 ± 2 ( . 0 4 5 3 5 )

.13 ±.0 9 or .04 to .22

7.49 aThe upp er and lowe r control l imits are

ÜCL = x + 3-^= = 155.9 + 3 ^ 2 = 155.9 + 5.77 = 161.67

L C L = x - 3 - ^ = = 1 5 5 . 9 - 3 ^ = 1 5 5 .9 -5 .7 7 = 150.13 y/5

b The control chart is constructed by plotting two horizontal l ines, one the upper

con trol limit and one the low er control l imit (see Figure 7.15 in the text) . V alúes of

x are plotted, and should remain within the control limits . I f not, the proces s should be

checked.

7.55 Calcúla te p = — — + -26 - 1 9 7 The upper and lower control l imitsk 30

for the p chart are then

UCL = p + 3 J P (1 P) = . 197 + 3 J - 1- - =. 197 + . 119 = .316

L C L = p - 3 l-P(1- = . 1 9 7 - 3.1' 19 7 (8 0 ^ = . 19 7 - . 1 19 = .0 78V n V 100

7.60 a C \ = = 6 samp les are p ossible.

b-c T h e 6 sam ples a long with the sample means for each are shown below.

Sample _________________ Observat ions ________________ x_

1 6 , 1 3.5

2 6 ,3 4.5

3 6 , 2 4.0

4 1, 3 2 . 0

5 1 , 2 1.5

6 3 ,2 2.5

d Since each of the 6 distinct valúes of x are equally likely (due to random

sampling) , the sampling distr ibution of x is given as

p ( T ) = - fo r x = 1 .5 ,2 ,2 .5 ,3 .5 ,4 ,4 .56

68



7.65

7.71

7.75

The s am pling dis t r ibut ion is shown be low.

1.5 2.0 2.5 3.0 3.5 4.0 4.5

*

e The popu lation mean is p = (6 + 1+ 3 + 2 ) /4 = 3 . Not ice that none o f the samples

of size n = 2 produce a valué o f x exactly equal to the population mean.

a To divide a grou p o f 20 people into two groups of 10, use Ta ble 10 in A ppen dix I.

Assign an iden tif ication n um ber from 01 to 20 to each person. Th en select ten two digit

num bers from the random num ber table to identify the ten people in the f irst group. (If

the number is greater than 20, subtract múltiples of 20 from the random number untilyou obtain a number between 01 and 20.)

b A lt hough it is not possib le to select an actu al ra n do m sam ple from th is hypo th eti ca l

popula ti on , th e re searcher m ust ob ta in a sa m ple th at behaves l ike a random sample . A

large datab ase o f som e sort should be u sed to ensure a fair ly represen tative sample.

c The researcher has actually selected a convenience sample-, however, it will

p ro bably behave like a sim ple ran do m sam ple , s in ce a p e rso n ’s en th us ia sm fo r a paid

jo b should not aff ect his re sponse to th is psychologic al experi m ent.

a Since each cluster (a city block) is censused, this is an exam ple o f cluster sampling.

b T his is a 1 - in -10 syste m ati c sa m ple .

c The wards are the strata, and the sample is a stratif ied sam ple.

d This is a l- in-10 sys tematic sample .

e This is a simple rando m sample fromthe popu lation o f all tax returns f iled in the

city of San Bernardino , California.

a The average propor t ion o f defect ives is

_ .04 + .0 2 H— + .03 ^ p = --------------------------- = .032

25

and the control l imits are

69

U CL = p + 3 j — — — = -032 +n ■/ 100

and L C L = p - 3 P(1 P) = . 0 3 2 - 3n

If subsequ ent sam ples do not stay w ithin the limits, UCL = .0848 and L C L = 0 , the

p ro cess should be checked .

b From par t a, we must have p > .0848 .

c An erroneou s conc lusión will have occurred if in fact p < .0848 and the sam ple has

p ro d u ced p = .15 by chance. On e can obtain an upper bound on the proba bility of this

p arti cu lar ty pe o f e rr o r by calc u la ti ng P ( p > .1 5 when p = . 0848) .

7.79 A nsw ers will vary from student to student. Paying cash for opinions will notnecessa rily produ ce a random sample o f opinion s of all Pepsi and Coke drinkers.

7.83 a Th e average propor t ion of inoperable com ponents is

_ _ 6 + 7 h h5 _ 75 _ ]0

P ~ 50(15) _ 75 0 ~~

and the control l imits are

and

If subseq uen t samples do not stay w ithin the limits , UCL = .2273 and L C L = 0 , the

pro cess should be checked.

7.87 a Th e theoretical mean and standard dev iation of the sampling distr ibution o f x when

ti = 4 are

p = 3.5 and a / y [ ñ = 1.708 / \[ Á = .854

b-c An swers will vary from student to student. The distr ibution sho uld be relatively

uniform with mean and standard deviation cióse to those given in part a.



8: Large-Sample Estimation

The margin o f error in estimation prov ides a practical up per boun d to the difference betw een a part ic u la r estím ate and th e param ete r w hic h it estim ate s. In th is ehapte r,

the margin of error is 1.96x (standard error of the estimator) .

Th e margin of error is 1.96 S E = 1.96-^L , where o can be estimated by the sample yjn

standard deviation s for large valúes o f n.

a 1 . 9 6 , — = . 5 5 4 b 1 .9 6, — = . 1 7 5 c 1 .9 6 ./—̂ — = .055 V 50 y 50 0 V 50 00

For the estím ate of p given as p = x /n , the margin o f error is 1.96 S E = 1.96. — .V n

Use the estimated valué given in the exercise for p.

a 1.96 = .0588 b 1 . 9 6 = .0898 c 1. 9 6 = .098\ 100 V 100 V 100

d 1.96 = .0898 e 1 .9 6 J— ^ — =.0 58 8V 100 V 100

f Th e largest margin o f error occurs w hen p = .5 .

The point est ímate o f p is x = 39.8° and the m argin of error with 5 = 17.2 and n = 5 0

is

1.96 S E = 1 . 9 6 = 1 .96-4= = 1,9 6 - 4 ¿ = 4 .768 yfn yjn V50

X a Th e point estímate for p is given as p = — = . lS and the m argin o f error is

n

approximately

1.96 J — = 1 .9 6 ,f 7 8 ^ 2 2 ^ = .026n V 1000

b Th e p oll’s margin o f error does not agree w ith the results of part a, because the

samp ling error was reported using the máxim um margin o f error using p = .5 :

1.96 E = 1.96 = .0 3 1 o r ± 3 .1 %\ n \ 1000

A p oint estímate for the mean length o f t ime is x = 19.3, w ith m argin of error

1.96 S E = 1 .9 6 -^ r « 1 . 9 6 - ^ = l . 9 6 - 5 ¿ = 1.86 yjn yj n >/30

.23

.27

.31

.35

The 90% confidence interval for/y given as

3c ± 1 .6 4 5 y¡n

w here <7 can be est im ated by the sam ple standard dev iat ion 5 for large v alúes of n.

a .84 + 1 .6 45 ./— = . 8 4 ± . 0 4 3 o r .79 7 < u < .883V 125 ^

J

3 44 — = 2 1 .9 ± .4 3 1 or 21.4 69 < /i < 22.331

50

c Intervals constructed in this m anne r wil l en do se the true valué o f p 90% of the

t im e in repeated sam pling. Henee, w e are fair ly confiden t that these pa rt icular

in tervals wi l l end ose p .

The width of a 95% confidenc e interval for p is g iven as 1 .9 6 -y r . Henee , yjn

a Wh en n = 100, the width is 2L 96j y = 2 ( i -9 6 ) = 3 -9 2 -

b W hen n = 20 0, the width is 2[ 1.96—¡Í£ = = 2(1.38 6) = 2.772 .I > / 2 0 0 j

c Wh en n = 40 0, the width is 2'■967 m ) - - 2 { -9 S ) - - L 9 6 -

With n = 40, x = 3 .7and s = .5 and a = .01 , a 99% conf idence in terval for ¡u is

approximated by

x ± 2 .5 8 -4 = = 3 .7 ± 2 .5 8 —¿ = = 3 .7 ± .2 0 4 o r 3 .4 9 6 < lí < 3.904

•Jn V 40In repeated sam pling, 99% of al l intervals construc ted in this m ann er wil l en do se / i .

Henee, we are fairly certain that this part icular interval con tains f j . (In orde r for this

to be true, the sample m ust be random ly selected.)

a The point est imate o f p is p = — = = . 136, and the approxim ate 95%" n 500

confiden ce interval for p is

p ± 1 .96 = . 136 ± 1 .96 | - 36 ( :864) = . 136 ± .030500or . 106 < /? < . ! 66 .

72

b In ord er to in cre ase th e accura cy o f th e confi dence in te rv al, you m ust decre ase its

width. You can accom plish this by (1) increasing the sample size n, or (2) decreasing

Za/2 by de creasing the co nfidence coeff icient.

8.39

8.43

a W hen estimating the difference /y, - / y 2, the (1 - a ) 100% co nfide nce ijiterval is

(3c. - x ,) ± za/2 — + — . Estimating o \ and o \ w ith s] and s2 , the approximate1 " , " 2

95% confidence interval is

l¡ 38 4 14( 1 2 .7 -7 .4 )± 1 .9 6 J— + — = 5 .3 ± .6 9 0 o r 4 .61 < / / , - f i 2 < 5.99 .

b S in ce th e valu é // , - / y , = 0 is not in th e confi dence in te rv al, it is not li kely th at

/y, = /y , . You should conc lude that there is a difference in the two po pulation m eans.

a Th e param eter to be estimated is /y , the mean score for the posttest for all BA CC

classes. The 95% con fidence interval is approximately

I ± 1 . 9 6 - ^ = 1 8 .5 ± 1 .9 6 -? = £ L = 1 8 .5 ± .8 2 4 o r 1 7 . 6 7 6 < /y < 1 9.3 24V365

b T h e p a ra m ete r to be esti m ate d is / y , th e m ean score for th e postt est fo r all

traditional classes. Th e 95% co nfidence interval is approximately

5 6 963c ± 1 . 9 6 4 = = 1 6 . 5 ± 1 . 9 6 - t = = 1 6 . 5 ± .7 9 0 o r 1 5 .7 10 < /y < 17 .2 90

yf n V298

c No w we are interested in the difference between posttest means, /y, - / y , , for

BA CC v ersus traditional classes. Th e 95% con fidence interval for /y, ~/y 2 is

approximately

( ,8 .5 -1 6 .5 , ± l . 9 6 j M * + « « lv 7 v 365 298

2.0± 1.14 2 or .858 < (/y , - / y 2)< 3.142

d Since the confidence interval in part c has two positive endp oints, i t does not

contain the valué /y, - / y 2 = 0 . H enee, it is not likely that the mean s are equal. It

appears that there is a real difference in the mean scores.

8.47 Refer to Exercise 8.18.

73

a The 95% conf idence interval for / /, - / / 2is approximately

(170 —160) ± 1.96.

1 0 ± 6.667 or 3 .333 < ( / /, - / / , )< 16 .667

b The 99% confidence in te rv al fo r is appro xim ate ly

- 1 5 ± 7 . 0 4 0 o r - 2 2 .0 4 0 < ( // , - / ¿ 2 ) < - 7 . 9 6 0

c N either of the intervals contain the valué ( /v ,- // 2) = 0 . I f ( / / , —//2) = 0 is

contained in the confidence interval, then it is not unlikely that H\ could equal / /2 ,

implying no difference in the average room rates for the two hotels . Th is wo uld be of

interest to the experimenter.

d Since neither confide nce interval con tains the valué //, = 0 , it is not likely

that the me ans are equal. Yo u should con clud e that there is a differenc e in theaverage room rates for the Marriott and Wyn dham and also for the R adisson and the

Wyndham chains .

x 337 x 37 48.51 a Calcúlate A = — = 1— = .42 and p 2 = — = - — = .5 8 . T h e a p p ro x im a t e 90 %

n, 800 ' n 2 640

confidence interval is

b The tw o bin om ia l sam ple s m ust be ra ndom and in dependent and th e sa m ple sizes

must be large enoug h that the distr ibutions of p xand p 2 are approximately normal.

Assum ing that fí le sam ples are random, these conditions are met in this exercise.

- .1 6 ± . 0 4 3 o r - .2 0 3 117

74

8.55 a W ith p. = — — = .45 and p , = -~ 2— = .51. The approxim ate 99% conf idence1 1001 1001

interval is

( A - p ; ) i 2 . 5 8 j M + M

.45 .55 ) .51 .49( .4 5 -.5 1 1 2 . 5 8 . — i ------^ + — i----

v y V 1001 1001

- . 0 6 1 . 0 5 8 o r - .1 1 8 < ( p , - p 2) < - .002

b Sin ce th e in te rv al in part a co n ta ins only negati ve valú es o f p , - p 2 , it is likely

that p| - p 2 < 0 P| < p 2 . Th is wo uld indicate that the proportion of adults who

claim to be fans is higher in November than in March.

r 120 x , 548.59 Ca lcúlate p, = — = ----- = .7 and P7 = — = ------ = .54 . The approximate 90%

1 n, 180 2 n 2 100

confidence interval is

( p , - p 2) ± l - 6 4 5 j ^ +

( .7 - .54) ± 1.645 1v ' V 180 100

.1 6 1 .0 9 9 o r . 0 6 1 < ( p , - p 2) < .2 5 9

Intervals constructed in this man ner w ill en do se the true valué of p , - p 2 95% of the

time in repeated samp ling. Hen ee, we are fairly certain that this particular interval

e n d o s e s p, - p 2 .

8.63 Follow the instructions in the My P ersonal Traine r section. The answ ers are show n in

the table below.

T ype of D ata O ne or Tw o

S a m p l e s

M a r g i n ofe r r o r

p o r a B o u n d ,B

Solve thisinequal i ty

Sam ple s i ze

Binomial One p ~ . 5 .051.96 ^ ^ < . 0 5

V n

n > 3 8 5

Quantitative O ne1 . 9 6 - i -

yjn

( 7 * 1 0 21.9641

« > 9 7

75

8.67 For the difference p x - /y , in the popu lation m eans for two quant i ta t ive p opulat ions ,

the 95% u pper confidence bound uses z ^ = 1.645 and is calculated as

( * , - J 2) + 1.545 + ̂ = ( 1 2 - 1 0 ) + 1 . 6 4 5 J | - + ̂ -\ n, n2 V50 50

2 + 2 .0 0 o r ( / / , - / / , ) < 4

8.71 In this exercise, the param eter o f interest is p¡ - p 2 , nx= n 2 = n , and B = .05 . Since

we have no pr ior knowledge about p xand p 2 . we assum e the largest possible

variation, which occurs if p x = p 2 = .5 . Then

za/2 x(s td er ror of p x- p 2) —— — => n > 1 08 5.7 8 o r « , = « , = 1086.05 1 2

8.75 The standard deviation is estimated as R / 4 = 104/4 = 26 , and .

2.58.1— + — < 5 => 2.58,1— + — < 5n, V n n

^ > 2 .5 8V Í3 52 ^ ^ > 3 59 9 g 0f ^ = ^ = 3 60

8.76 a Fo r the difference /y, - f i2 in the population means this year and ten years ago, the

99% lower conf idence bound uses z 01 = 2.33 and is ca lcula ted as

K s,2 _ 2 5 2 2 8 :

( x - x , )- 2 .3 3 í— + — =(73 —63) —2.33A/------+ ■V 1 ^ nx n2 ’ \ 4 00 4 00

1 0 - 4 . 3 7 o r ( jjx - / / , ) > 5.63

b S in ce th e dif ference in th e m eans is posit ive, you can co n c lu d e th at th ere has been

a decrease in the average per-capita beef consumption over the last ten years.

8.83 a The point es tímate of p is x = 29.1 and the margin o f error in estimation with

5 = 3.9 an d n = 64 is

1.96(7, = 1 .96-?= * 1.96 -7= = 1.963.9

y fñ y fñ y¡64, b T he appro xim ate 90% confi dence in te rv al is

= .9555

x ± 1 .6 4 5 -4 = = 2 9 . 1 ± 1 . 6 4 5 - ^ L = 29 .1 ± . 8 0 2 o r 2 8 .2 9 8 < p < 29.902 yf n V64

76

Intervals constructed in this man ner end os e the true valué of /y 90% o f the t ime in

repeate d sam pling. Th erefore , we are fairly certain that this particular interval

e n d o se s / y .

c The approximate 90% lower conf idence bound is

1 - 1 . 2 8 - ^ = 2 9 . 1 - 1 . 2 8 ^ = 28.48 o r /y > 28.48-Jn V64 _

d W ith B = .5, o ~ 3 . 9 , a nd 1 - a = .95 , we mu st solve for n in the following

inequality:

cr 3.9\ . 9 6 —j= \. 9 b —j= < . 5

■Jn yjn

4 ñ > 15.288 => n > 233.723 or n > 234

8.87 Assum ing máximum var ia t ion wi th p = .5 , solve

1.645J - ^ < .0 2 5

4~n >. i W M =3 2 , n > 1082 .41 o r ai > 1083.025

8.91 a Define sam ple #1 as the samp le o f 482 w omen and samp le #2 as the sample of

356 men. The n p, = .5 and p : = .75.

b T he approxim ate 95% confid ence in te rv al is

( P \ - M ± 1.96 +n, n 2

( .5 - .7 5 ) ± 1 .9 6 M +v ’ \ 482 356

- .2 5 ± . 0 6 3 o r - .3 1 3 < ( p , - p 2) < - .1 87

c Since the valué p, - p 2 = 0 is not in the confidence interval , i t is unlikely that

p l = p 2 . You should not conclude that there is a difference in the proportion of

wo me n and men on Wall Street who have children. In fact, since all the probable

va lúes of p , - p 2 are negative, the proportion of men o f Wall Street who have

children appears to be larger than the proportion o f women.

8.95 Assume that o = 2.5 and the desired bou nd is .5. Then

1 . 9 6 n > 9 6.0 4 or n > 9 7 y¡n y¡n

77

8.99

8.103

8.107

8.111

8.115

a If you use p = .8 as a conservative est íma te for p , the margin o f error is

approximately

± 1 . 9 6 . ® ^ = +.029V 750

b To reduce the m argin o f error in part a to ± . 0 1 , sol ve for « in the equ ation

1 . 9 6 . p í ^ = , 0 1 => V ^ = L 9 6 ( '4 > = 7 8.4 => « = 6 14 6 .5 6 o r « = 6147V « .01

It is assumed that p = .2 and that the desired boun d is .01. Hen ee,

1 .9 6 J — < .01 => yj n > í = 42 .72.01

« > 1 8 2 4 . 7 6 o r « > 1 8 2 5

a The approximate 95% conf idence in terval for p is

í 529I ± 1 . 9 6 - = = 2 .9 62 ± 1 .9 6- = = = 2 .962 ± . 125

yjn V69

or 2 .837 < p < 3.087 .

b In orde r to cut the interval in half, the sam ple size m ust increas e by 4. If this is

done, the new half-width of the confidenc e interval is

l . 9 6 - í = = - ( l . 9 6 - í \ . y[4n 2 ^ y fñ

Henee, in th is case, the new samp le s ize is 4( 69 ) = 276 .

The approximate 98% confidence interval for p is

T ± 2 .3 3 -4 = = 2 .7 05 ± 2 . 3 3 ^ 2 = 2 .7 05 ± .0 11V« V36

or 2 .694 1 .9 6,p í í ! < .08«

r 1 .9 6 Í2 ( \8 )V« > ----------------- = > « > 9 . 8 o r « > 9 6 . 0 4

.08or « > 97.

78



8.117 Use the Interpret ing Co nf ídence Intervals applet. A nsw ers will vary, but the

w idths of all the intervals should be the same. M ost of the simu lations will show

betw een 8 and 10 in terv als th at w ork corr ectly .

8.121 Use the Ex ploring Conf ídence Intervals applet.a-b M ove the slider on the r ight side o f the applet to change the samp le size.

Increasing the sam ple size results in a sma ller standard error and in a narrower

interval.c By increasing the sam ple size n, you obtain more information and can obtain this

more precise es tímate o f (J without sacrif icing confídence.

79

9: Large-Sample Tests of Hypotheses

a Th e cri t ical valué that separates the reject ion and nonreject ion regions for a r ight-

tai led test based o n a z-statist ic w ill be a valué of z (cal led za ) such that

/ >( z > z a ) = tf = .01 . T hat is, z 0I = 2.33 (see the figure below). The nuil hypothesis

H0 will be rejected if z > 2 .33 .

b F or a tw o-ta il ed te s t w ith a = .05 , the critical va lué for the rejection región cuts

off oc¡2 = .025 in the tw o tai ls o f the z distributio n in F igu re 9 .2, so th at z 025 = 1.96 .

The nuil hypothesis H0 will be rejected if z > 1.96 or z < -1 .96 (w hich you can also

write as Id > 1.9 6 ).

c Sim ilar to part a, with the reject ion región in the low er tail o f the z distribution.

The nuil hypothesis H0 wil l be rejected i f z < -2 .3 3 .d Sim ilar to part b, w ith a / 2 = .005 . Th e nuil hyp othes is H0 will be rejec ted if

z > 2.58 or z < -2 .5 8 (w hich you can also wri te as |z| > 2.58 ).

81



In this exercise, the parameter of interest is fJ , the pop ulation mean. The objective of

the experimen t is to show that the mean ex ceeds 2.3.

a W e want to prov e the alternative hyp othe sis that /v is, in fact, grea ter then 2.3.

Henee, the alternative hypo thesis is

H a : / y > 2 . 3

and the nuil hypothesis is

H 0 : / / = 2.3 .

b T he best estim ato r fo r f i is the samp le averag e x , and the test statistic is

I z ñ ." a / J ¡

w hich represents the distance (mea sured in units of standard dev iations) from x to the

hyp othesize d m ean /y . H enee, if this valué is large in abso lute valué, one of two

conclusions may be drawn. Either a very unlikely even t has occurred, or the

hyp othesize d mean is incorrect. Re fer to part a. If a - .05, the critical valué o f z that

separates the rejection and non-rejection regions w ill be a valué (denoted by z0 ) such

that

P ( z > zn ) = a = .05

That is, z0 = 1.645 (see below). H enee, H0 will be rejected if z > 1-645 .

a - 05

Reject H0

c The standard error o f the mean is found using the sample standard deviation 5 to

approxim ate the pop ulation standa rd deviation c r :

nr, cr 5 .29 _

d To cond uct the test, calcúlate the valué of the test statistic using the inform ation

con tained in the sam ple. No te that the valué of the true standa rd deviation , o , is

approximated using the sample standard deviat ion 5 .

a ¡ 4 ñ s / y f ñ .049

The observed valué of the test statistic, z = 2.04 , falls in the rejection región and ti

nuil hypothesis is rejected. Th ere is sufficient evidenc e to indíca te that p > 2 .3 .

9.11 a In order to make sure that the average w eight was one po und, you wo uld test

H 0 : / / = 1 v ers us H a : p * 1

b -c T he te st st atist ic is

z = x-Mo a *~Po _ 1 . 01 -1 = 33

a / y f ñ s / y f ñ .1 8/ >/35

with p-value = ^ ( |z | > -33) = 2(.3 70 7) = .7414 . Since the /?-value is greater than .(

the nuil hypothesis should not be rejected. The m anage r should repo rt that there is

insufficient evidence to indícate that the mean is different from 1.

9.15 a The hypothesis to be tested is

H0 :/y = 110 versus H a : /y < 110

and the test statistic is

x - H o = 1 0 7 - 1 1 0 _

O \ [ ñ s / y f ñ 13/Vi 00

with p-value = P ( z < -2 .31 ) = .0104. To draw a conclusión f rom the p - valué, use t

gu idelines for statistical significance in Sec tion 9.3. Since the p-v alue is betw een .(

and .05, the test results are sign ifícant at the 5% level, but not at the 1% level.

b I f a = .05 , H0 can be rejected and you can conclude that the average score

improve m ent is less than claimed. This w ould be the most ben eficial way for thecompetitor to State these conclusions.

c If you worked for the Princeton Review, i t would be more beneficial to concludi

that there was insuf f icient evidence at the 1% level to conclud e that the ave rage scoi

improve m ent is less than claimed.

9.19 The hypothesis of interest is one-tailed:

H 0 :/A = 0 v ersu s H a : / / , - / / 2 < 0

The test statistic, calculated under the assumption that // , —Jl^ = 0 , is

83

with the unknown cr2and a \ estimated by s f and s \ , respect ively. The s tudent can

use one o f two m ethods for decis ión m aking.

p-value approach: Calcúlate p-value = P ( z < -1 .3 3 ) = .0918 . S ince this p-value is

grea ter than .05, the nuil hy po thesis is not rejected. Th ere is insufficient evid enc e to

indicate that the mean for population 1 is smaller than the m ean for population 2.

Critical valué approach; The rejection región with a = .05 . is z < -1.6 45 . S ince

the observed valué o f z does no t fall in the rejection reg ión, H 0 is not rejected. Th ere

is insuff icient evidence to indicate that the mean for popu lation 1 is smaller than themean fo r population 2.

a The hypothesis o f interest is two -tailed:

H 0 : p , - p 2 = 0 versus H a :p , - p 2 * 0

The test s tatistic, calculated un der the assumption that / / , - p , = 0 , is

w ith p-valu e = P ( |z | > 2.26 ) = 2 ( .0 1 19) = .0238 . Since the p-va lue is less than .05,

the nuil hypothesis is rejected. There is evidence to indicate a difference in the mean

lead levels for the two sections o f the city.

b From Section 8.6, the 95% confidenc e interval for Pj - p , is approximately

c Since the valué p, - p 2 = 5 or p 1- p , = -5 is not in the conf ídence interval in par t

b, it is not likely that the difference will be more than 5 ppm, and henee the statistical

signif ícance o f the difference is not of practical im portance to the engineers.

a Th e hy pothesis o f interest is two-tailed:

H 0 : p , - p 2 = 0 versus H a : p , - p 2 ^ 0

-1 .9 ± 1 .6 5 or - 3 .5 5 < ( p , - p , ) < - . 2 5

9.29

9.33

B - ^ ) - 0 ,9 4 - 2 .8 = _ 3 1 g

1-22 2.82

n, + n2 V 36 + 26

with p-va lue = P ( |z | > 3 .18) = 2( .0007 ) = .001 4. Since the p -\a lue is less than .05,the nuil hy poth esis is rejected. Th ere is evide nce to indícate a differen ce in the mean

concentrations for these two types o f s ites,

b T he 95% confid ence in te rv al fo r/y , - / y , is approxim ate ly

(*i “ * 2 ) ± 1-96 + ~V "i n2

( . 9 4 - 2 . 8 ) ± 1 . 9 6 J — + — v 1 V 36 26

-1 .8 6 + 1.15 or —3.01 < (//, —// 2) < —.71Since the valué p ^ - p 2 = 0 d oes not fal l in the interval in part b, it is not likely that

/y, = p 2 . Th ere is evidence to indícate that the mean s are different, con firm ing the

conclusión in part a.

a Th e hyp othesis of interest is two-tailed:

h o : ^ i - ^ 2 = 0 v e r s u s H a -Mi-Mi 5 6 0


_ (x , - x2) - 0 _ 9 8 .1 1 -9 8 .3 9 _ „ 22 j s [ + s [ I. 72 | ,742

n, n2 V65 65

with p-value = P(\z \ > 2 .22 ) = 2 ( 1 - .9 86 8) = .0264 . S ince thep -va lue i s be tween .01

and .05, the nuil hypo thesis is rejected, and the results are significan t. Th ere is

evidence to indicate a difference in the mean tem peratures for men v ersus wom en.

b Since the p-value = .0264, we can reject H0 at the 5% level (p-value < .05), but not

at the 1% level (p-value > .01). Using the guidelines for significance g iven in Section

9.3 o f the text, we declare the results statistically s ignif icant , but not highly

sig nif ic ant.

a The two sets of hypothesis both involve a different binom ial param eter p:

H 0 : p = .6 versus H a : p .6 (part c)

H 0 : p = .5 ve rsu s H a : p < .5 (p art b)

x 35b For the second test in part a, x = 35 and n = 75 , so that p = — = — = .4667 , the

n 75

test statistic is

85

z = P Z P g = A 6 V - ¿ = _ 5S

M o -5( '5)

n V 75Since no valué of a is specif ied in advance, we calcúlate

p -v alu e = P ( z < - .5 8 ) = .281 0. S ince this p-value is greater than .10, the nuil

hyp othesis is not rejected. The re is insuff icient eviden ce to contradict the claim.

x 49c Fo r the first test in part a, x = 49 and n = 75 , so tha t p = —= — = .6533 , the te st

n 75

statistic is

_ _ P - P o _ .6 5 3 3 - . 6 _ n i

M o /.6 (.4 )

75

with p-v alue = P ( |z | > .94) = 2( . 1736) = .3472 . Since the p-va lue is greater than .10,

the nuil hyp othesis is not rejected. Th ere is insuff icient evidenc e to con tradict theclaim.

The hypo thesis o f interest is

H 0 : p = .45 vers us H a : p * .45

x 32W ith p = —= — = .4 , the tes t s ta tis tic is

ai 80 z = P S P ^ = - 4 0 - .4 5 - 90

Po lo .4 5 (.55)

" V 80

The rejection región is two-tailed a = .01 , or |z¡ > 2.58 and H 0 is not rejected. Th ere

is insuff icient evidence to dispute the newspaper’s claim.

The hypo thesis of interest is

H 0 : p = .40 versu s H a : p * .40

x 114with p = — = ------ = .38 , the test s tatistic is

ai 300

P - P o _ -3 8 - . 4 0 _ n]

M o 1.40 (.60 )

300

The re ject ion región w kh a =.05 is | z |> 1.96 and the nuil hypothesis is not rejected.

(Alternatively, we cou ld calcúlate p-va lue = 2 P ( z < - .71 ) = 2( .2389) = .477 8. S ince

this p-va lue is greater than .05, the nuil hypo thesis is not rejected.) Th ere is

insuff icient evide nce to indicate that the proportion o f hou seholds w ith at least onedog is different from that reported by the H um ane Society.

86

9.45

then

9.47

9.51

a Th e hypothesis of interest is:

H 0 : P\ - p 2 = 0ve rsu s H a : p, - p 2 < 0

Calcúlate p , = .36 , p , = .60 and p = + n2 P i _ 18 + 3 0 = 4 3 . T he test statistic is

n, + n2 50 + 50

T - P ' - P > - -3 6 - - 6 0

l - f l + j J l V-4 8 (-5 2 ) ( 1/ 5 ° -+ 1/ 5 0 )

V U i n2 J

Th e reject ion región, w ith a = .05 , is z < -1 .645 and H 0 is rejected . Th ere is

evidenc e o f a difference in the proport ion o f survivors for the two groups.

b F rom Section 8.7 , the appro xim ate 95% confíd ence in te rval is

( Á - P : ) ± 1 .9 6 M + MV ni n2

( .3 6 - .6 0 ) ± 1 . 9 + Mv 7 V 50 5 0

- .2 4 ± .1 9 o r - .4 3 < ( p , - p , ) < - . 0 5

The hypothesis o f interest is

H 0 : p, - p, = 0 versus H a : p, - p 2 ^ 0

Ca lcúlate p, = — = .214 , p 1 = — = .25 , and p = + * 2 = + ^ = .227 .56 - 32 n, + n 2 56 + 32

Th e test statistic is then

T=h z h - -2 1 4 ~ -2 5 - ™ yj.2 21 ( .773) ( l / 56 + 1 /32)í 1 O— + —

n. n ,v i 2 y

Th e rejection reg ión, w ith or = .05 , is |z| > 1.96 and H 0 is no t rejected . Th ere is

insufficient evidenc e to indicate a difference in the p roport ion o f red M & M s for the

p la in and peanu t vari eti es. T hese re sults m atc h th e conclu sio ns o f E xercis e 8.53.

Th e hy pothesis o f interest is

H 0 : p, - p 2 = 0 versus H a : p, - p , > 0

93 119 x + x 93 + 119C a lc úla te p, = — = . 7 6 9 , p 0 = ------ = .5 9 8 , and p = — - = ---------------= .6625 .

121 199 n ,+ n 2 121 + 199

The test statistic is then

^/.6625 (.3 37 5) (1/121 + 1/199)

z = , -7 6 9 - 5 9 8 = 3 . 1 4I

1 1w — + — « 2 /

87

9.55

9.59

9.63

9.67

with p-value = P { z > 3.14) = 1- .9 99 2 = .0008 . Since the p-valu e is less than .01, the

results are repo rted as highly sign ificant at the 1% level o f significan ce. The re isevidence to confirm the researcher 's conclusión.

The power of the test is 1 - / ? = ^( re jec t H0 when H0 is fa ls e) . As p gets farther

from p 0 , the pow er of the test increases.

a-b Since it is nec essary to próve that the averag e pH level is less than 7.5, the

hypo thesis to be tested is one-tailed:

H n : p = 7.5 versus H a : p < 7.5

d Th e test statistic is

1 Z _ÍL x x - E = - .2

a / 4 7 t s / J ¡ .2/V3Ó

and the rejection región w ith a = .05 is z < -1.6 45 . The observed valué ,

z = -5 .4 7 7 , falls in the rejection región and H0 is rejected. W e conclude that the

average pH level is less than 7.5.

Let p, be the proportion of defectives produced by m achine A and p, be the

p ro port io n o f defe ctiv es p ro du ced by m achin e B. T he hypoth esis to be te ste d is

H(>: P\ ~ Pi - 0 ve rsus Ha : p , - p 2 * 0

Calcúla te p , = = .08 , p , = —— = .04, and p = — - .06 . The20 0 200 K ni + n 2 200 + 200

test statistic is

then

, •Q8~ 04 = 1.684 N 0 6 ( .9 4 ) (V 200 + 1 /2 0 0 )

\pq1 1

— H-----

n. n 2

T h e rejection región, with a = .05, is |z| > 1.96 and H0 is not rejected. Th ere is

insuff icient evidenc e to indícate that the mach ines are performing differently in terms

o f the percentage o f defect ives being produced.

The hypo thesis to be tested is

H n : p , - p : = 0 v ers us H a : p , - p : > 0


( I , - T 2 ) - 0 _ 1 Q -8 _ /l

n, n , V 40 40

88

The reject ion región, w ith a = .05, is one -tailed or z > 1 645 and the nuil hypothesis

is rejected. There is sufficient evidence to indícate a difference in the two means.

Henee, we conclude that diet I has a greater mean weight loss than diet II.

9.71 a The hypothesis to be tested is

H 0 : / y , - / / 2 = 0 versus H a > 0


The rejection región, with a = .05, is one -tailed or z > 1.645 and the nuil hypo thesis

is rejected. T here is a difference in mean yield for the two typ es o f spray.b An approximate 95% confidence interval for/y, - is

The rejection región w ith a = .01 is z > 2.33 . Since the observ ed valué, z = 2.19 ,

doe s not fall in the rejection región and H 0 is not rejected. Th e data do not provide

sufficient evidence to indicate that the mean ppm o f PCB s in the po pulat ion o f game

b irds exceeds the F D A ’s re com m ended lim it o f 5 ppm .

9.79 The hypothesis to be tested is

13 + 5.88 or 7.12 < (// , - f r ) < 18.88

9.75 The hypothesis to be tested isH 0 : // = 5 versus H a : // > 5


_ _ x Mo x Mo _ 7.2 5 _ . . .

a / J ñ s / ^ J ñ 6 .2 /V38

H 0 : P\ - Pi ~ 0 vers us H a : p x - p 2 * 0

Calcúlate p , = — = .40 , p , = -1 6124 5

. _ x, + x 2 _ 6124(.4) + 5512(.37)

^ n i + n 2 11636

= .40 , p 2 = — = .37 , and5512

1) + 551 2(.37 ) _ ■phe test statistic is then11636

89



z =• P \ - P l . 4 0 - . 3 7

>/ .386 ( .61 4)( l /61 24 + l /5512 )= 3.32

The rejection región for a = .01 is | z \> 2.58 and the nuil hyp othesis is rejected.

Th ere is sufficient evidenc e to indícate that the percen tage o f studen ts who are fluent

in English differs for these two districts.

9.83 a The param eter of interest is / / , the average daily wage o f wo rkers in a given

industry. A samp le o f n = 40 w orkers has been drawn from a part icular company

within this indu stry and x , the sam ple average, has been calculated . Th e objective is

to determine w hether this com pany p ays wages different from the total industry. That

is, assume that this sample o f forty wo rkers has been drawn from a hypo thetical popula tion o f w ork ers . D oes th is popu la ti on have as an average wag e / / = 54 , o r is

/ / different from 54 ? Thu s, the hy pothesis to be tested is

H0 : ¡x = 54 versus H : / / * 54 .

b-c T he te st sta ti st ic is

x - M 5 1 . 5 0 - 5 4 z ~

s / y /ñ 11 .88 /740= -1 .331

and the Larg e-S am ple T es t of a P op ula t io n M ean applet g ives p-va lue = . 1832 .

(Using Table 3 w ill produce a p - \ a lue o f . 1836.)

d Since a = .01 is sma ller than the p-valu e, . 1832, H0 can not b e rejected and w e

canno t conclude that the com pany is paying w ages different from the industry

average.

e Since n is greater than 30, the Central Limit Theorem will guarantee the norm ality

o f x regardless of w hether the original population w as normal o r not.

90

10: Inference from Small Samples

10.1

10.5

10.9

Refer to Table 4, Appendix I , indexing d f a long the left o r r ight m argin and ta across

the top.

a tM = 2.015 with 5 d f b t 02S = 2.306 with 8 d f

c / 10 = 1.33 0 w ith 18 d f c t 02i » 1 .96 w ith 30 d f

a U sing the form ulas given in Ch apter 2, calcúlate X * (. = 70.5 and X * 2 = 49 9.27 .

Then

y . Z i . T O í . ™ ,

n 10

4 9 9 . 2 7 - ^/ = --------------2— = ------------------1 ^— = . 24 9 44 4 a n d 5 = .4 9 94

n - \ 9

b W ith d f = n -1 = 9 , the appropr ia te valué o f t is / 01 = 2.821 (from Ta ble 4 ) and

the 99% upper one-s ided conf idence bound is

c 19 4 9 4 4 4 x + t 0i- f = => 7 .05 + 2 .8 2 1J: => 7.05 + .446

01 V io

or /y < 7 .4 96 . Intervals constructed using this procedure wil l end os e /y 99% o f the

time in repeated sam pling. H enee, we are fairly certain that this particular interval

e n d o s e s / y .

c The hypothe sis to be tested is

H 0 : /y = 7.5 versu s H a : /y < 7.5


7 0 5 - 7 .5 ,

s / J ñ 1.249444

10

Th e re ject ion región with a = .01 and « - 1 = 9 degrees o f f reedom is located in the

lower ta il of the /-dist ribut ion and is found f rom Table 4 as / < - / 01 = -2.8 2 1 . S ince

the observ ed valué o f the test s tatistic falls in the rejection región, H 0 is rejected an d

we conclude that /y is less than 7.5.

d N otice that the 99% up per one-sided confidence bou nd for /y does not include the

valué /y = 7.5 . Th is would con firm the results o f the hy pothe sis test in pa rt c, in

w hich w e conclud ed that /y is less than 7.5.

a Sim ilar to previous exercises. The hypothesis to be tested is

H 0 :/ y = 1 00 v ers us Ha : / y < 1 0 0

91

Calcúlate x = — Ln

Ijc,. _ 1797.095

n ~ 20= 89.85475

165,697.7081-(1797.095 )2

= 222.115 0605 and 5 = 14.9035n -1

The test statistic is

19

V2Ó

The cri t ical valué of t with or = .01 and n - \ = 19 degrees o f freedom is

r01 = 2 .539 and the reje ctio n reg ión is t < -2.53 9 . The nuil hypothesis is rejected and

we conclude that /J is less than 100 DL.

b T he 95% upper one-s id ed confi dence bound, based on n -1 = 19 degrees of

freedom, is

x + í 0 => 8 9.8 54 75 + 2 .5 3 9 14' 9Q^ 12 5 11 => / / < 9 8.3 16 yj n V 20

This confirms the results o f part a in w hich we con cluded that the mean is less than

100 DL.

10.13 a Th e hypo thesis to be tested is

H 0 : n = 25 vers us H a : ¡u < 25


The cri t ical valué o f t with a = .05 and n - 1 = 20 degrees of f reedom is

t 05 =1 .725 and the rejection región is t < -1 .7 2 5 . Since the observed valué does falls

in the rejection región, H0 is rejected, and we conclude that pre-treatment mean is lessthan 25.

b The 95% confid ence in te rv al based on d f = 20 is

o r 2 3 . 2 3 < 29.97 .

c The pre-treatment mean looks considerably sm aller than the other two means.

10.17 Refer to Exercise 10.16. If we use the large sam ple method o f Ch apter 8, the large

sample con fidence interval is

=> 2 6.6 ± 2 . 0 8 6 - = => 26 .6 ± 3 . 3 77.4

V 2 l

=> 246.96±12.98

92

10.19

10.25

10.27

or 233.98 3 0 .

s / y j n

a _ (/,, + ( „ , - 1 ) k _ 9 ( 3 .4 ) + 3 ( 4 .9 ) 2 7 ? ;

n ] + n 2 — 2 10 + 4 - 2

b + ( n 2 - \ ) s ; l l (1 8 ) + 2 0 ( 2 3 ) ^

nt + n 2 —2 12 + 21 — 2

a Th e hypo thesis to be tested is

H 0 : // , ~ n 2 = 0 v ers us H a : p , - p 2 * 0

From the M in i ta b printout , the fol lowing inform ation is available:

^ = .896 s f = (.4 0 0 )2 n, = 1 4

x 2 = 1 . 1 4 7 s \ = ( 6 79)2 rc2 = 11


(T, - x 2 ) - 0t = U = - 1 .1 6

\2

= 2.1

«2

T h e rejection región is two-tailed, based on n, + n 2 - 2 = 23 degrees o f f reedom. W i th

a = .05 , from Table 4, the rejection región is \t\ > tm5 = 2.06 9 and Hq is not rejected.

Th ere is not enough e vidence to indícate a difference in the popu lat ion m eans.

b It is not necessary to bound the p -va lu e usin g T able 4, sin ce th e exact p -v a lu e is

given on the p rintout as P-Value = .260.

c If you check the ratio of the two variances using the rule of thumb g iven in thissection you will find:

l a r g e r r _ ( .6 7 9)2

sm aller s2 ( .400 )2

w hich is less than three. The refore, i t is reasonable to assum e that the two popu lat ion

variances a re equal.

a Ch eck the ratio of the two variances using the rule of thum b given in this section:

J a r g e r r _ _ 2/78095 _ lf i2 2

smaller 5 .17143

w hich is greater than three. Therefore, i t is not reasona ble to assum e that the two

popu la ti on varia nces are equal.

b Y ou shou ld use the unpoole d t test with Satterthwaite’s approximation to the

degree s of freedom for testingH 0 : p , ~ n 2 = 0 versu s H a : //, - ¡i2 * 0

93

10.35

10.37

For a on e-tai led test with d f = «, + « , - 2 = 18 , t he p -va lue can be bounded us ing

Ta ble 4 so that p-value > .10 , and H 0 is not rejected. Th ere is insu fficient evidence to

indícate that sw imm er 2’s average t ime is sti ll faster than the averag e t ime for

swimmer 1.

a Th e test statistic is

d - n d . 3 - 0t =

J y f ñ= 2.372

wi th « -1 = 9 degrees of f reedom. The p-value is then

1 P(\t\ > 2 .372) = 2 P ( t > 2 .372) so that P ( t > 2 . 3 7 2 ) = - p - va lu e

Since the valué t = 2.372 fal ls betwee n two tabled entr ies for d f = 9 ( f 025 = 2.26 2 an d

f01 = 2.8 21 ), you can co nclu de that

.01 < —p-v alue < .0252

.02 < p-v alue < .05

Since the p-va lue is less than a = .05 , the nuil hypo thesis is rejected and we conclud e

that there is a difference in the two po pulat ion m eans.

b A 95% con fid ence in te rv al fo r p 1 - p , = p^ is

d ± t a a -^= => 3 ± 2.262. — => ,3±.286

025 V IOo r ‘.0 1 4 < ( p , - p 2 ) < .5 8 6 .

c Using sf¡ = .1 6 and B = . 1, the inequali ty to be solved is approxim ately

1 . 9 6 - ^ < . l■Jn

r 1 .9 6V 46 f n > -------------- = 7.84

.1« > 6 1 . 4 7 o r « = 6 2

Since this valué of n is gre ater than 30, the samp le size, « = 62 pairs, w ill be v alid.

a I t is necessary to use a paired-difference test , s ince the two sam ples are not

rando m and indepen dent . The hyp othesis o f interest is

H 0 : P] - P 2 = 0 or H0 : p d = 0

H a : p , - p 2 * 0 or H a : p ^ 0

The table of differences, along w ith the calculation o f d and s] , is presented below.

d¡ .1 .1 0 .2 -.1 I d, = .3

d f .01 .01 .00 .04 .01 Z d f = .07

95

10.41

d = ^ - = - = . 06n 5


and

. 0 7 -(■3):

JJ _ = .013

d - j U j . 0 6 - 0 ,t = — = =1 .177

s j y j n I m 3

with n - 1 = 4 degrees o f freedom . Th e rejection región w ith a = .05 is

H > ¿025 = 2 .77 6, and H0 is not rejected. W e cannot conclude that the m eans are

different.

b T he p -va lu e is

Pf l íl > 1 .177) = 2 />( f >1 .177 ) > 2( .10 ) = .20

c A 95% con fidenc e interval for // , —// 2 = / id is

¿ ± '0 2 5 - fv n

.06 + 2.776..013

.06 ± .14 2

o r - . 0 8 2 < ( / / , - / / , ) < .2 02 .

d In orde r to use the paired-difference test, it is necessary that the n paired

obse rvations be random ly selected from norm ally distr ibuted populations.

a Each sub ject wa s presented w ith both signs in random order. If his react ion timein general is high, bo th respon ses will be high; if his reaction time in gene ral is low,

both responses w ill be lo w . T he la rge vari abil ity fr om subje ct to subje ct will mask

the variabil ity due to the difference in sign types. The pa ired-difference design will

eliminate the subject to subject variability.

b The hypo th esis o f in te rest is

h o : ^ i - / /2 = ° or H (): / rd = 0

H a : / r , - / r 2 * 0 o r H a : / / r f* 0

The table of differences, along with the calculat ion o f d and s] , is presented below.

Driver 1 2 3 4 5 6 7 8 9 10 Totalsd, 122 141 97 107 37 56 110 146 104 149 1069

j =I 4 =1069 =106.9n 10

Z d ? - .( £ 4 ) 2

126,561 -(1069)2

10“ 1n - \


= 1364. and = 36.9458

J - ^ = 1 0 6 .9 -0

sd /4~n 36.9458

Vio

96

10.45

10.49

Since t = 9.150 w ith d f = n -1 = 9 is greater than the tabled v alué t m ,

p -valu e < 2 (.0 0 5 ) = .01

for this two tailed test and Ho is rejected. W e canno t conclude that the m eans are

different.c The 95% conf idence interval for / i, - / / , = fi d is ~~~

d ± t m, - ^ = => 106.9 ± 2 .2 6 2 --6-^i 5-8 => 106.9 ±2 6.4 28' yfn Vio

o r 8 0 . 4 7 2 < ( f i x ~ n 2 ) < 133 .328 .

a Use the M in i ta b pr intout given in the text below . Th e hypo thesis of interest is

H 0 ' V a ~ M b = 0 H a : ^ - / y B > 0

and the test statistic isd - n d 1 . 4 8 7 5 - 0

r - ^ 7 X - T 4 9 Í 3 T - 2 -82

V8

Th e p-valu e sho wn in the printout is p-valu e = .013 . Since the p-valu e is less than

.05, H0 is rejected at the 5% level o f signif icance. W e conclude that assessor A gives

higher assessments than assessor B.

b A 95% lo w er one-s id ed co nfi dence bound for p , - / / 2 = n d is

- s 1 4 9 1 3 4

d - 105 —j= => 1 .487 5-1 .895 ' => 1 .48 75 - .999V» V8

or ( // , - / / 2 ) > .4885 .

c In order to apply the paired-difference test, the 8 prop erties must be random ly and

independently selected and the assessments must be normally distr ibuted.

d Yes. I f the individual assessments are normally distr ibuted, then the mean of four

assessments will be norm ally distr ibuted. Henee, the difference x A —x will be

normally distr ibuted and the t test on the differen ces is valid as in c.

Fo r this exe rcise, s 2 = .3214 and n = 15 . A 90% co nfidence interval for c r will be

( n - l ) s 2 2 ( m —1)< o~ <

Xa/ 2 X(i-a/2)

where Xa /i represents the valué o f x~ such that 5% o f the area un der the curve

(shown in the figure on the next pag e) lies to its right. Sim ilarly, x]\-a¡i) he the

X~ valué such that an area .95 lies to its right.

97

10.53

Henee, we have located o ne-ha l f of a in each tail o f the distribution . Index ing

X 2m and x l s with n -1 = 14 degrees o f f reedom in Table 5 yields

X ls = 23.6848 and x l s = 6.57063

and the confíden ce interval is

14 ( .3214) , 14Í .3214)

— -------- - < a < — ---------- o r . 1 9 0 < <T 2 < . 6 8 523.6848 6.57063

a The hypo thesis to be tested is

H 0 :/¿ = 5 H a : / y * 5

T x 19 96C alcú late J = ± ^ = — - = 4.99

n 4

9 9 . 6 2 2 6 - .( 1 9 % >2

s~ = ---------------2— = -------------------- = = .0074

n - 1 3and the test statistic is

4 . 9 9 - 5t = . ^ , = - .23 2

s /y fn .0074

4

The reject ion región with a = .05 an d n -1 = 3 degrees o f f reedom is found f rom

Table 4 as~~|f| > tms = 3.182 . Since the observed valué o f the test statistic does not

fall in the rejection región , Ho is not rejected. Th ere is insufficient evide nce to show

that the me an differs from 5 mg/cc.

b The m an ufactu rer c la im s th a t th e ra nge o f the po te ney m easurem ents will equal .2.

Since this range is given to equal 6cr . we know that o ~ .0333 . Th en

H 0 :<72 = ( .03 3 3)2 =.0011 H 3 :c72 >.0011

98

, ( n - l ) í 2 3 Í.0 07 4 ) y = 4 J - = - ± -------- = 20.18

a - .0011

and the one-tailed rejection región with a = .05 and n - 1= 3 degrees of f reedom is

r > * 2 5 = 7.81

H0 is rejected; there is sufficient evidence to indicate that the range of the potency

will exceed the m anu facturer’s claim.

10.57 The hypothesis of interest is H 0 : a = 150 H a : a < 150

Calcúlate

(n - 1)s 2 = I * 2 - = 9 2 , 3 0 5 , 6 0 0 - = 662,232.8

and the test statistic is y 2 = —— íp—- = ^> 2,232 .8 = 29.43 3. The o ne-tailed rejection<t(; 150*

región with or = .01 and n - 1= 19 degrees o f f reedom i s # 2 < Z .99 = 7.63 273 , and H 0

is not rejected . Th ere is insu fficient evide nce to indicate that he is mee ting his goal.

10.61 The hypothesis of interest is H 0 : o f = o f versus Ha :a2 * of


s 2 712 F = A r = — 7 = 1.059 .

s¡ 69Th e crit ical valúes of F for various valúes o f a are given below using df¡ = 15 and

d f 2 = 14 .

a .1 0 .05 .025 .01 .005

Fa 2.01 2.46 2.95 3.66 4.25

Henee,

p -valu e = 2 P ( F > 1.059) > 2 (. 10) = .20

Since the p -\a lu e is so large, H0 is not rejected. Th ere is no evidenc e to indicate that

the variances are different.

10.65 For ea ch o f the three tests, the hyp othesis of interest is

H 0 : of = of ve rsu s H a : of * of

and the test statistics are

F 1.03 F = 4 = 4 ^ = 2.01 and F = ¿ = ^ = 14.29S-s-2 3 .922 ' s i 3 .4 92 ' 5 ,2 4 .4 7 2

The cri tical valúes of F for various valúes of a are given on the following page,

using d f = 9 and d f2 = 9 .

99

10.69

10.73

10.76

a .10 .05 .025 .01 .0052.44 3.18 4.03 5.35 6.54

Henee, for the first two tests,

p-va lu e > 2 (. 10) = .20

wh ile for the last test,

p-va lu e < 2 (.0 05) = .01

Th ere is no e vidence to indícate that the variances are d ifferent for the f irst two tests,

but H0 is reje cte d fo r th e th ird vari able . The tw o-s am ple r- te st w ith a poo le d estím ateo f o 1 cannot be used for the third variable.

Paired observations are used to est ímate the difference between two populat ion means

in preference to an est imation based on independent random samples selected from

the two populat ions beca use o f the increased information caused by blocking the

observations. W e expect blocking to create a large reduction in the standard

deviat ion, i f differences do ex ist amon g the blocks.

Paired observations are not alway s preferable. The de grees of freedom that are

available for est im ating o 2are less for paired than for unpaired observations. If there

were no difference between the blocks, the paired experiment would then be less benefi cia l.

Since i t is necessary to determine w hether the injected rats drink m ore w ater than

noninjected rates, the hyp othesis to be tested is

H 0 : p = 22.0 H a : p > 22.0


f = 3 1 . 0 - 2 2 . 0 = 5 9 8 5

s/ 'y jn 6 2

Using the cri tical valué approach , the rejection región with a = .05 and

n - 1 = 16 degrees of freedom is located in the uppe r tai l of the r-distribution and is

found from Table 4 as t > r05 = 1.746 . Sinc e the ob ser ve d valu é of the test statistic

falls in the rejection región, H0 is rejected and we conclude that the injected rats do

drink more w ater than the noninjected rats. The 90% confidence interval is

I ± 7 ;, ~ => 3 1 . 0 ± 1 . 7 4 6 4 ¿ 3 1 .0 ± 2 .6 2 5 yf n V Í7

or 28.375 < p < 33.625 .

The student may use the rounded valúes for x and s given in the display, or he may

wish to calcúlate x and s and use the more ex act calculat ions for the confidence

intervals. The ca lculat ions are shown on the next page.

100

10.79

_ I jc¿ 1845a x = L = --------= 184.5

n 10

3 4 4 , 5 6 7 - í ^s 2 = ---------- *— = ---------------------------------= 462.7222

n - 1 9

5 = 21.511 and the 95% co nfidence interval is

5 _ ____________ 21.511 x ± t m - = => 1 .845 ± 2 .262 — ^= — => 184 .5 ±15 .4

yj n -s/l 0

or 169.1 '

b I = ^ Í L = — = 73.0 52 = — -5— = —— = 24.8889n 10 n - 1 9

s = 4.989 and the 95 % confidence interval is

j 4 989 x ± t m. - 7= => 7 3 . 0 ± 2 . 2 6 2 - = => 73 .0 ±3 .5 7

J ü VTÓ

or 69.43 < /J < 76.57 .

_ I x 25.42c x = i- = --------- = 2.542

n 10

S x j _ ( £ ü ) _ 6 5 . 8 3 9 8 _ ( 2 5 ' 4 2 )

^ ------= .13579556n - 1 9

5 = .3685 and the 95% confidence interval is

x ± tms => 2 . 5 4± 2 . 2 6 2 = > 2 .5 4 ± .2 6Vn VlO

or 2.28 < f i < 2.80 .

d No. Th e relationship between the confidence intervals is not the same as the

relationship between the original m easurements .

U se the com puting formulas or you r scientific calculator to calcúlate

_ I x , . 322.1v = -----= ----------- = 24.777

n 13

„ 2 ( I ± ) 2 (322.1)2Ijc ,2 - V 8 114.59 — } —

52 = --------------- '-1— = — — = 1 1.1 61 9n - 1 12

5 = 3.3409 and the 95% confidence interval is

101

c t . "unoI ± t m5 —= 24 .777 ± 2 . 1 7 9 ^ = ^

yjn v i 324 .777 ±2 .019

o r 2 2 .5 7 8 < m < 26 .796 .

a The range o f the f irst samp le is 47 while the range o f the second samp le is only

16. Th ere is proba bly a differen ce in the variances.

b The hypoth esis o f in te re st is

H (l : ay = o \ ver sus Ha : a ,2 * a \

Calcúlate s i =

1 7 7 , 2 9 4 -(838)2

= 577.6667

1 9 2 , 3 9 4 - M

*,2= 6 = 29.6

and the test statistic isF = ii_ = 5TL666T_ _

s i 29.6

Th e cri tical valúes with d f x = 3 an d d f2 = 5 are shown be low from T able 6 .

a .10 .05 .025 .01 .005

Fa 3.62 5.41 7.76 12.06 16.53

Henee,

p -v a lu e = 2 P ( F > 19.516) < 2(.0 05) = .01

Since the p-value is smaller than .01, H0 is rejected at the 1% level of significance.There is a difference in variability.

c Since the Stu den t’s t test requires the assump tion o f equal variance, i t would be

inapp ropriate in this instance. You should use the unpooled t test with Satterthwaite’sapproximation to the degrees of freedom.

A paired-difference test is used, since the two sam ples are not random and

indepen dent . The hypo thesis of interest is

H 0 : A ~ = 0 H a : / i l - l i 2 > 0

and the table o f differences, along w ith the calculat ion o f d and s] , is presented

belo w .

Pair 1 2 3 4 Total s

4 -1 5 11 7 22

d- = M = ̂ = 5.5

n 4 sd = 5


Z d > -( 1 4 )2

1 9 6 -(22)2

n - 1

= 25 and

102

10.91

10.97

d - n (i 5 . 5 - 0 I = y-£= = = 2.2

s J y J n A

The one-tai led p-value with d f = 3 can be bounde d betw een .05 and .10. Since thi

valué is greate r than .10. H0 is not rejected. Th e results are not significan t; there is

insufficient evidence to indicate that lack of school experience has a d epressing efl

on IQ scores.

Th e object is to determine w hether or not there is a difference betwe en the m ean

responses for the two different st imuli to which the people have been sub jected. T

samples are independently and randomly selected, and the assumptions necessary 1

the t test of Sec tion 10.4 are met. Th e hypo thesis to be tested is

H o ; A i - / / 2 = 0 H a : //, - f ¿ 2 * 0

and the preliminary calculations are as follows:

15 21T, = — = 1.875 and x 2 = — = 2.625

(15)2 (21)23 3 - - — — 61 - - —

= .6964 3 and s22 = = .8392 9

Since the ratio o f the varian ces is less than 3, you can use the p ooled t test. Th e

poole d estim ato r o f <r: is calc u la te d as

2 (rh - \ ) s ; + ( n 2 - \ ) s ¡ 4.875 + 5.875= .7679

n , + n 2 - 2 14


. .. . 1 . 8 7 5 - 2 .6 2 5 , m

1.7679 ÍI+H

Ui n 2J V

o o

o c

Th e two -tailed reject ion región with a = .05 and d f = 14 is |r| > t m¡ = 2.1 45 , and 1

is not rejected. Th ere is insufficient evide nce to indicate that there is a difference i

means.

It is possible to test the nuil hypo thesis H 0 : tJ,‘ = cr2a gain st any one o f three

alternative hypotheses:

(1) H a io f * o \ (2) H a : <t2 < o \ (3) H a : o \ > o \

a The first al ternat ive wou ld be preferred by the m anage r of the dairy. He doe s n

know anything about the variabil ity o f the two m achines and would wish to detectdepa rtures from equ ali ty o f the type o \ < o \ o r o \ > o \ . These alternatives are

implied in (1).

103



b T he sale sm an fo r com pany A w ould p re fe r th at the experim en te r sele ct th e second

alternative. Re jection of the nuil hypothesis w ould im ply that his m achin e had

sm aller variability. M oreo ver, even if the nuil hypothesis were not rejected, there

wo uld be no evidence to indicate that the variabili ty o f the compa ny A m achine wasgreater than the variabili ty o f the compa ny B m achine.

c Th e salesm an for com pany B would prefer the third al ternat ive for a similarreason.

101 A pa ired -diff ere nc e an alys is is used. To test H (): //, —// 2 = 0 versus H a : /v, > 0 ,

w here is the mea n reaction time after injection and // , is the m ean reac tion time

before in je ction, calc úla te the d if ferences ( x 2 - x , ) :

6, 1,6, 1

( W 7 4 _ M l

Then d = ^ = — = 3.5 = — * — = ------------ 4— = g .33 andn 4 d n - 1 3

sd = 2.88 675 and the test statistic is

d - n d 3 . 5 - 0t = — = -------------- = 2 425

s J J i 2.88675 '

Fo r a one-tai led test with d f = 3 , the rejection región with a = .05 is t > tM = 2.35 3 ,

and H 0 is rejected . W e conc lude that the drug significantly increases w ith reaction

time.

.105 Th e un derlying populat ions are rat ings and can only take on the f inite num ber of

valúes, 1, 2, ..., 9, 10. N either population has a norm al distribution, but both are

discrete . Further, the sam ples are not indepe nden t, since the same person is asked to

rank each car design. Henee, two of the assumptions required for the S tuden t’s t test

have been violated.

.109 The M in i tá b p rintout below sho ws the sum ma ry stat ist ics for the two samples:

Descriptive Statistics: Method 1, Method 2Variable N Mean SE Mean StDev

Method 1 5 137.00 4.55 10.17Method 2 5 147.20 3.29 7.36

Since the rat io of the tw o sam ple variances is less than 3, you can use the pooled t test

to com pare the tw ojne thod s o f measurement , using the rem ainder of the M in i tab

pri n to ut on the next pa ge:

104



Two-Sample T-Test and Cl: Method 1, Method 2Difference = mu (Method 1) - mu (Method 2)Estímate for difference: -10.200095% CI for difference: (-23.1506, 2.7506)

T-Test of difference = 0 (vs not =): T-Value = -1.82 P-Value = 0.107 DFBoth use Pooled StDev = 8.8798 —

The test statistic is t = -1 .8 2 w ith p-value = . 107 and the results are not significant

Th ere is insufficient evidence to declare a difference in the two popu lat ion m eans.

10.113 The hyp othesis to be tested is

The cri t ical valué of t w ith a = .01 an d n - 1 = 9 degrees of freedom is í0l = 2.821

the reject ion región is t > 2.821. Since the observed va lué falls in the reject ion

región, H0 is rejected. The re is sufficient evidenc e to indicate that the ave rage

num ber o f calories is greater than advert ised.

10.119 Use the Interpreting Confidence Intervals applet . An sw ers will vary from studt

to student. Th e w idths of the ten intervals will not be the sam e, since the valué o f

chan ges w ith eac h new sam ple. Th e student should f ind that approxim ately 95% i

the intervals in the first app let contain // , wh ile rou ghly 99 % o f the intervals in tf

second applet contain / / .

10.123 Use the Tw o Sam ple t Test: Independent Sam ples applet . The hyp othesis to b(

tested concerns the differences between mean recovery rates for the two surgical

p ro cedures. L et // , be th e popu la tion m ean fo r Pro cedure I and f i 2 be th e popula ti

m ean for Procedure II . The hyp othesis to be tested is

Since the rat io of the variances is less than 3, you can use the po oled t test. En ter

appropriate statistics into the applet and you will find that test statistic is

with a two-tai led p-value of .0030. Since the p-va lue is very sm all, H0 can be reje

for any valué of a g reater than .003 and the results are judg ed highly significant.

The re is sufficient evidence to indicate a difference in the me an recov ery rates fo r

two procedures.

H 0 : n = 280 versus H a : a > 280


Vio

H0 : A, - A2 = 0 vers us H a : A, - A2 ^ 0

105



11: The Analysis of Variance

1 1 . 1

11.4

11.7

In comp aring 6 popu lat ions, there are k -1 degrees o f f r eedo m fo r tr ea tmen t s and

n = 6 (10) = 60 . The AN OV A table is shown be low.

Source d f Treatments 5

E rror 54

Total 59

Sim ilar to Exercise 11.1. W ith n = 4 (6 ) = 24 and k = 4 , the sources o f variat ion

associated d f are show n below.Source d f Treatm ents 3

Erro r 20

Total 23

The fol low ing prel iminary calculat ions are necessary:

7¡ = 1 4 7; = 1 9 7¡ = 5 G = 3 8

( 3 8 ) :a C M =

14

= 103.142857

Tota l SS = S x l - CM = 32 + 22 + • • • + 22 + 12 - CM = 130 -10 3.1 42 85 7 = 26 .8í

S S T = Z - — C M = - + — + - — C M = 1 1 7 .6 5 -1 0 3 .1 4 2 8 5 7 = 14.5071n, 5 5 4

, SS T 14.5071a nd M S T = -------= ------------- = 7.2536

k - 1 2

c By sub trac tion, SS E = To ta l S S -S S T = 26 .8571 -14 .5071 = 12 .3500and the

degrees of f reedom, by subt ract ion, are 1 3 -2 = 11. Then

M S E = SS E = 1 2 3 5 0 011 11

d The information obtained in parts a-c is Consolidated in an A N O V A table.

Source d f SS M S

Treatm ents 2 14.5071 7.2536

Error 11 12.3500 1.1227

Total 13 26.8571

e The hypothesis to be tested is

H 0 : //, = = fiy versus H a : at least one pair of m eans are differen t

107

1.10

.1 3

f The reject ion región for the test statistic F = ---------= —;— — = 6.46 is based on anMSE 1.1227

F-distr ibution w ith 2 and 11 degrees of freedom. The cri t ical valúes of F for

boundin g th e p -valu e fo r th is one-t a iled te st are show n belo w .

a .10 .05 .025 .01 .005

2.86 3.98 5.26 7.21 8.91

Since the observed valué F = 6.46 is be tw een F 01a nd F 025 ,

.01 < p -va lue < .025

and H0 is rejected at the 5% level o f signif icance. There is a difference amo ng the

means.

a The following prel iminary calculat ions are necessary:

7; = 3 8 0 T2 = 199 F, = 261 G = 840

( I jÜ 2 (8 4 0 )2C M = — = --------- = 64,145.4545

n 11

T o ta l SS = I x¡ - C M = 6 5 ,2 8 6 - C M = 1 14 0.5 45 5

SST = Z —— C M = ^ 5 _ + — + — — C M = 641.87883n, 5 3 3

Calcúla te MS = S S / d f and consolídate the information in an ANOVA table.

Source df SS M S

Treatm ents 2 641.8788 320.939E rror 8 498 .6667 62.333

Total 10 1140.5455

b T he hypoth esis to be te ste d is

H 0 : //, = fi2 = p , versus H a : at least one pair of me ans are different

and the F test to detect a difference in m ean student response is

F = M 5 I = 5 , 5 .M SE

The reject ion región w ith a = .05 and 2 and 8 d f i s F > 4.46 and H0 is rejected.

The re is a signif icant difference in m ean respon se due to the three different methods.

a W e would be reasona bly confiden t that the data sat isf ied the norm ality assum ption

because each m easure m ent re presen ts th e avera ge o f 10 continuous m easure m ents .

Th e Central Limit Th eorem assures us that this mean w ill be app roximately norm ally

distributed.

b W e have a com ple te ly random iz ed desig n w ith fo ur tr eatm ents , each con ta in in g 6

m easuremen ts. The analysis o f variance table is given in the M in i ta b printout . The F

test is

f = M S T = ^ 5 8 0 = 5 7 38

MSE .115

108

11.17

w ith p-va lue = .000 (in the colum n m arked “P” ). Since the p-va lue is very small i

than .01), H0 is rejected . Th ere is a signifícant difference in the me an le af length

among the four locations with P < .01 or even P < .001.

c The hypothesis to be tested is H0 : p, = p 4 versus H a : p, * p 4 and the teststatis tic is ~~~

x , - x á 6 . 0 1 6 7 - 3 . 6 5t =

M SE1 1

n, n.

= 12.091 1

6 6

The p-value with d f = 20 is 2 P ( t > 12.09) is bounded (using Tab le 4) as

p-valu e < 2 ( .0 0 5 ) = .01

and the nuil hypo thesis is reject. W e con clude that there is a differe nce b etw een t

means.d The 99% conf idence interval for p , - p 4is

1 1 — + — ( x ¡ - x 4 ) ± r m ^ M S E

(6 .0 16 7 - 3 .6 5) ± 2 . 8 4 5 ^ . 1 1 5 ^ + ^ )

2 .367 ± .557 o r 1 .810 < p , - p 4 < 2 .924

e W hen conduc ting the t tests, remem ber that the stated con fidence co efficients

based on ra ndom sam pling. If you lo oked at the data and on ly com pared th e la rg eand smallest sample means, the randomness assumption would be disturbed.

a Th e design is a com pletely random ized design (four independ ent samp les).

b T he fo llow in g pre lim in ary calc ula tions are necess ary :

7; =1211 T 2 = 1074 r 3 =11 58 T4 =1 243 G = 4686

CM =( I * , ) 2 (4 6 8 6 )2

20= 1,097,929.8

T o ta l S S = - C M = 1 ,1 0 1,8 6 2 - C M = 3 93 2.2

SST = Z— — CM =n; 5

12112 10742 11582 12432+ - - C M = 3 27 2.2

5 5 5

Calcú la te MS = S S / d f and consolídate the information in an ANOVA table.

Source d f SS MS

Treatments 3 3272.2 1090.7333

Error 16 660 41.25

Total 19 3932.2

The hypothesis to be tested is

H0 :p , = p 2 = p 3 = p 4 versus Ha : a t least one pai r of m eans are d i f fer

109



and the F test to detect a difference in average price s is

M S T o a F = ---------= 26.44 .M SE

The rejection región w ith a = .05 and 3 and 16 d f is approximately F > 3.24 and Ho isrejected. There is enough ev idence to indicate a difference in the average p rices for

the four States.

11.21 a c o = q0, ( 4 , 1 2 ) - ^ = 4 . 2 0 - ^ = 1.878sV5 V 5

b c o= q m ( 6 , 1 2 ) - ^ = 6 . 1 0 - ^ = 2.1567sV 8 V 8

11.25 The d esign is completely random ized w ith 3 treatm ents and 5 replications pertreatment. The M in i ta b printout below shows the an alysis of variance for this

experiment.

One-way ANOVA: mg/dl versus LabSource DF SS MS F PLab 2 42.6 21.3 0.60 0.562Error 12 422 . 5 35 .2Total 14 465.0

S = 5. 933 R-Sq = 9.15% R-Sq(adj)

Individual 95% CIs For Mean Based on

Pooled StDevLevel N Mean StDev . . +------........+----------+--------- +-----

1 5 108.86 7.47 (--------- -)2 5 105.04 6 . 01 (-------3 5 105.60 3 .70 (------ ---)

100.0 104.0 108.0 112.0Pooled StDev =5.93

Tukey 95% Simultaneous Confidence IntervalsAll Pairwise Comparisons among Levels of LabIndividual confidence level = 97.94%

Lab = 1 subtracted from:Lab Lower Center Upper

2 -13.824 -3.820 6.1843 -13.264 -3.260 6.744

14.0 -7.0 0.0 7.0

Lab = 2 subtracted from:Lab Lower Center Upper +--------- +----------+----------+-----

3 -9.444 0.560 10.564 (...... -..... -*...............)

-14.0 -7.0 0.0 7.0

110



11.29

11.33

a The analysis o f var iance F test for H 0 : //, = //2 = //? is F = .60 with

p -va lu e = .562 . T he re sults are not sig n if ic an t and H 0 is not reje cte d . T here is

insufficient evidence to indícate a difference in the treatme nt means.

b Sin ce th e tr eatm ent m eans are not sig n if ic an tly dif ferent, th ere is no need to us eT uk ey ’s test to search for the pairw ise difference s. NotiCe that all three intervals

generated by M in i tab contain zero, indicating that the pairs cannot be judg eddifferent.

Re fer to Ex ercise 11.28. The given sum s of squares are inserted and m issing entries

found by subtraction. The me an squares are found as M S = S S / d f .

Source d f SS M S F

Treatments 2 11.4 5.70 4.01

Blocks 5 17.1 3.42 2.41Error 10 14.2 1.42

Total 17 42.7

Use M in i tab to obtain an ANOVA printout , or use the fol lowing calculat ions:

( I * y ) ( 1 1 3 ) “CM = 4 ^ = 1064.08333

n 12

To ta l S S = X x 2 - C M = 6 2 + 1 0 2 +••■ + 14 2 - C M = 1 2 1 3 - C M = 1 4 8 .9 1 6 6 7

T 2 2 2 2 + 3 4 2 + 2 7 2 + 3 0 2S ST = I - C - C M = - ~U - C M = 25.583333 3

SSB = I í - - C M = 3 3~ + 2 5 2 + 5 5 ~ - C M = 1 2 0.6 66 67 an d4 4

SSE = To ta l SS - SST - SSB = 2 .6667

Calcúlate MS = S S / d f and consol ídate the informat ion in an A N OV A table.

Source d f SS M S F

T reatm ents 3 25.5833 8.5278 19.19

Blocks 2 120.6667 60.3333 135.75Error 6 2.6667 0.4444

Total 11 148.9167

a To test the difference amon g treatm ent means, the test stat istic is

P _ M S T _ ^ 5 2 8 _ ]9 19

MSE .4444

and the reject ion región w ith a = .05 and 3 and 6 d f is F > 4 .76 . There i s a

significant difference am ong the treatm ent m eans.

b T o te s t th e d if fere nce am ong b lo ck m eans, th e te st sta ti sti c is

F = M S B = 6 0 3 3 3 3

MSE .4444

111

1.37

and Ihe rejection región with a = .05 and 2 and 6 d f \s F > 5.14 . Th ere is a

significant difference am ong the block m eans.

c With k = 4 , d f = 6 , n, = 3 ,

ÍA VÑÍSE 1.4444a)=<im (4*6) r = 7.03 J — r— = 2.71

yjn, V 3

The ranked means are shown below.

7.33 9.00 10.00 11.33

-*l - 4̂ X2

d Th e 95% confidence interval is

( x A - x B) ± t M SE

( 7 .3 3 3 - 1 1 .3 3 3 ) ± 2 .4 4 7 ^ .4 4 4 4

- 4 1 1 . 3 3 2 o r - 5 .3 3 2 < / i A - f i B < - 2 .6 6 8

e Since there is a significant difference amo ng the block m eans, blocking has been

effective. Th e variation due to block differences can be isolated using the rando m ized

blo ck desig n.

Sim ilar to previous exercises. The M in i tab printout for this random ized blockexperiment is shown below.

Two-way ANOVA: Measurements versus Blocks, ChemicalsSource DF SS MS F PBlocks 2 7.1717 3.58583 40 .21 0.000Chemicals 3 5.2000 1.73333 19.44 0.002Error 6 0.5350 0.08917Total 11 12 . 9067S = 0.2986 R-■Sq = 95. 85% R-:Sq (adj) = 92.40%

Individual 95% CIs For Mean Based on PooledStDev

Blocks123

Mean10.875

12.70012.225

10.50 11.20 11.90 12.60

Individual 95% CIs For Mean Based onPooled StDev

Chemicals1234

Mean11.400012.3333

11 .'ZhOO12.8000

1 1 . 2 0 11. 90 12 .60 13 .30

112

Bo th the treatm ent and b lock m eans are signif icantly different. Since the four

Chemicals represent the treatments in this experiment, T uk ey 's test can be used to

determ ine w here the differences lie:

, . > / M S É 1 .0 89 17 -

m= q<* = y ~ r ~ =

The ranked means are show n below. 11.20 11.40 12.33 12.80

X, Xj x^

The Chemical falls into two signif icantly different groups - A and C versus B and D.

11.41 A random ized block design has been used with “estim ators” as treatm ents and

“construction jo b ” as the block factor. The a nalysis of variance table is found in the

M in i ta b pr intout below.Two-way ANOVA: Cost versus Estimator, JobSource DF SS MS F PEstimator 2 10. 8617 5.4308 7.20 0.025Job 3 37. 6073 12.5358 16.61 0.003Error 6 4 .5283 0.7547Total 11 52 .9973S = 0.8687 R-Sq = 91.46% R-Sq(adj)

Individual 95% CIsPooled StDev

= 84.34%

For Mean

Estimator Mean ------- +----------+ _

A 32 ..6125 (-------- *---------)

B 34 .. 8875 (--------

C 34..1875 (----- --

★-----

3 2 . 4 3 3 . 6 3 4 . 8 3 6 . 0

Both treatments and blocks are signif icant. The treatment m eans can be further

compared using Tukey’s test with

a> = q m ( 3 , = 434 = 1 -885


32.6125 34.1875 34.8875

Estimators A and B show a signif icant difference in average costs .

11 .45 a -b There a re 4 x 5 = 20 t rea tments and 4 x 5 x 3 = 60 total observat ions .

c In a factorial experimen t, variation due to the interaction A x B is isolated from

SSE. The sources of variation and associated degrees of freedom are given on the

next page.

113



Source df

A 3

B 4A x B 1 2

Error 40

Total 59

49 a Based on the fact that the mean respon se for the two levels o f factor B behav es

very differently dependíng on the level of factor A under investigation, there is a

strong interaction present between factors A and B.

b T he te st st ati st ic for in te racti on is

F = M S(A B ) /M SE = 37 .85 wi th p -va lue = .000 f rom the M in itab printout. Th ere is

eviden ce of a significant interaction. Th at is, the effect of factor A depen ds upon thelvel of factor B at which A is measured.

c In light of this type of interaction, the main effect means (averaged ov er the levels

o f the other factor) differ only slightly. Henee, a test o f the main -effect terms

produces a non-s ig nif ic ant re su lt .

d No. A significant interaction indicates that the effect of one factor dep end s upon

the level of the other. Each factor-level combination should be investigatedindividually.

e A nsw ers will vary.

53 a The design is a 2 x 4 factorial experiment with r = 5 replications. T here are two

factors, Gender and School, one at two levels and one at four levels.

b T he analy sis o f vari ance ta ble can be fo und usin g a Com pute r prin to ut o r the

following calculations:

Schools

Gender 1 _____ 2 ______ 3 ______ 4 Tota l

M ale 2919 3257 3330 2461 11967

Fem ale 3082 3629 3344 2410 12465

Total 6001 6886 6674 4871 24432

? 4 4 3 2 2C M = ----------- = 14923065.6

40Tota l SS = 15281392 - CM = 358326.4

20

6 00 12 + 68 862 + 66 742 + 48 712- C M = 24 67 25 .8iC =

10

- S S G - S S ( S c ) - C M = 10574.9

114

Source d f SS M S F

G 1 6200.1 6200 .100 2.09

Se 3 246725.8 82241.933 27.75G x S c 3 10574.9 3524.967 1.19

E rro r 32 94825.6 2963.300

Total 39 358326.4

c Th e test statistic is F = M S (G S c) /M S E = 1 .19 and the reject ion región is

F > 2 .92 (with a = .05 ). Al ternately , you can bound the p-value > .1 0 . H en ee ,?

is not rejected. Th ere is insuffícient evidence to indícate interact ion betw een gende

and schools.

d You can see in the interact ion plot that there is a small differenc e betw een the

averag e scores for male and fem ale students at schoo ls 1 and 2, but no difference te

e Th e test statistic for testing gen der is F = 2 .09 wi th F os = 4 .1 7 (or p-value > .11

The test statistic for schools is F = 27.75 with F 05 = 2.92 (or p-value < .005 ) . The

is a significant effect due to schools.

Using T ukey ’s method of paired compar isons wi th a = .01, calcúlate

„ = í „ , ( 4 , 3 2 ) ^ 4 , o J f L 82.63 yjn, V 10

The ranked m eans are shown below.

487.1 600.1 667 .4 688.6

*4 *3 x2

115



11.59 The objective is to determine w hether or not mean reaction time differs for the f ive

stimuli. The fou r peo ple used in the experim ent act as block s, in an attem pt to isolate

the variation from person to person. A random ized block design is used, and the

analysis of variance table is given in the printout.

a The F statistic to detect a difference due to stimuli is

f = ̂ I = 2 7 .7 8

M SE

with p-value = .00 0. Th ere is a signif icant difference in the effect of the f ive stimuli.

b T he tr eatm ent m eans can be fu rt her com pared usin g T u k ey ’s te st w ith

, c i M > /M S É „ C1 ¡.00708 i n nú) = q os(5,12 ) — — = 4 .5 l J — — = . '9 0

V"r V 4

The ranked means are shown below.E A B D C

.525 .7 .8 1.025 1.05

c The F test for blocks produ ces F = 6.59 with p-value = .007 . The block

differences are significant; blocking has been effective.

11.63 Th is is similar to previous exercises. Th e com plete AN OV A table is shown below.

Sourced f

SS MS F

A 1 1.14 1.14 6.51

B 2 2.58 1.29 7.37

A x B 2 0.49 0.245 1.40

Error 24 4 .20 0.175

Total 29 8.41

a The test s tatistic is F = MS ( A B )/M S E = 1.40 and the rejection región is

F > 3.4 0. There is insuff icient evidence to indicate an interaction.

b U sin g T able 6 w ith d f = 2 and d f 2 = 24 , the following critical valúes are

obtained.

a .10 .05 .025 .01 .005

2.54 3.40 4.32 5.61 6.66

The observed valué of F is less than F 1(), so that p-v alu e > . 10 .

c Th e test statistic for testing factor A is F = 6.51 with F ()5 = 4.2 6. The re is

evidence that factor A affeets the response.

d Th e test statistic for factor B is F = 7.37 with Fm = 3.4 0. Factor B also affeets

the response.

11.67 a The design is a random ized block design, with week s representing blocks and

stores as treatments.

b T he M in i tab Computer printout is shown on the next page.

116



11.71

Two-way ANO VA: Total ve rsu s Week, StoreSource DF SS MS F PWeek 3 571.71 190.570 8 .27 0.003Store 4 684.64 171.159 7 .43 0 .003Error 12 276.38 23.032

Total 19 1532.73S = 4.799 R-Sq = 81.97% R-Sq(adj) = 71.45%

c The F test for treatments is F = 7.43 with p-value = .003 . Th e p-value is small

enoug h to allow rejection o f H0. There is a signif icant difference in the average

w eekly totals for the f ive supermarkets.

d Wi t h k= 5, df = \2, n = 4 ,

t e , ^ V M S E , (2 3.0 32a> = qtí5 (5,12) - j = = 4 .5 i y — - — = 10.82


1 5 4 3 2

240.23 249.19 252.18 254.87 256.99

a The des ign is a comp lete ly random ized des ign with three samples , each having a

dif ferent num ber o f measurements .

b Use the computing formulas in Section 11.5 or the M in i ta b pr intout below.

One-way ANOVA: Iron versus SiteSource DF SS MS F P

Site 2 132.277 66.139 126.85 0.000Error 21 10.950 0.521Total 23 143.227S = 0.7221 R-Sq = 92.36% R-Sq(adj) = 91.63%

Th e F test for treatm ents has a test s tatistic F = 126.85 w ith p-va lue = .000. Th e nuil

hyp othesis is rejected and we conclude that there is a signif icant difference in the

average p ercentage o f iron ox ide at the three sites.

c Th e diagno stic plots are show n on the next page. There appears to be no violation

o f the norm ality assum ptions; the variances may be unequal, jud gin g by the differ ing

bar w id th s above and belo w th e cen te r line.

117



Normal Probability Plotof the Residual!( respo nse is I ron)

118



s ** = Z x ? ~ ^ X¡^ = 21,066.82n

2

b-c

= 88-800033 = ft0()42

S_„ 21066.82

and a = y - b x = 0 .41722 -0 .004215 16(94 .53 33) = 0 .187

The least squares line is y = a + b x = 0 .187 + 0 .0042*. The graph o f the least squares

l ine and the nine data points are show n below.

d Wh en x = 100, the valué for y can be pred icted using the least squares l ine as

y = 0.187 + 0.0042(100) = 0.44

Calcúlate

an d

SSR = ( ^ = m 0 3 3 3 3 ) l = 3743064

S„ 21,066.82

( 5 l 2

S S E = T o ta l SS - S S R = S - ± - 2 - L = .3747976 - .374306417 = .00049118$xx

The A NO VA table with 1 d f for regression and n - 2 d f for error is show n on the next

page. R em em ber th at th e m ean square s are calc ula te d as M S = S S / d f .

120



Source di SS M S

R egression 1 .3743064 .374306

Error 7 .0004912 .0000010

Total 8 .3747976

12.13 a The scatterplot gene rated by M in i ta b i s shown below. The assumption of

b C alc ú la te

I - * = 1 1 9 2 I * = 7 2 5 = 59 ,3 2 9

£ x 2 = 9 6 ,9 9 0 £ y 2 = 3 6,4 61 n = 15

Then

5 = Y x y -Í5^1í5iil = 1710.6667

,2

S a = y x f = 2265.7333

n2

1419.3333

b = - 2 - = --------= .75502S„ 2265 .7333

and a = y - b x = 4 8 . 3 3 3 3 - ( 0 . 7 5 5 0 2 ) ( 7 9 . 4 6 6 7 ) = - 1 1 . 6 6 5

(using full accurac y). Th e least squares line is

y = a + fox = -11 .66 5 + 0.755*.

c W hen x = 85, the valué for y can be predicted using the least squares line as

121



y = a + bx = - \ \ .665 + .755( 85) = 52 .51 .

12.17

12.21

a The hypothesis to be tested is

H0 : (1 = 0 versus H a : /? * 0

and the test statistic is a Student's t, calculated as

b - / 3 0 _ 1 .2 - 0

Vmse J s 2 V 0-533/10

The cr it ical valué o f t is based on n - 2 = 3 degrees o f freedom and the rejection

región for a = 0.0 5 is |?| > t m = 3.182 . Since the observed valué o f t falls in the

rejection región, we reject H0 and conclude that /? * 0 . Th at is , x is useful in the pred ic ti on o f y.

b From the AN O VA table in Exercise 12.6, calcúlate

f = M S R = J ± 4 _ = 2 7 0 ()

MSE 0.5333

which is the square of the t statistic from p art a: r = (5 . 2 0 ) 2 = 2 7 . 0 .

c The cr itical valué of t from part a was t m5 = 3.182 , while the critical valué of F

f rom part b with d f = 1and d f 2 = 3 is F 05 = 10.13 . N otice th at the relation ship

betw een th e tw o cri ti cal valú es is

F = 10.13 = (3 .182 )2 = t 2

a Th e depen dent v ariable ( to be predicted) is >’= cost and the indepen dent va riable

is x = distance.

b Preliminary calculations:

: = 2 1 , 5 3 0 = 5 0 5 2 ^ x j , = 7 , 5 6 9 ,9 9 9

Y , x f = 3 7 , 7 6 3 , 3 1 4 = 1 ,6 9 5 ,9 3 4 n = 18

Then

x iy¡ ~ ̂ ) ( ^ ^ } = ̂ 5 2 7 ^245.667

\ 2

=12,011 ,041 .78n

S /, = - 5 1 = 0 ^ 2 7 1 5 3

a = y - b x = 280.6667 -0.1271 53(119 6.1111) = 128.57699

and the least squares line is y = a + bx = 128.57699 + 0.127153*.

122



c Th e plot is show n below . Th e line app ears to fit well through the 18 data poii

12.25

Fitted Line Ploty = 128.6 +0.1272 x

s 72.3755

R-Sq 69.9%

R-Sq(adJ) 68.0%

d Ca lcúlate Total SS = S yy= ' £ y ? - ̂ i L = 1 ,695 ,934 - = 278 ,006 .n 18

Then

(5 „ ) (1 ,527 ,245 .667)"S S E = 5 = 2 7 8 , 0 0 6 - — ------ ------------- ¿- = 83,811.41055

” S 12,011,041.78

SS F 83 811 41055and MS E = -------- = — ------ 1---------- = 5238.213. The hypothesis to be tested is

n - 2 16

H 0 : [3 = 0 versus H a : /? * 0


O 197151 — 0

= 6.09t _ b - P Q 0 . 1 2 7 1 5 3 - 0

^ M S E / S „ V5238-213/12,011,041.78

The cri tical valué of t is based on « - 2 = 16 degrees of f reedom and the rejectionregión for a = 0.05 is |/| > t m = 2.1 20 , and H0 is rejected. Th ere is evide nce at tf

5% level to indícate that x and y are linearly related. T hat is, the regress ion m odel

y = a + f i x + e is useful in predicting co st y.

a Th e scat terplot generated by M in i tab is show n on the next page. Th e

assum ption o f l ineari ty is reasonable.

123

Scatter plot of Final Exam vs Pos ttest

10 0 •• •

• • •90 •

•• • •80 •

E •

5

i ”•

60• •

50•

4 0 -

7 0 7 5 8 0 8 5 9 0 9 5 10 0

Posttest

b U sin g th e M in i ta b printout, the equation of the regression line isy = - 2 6 .8 2 + 1 .2 6 1 7 * .

c T he hypothesis to be tested is

H n : p = 0 versus H a : /? * 0

and the test statistic is a Student’s t, calculated as

t = - r b ~ P Q = = 1.49 Vmse f s „

with p-value = .000 . Since the p-va lue is less than a = .01, we reject H0 and

con clude that /? * 0 . Th at is, final exam scores and po sttest score s are linearly

related.

d Th e 99% confidence interval for the slope /? is

b ± t a / 2 ylM S E / S a => 1.2617 ± 2.878(0.1253) => 1.2617 + 0.4849

or 0.7768 < /? < 1.7466 .

2.29 Use a plot of residuals versus f its. The plot should appear as a random sc at ter of poin ts , fr ee o f any pattern s.

2.33 a If you look careful ly, there appea rs to be a sl ight curve to the f ive points.

b T he fi t o f th e regressio n line, m easure d as r 2 = 0.959 indicates that 95.9% of the

overal l variat ion ca n be e xplained b y the straight l ine m odel .

c W hen we look at the residuals there is a strong curvil inear pat tern that has not

been expla in ed by the str aig ht line m odel. T he re la ti onsh ip be tw een tim e in m onth s

and num ber of beok s appears to be curvil inear .

2.39 a Although very sl ight , the student m ight notice a sl ight curva ture to the data

poin ts . b T he fit o f th e li near m odel is very good, assum in g th at th is is indeed the correct

mo del for this data set.

124



12.42

c Th e normal probability plot follows the correct pattern for the assum ption of

norm ality. Ho wev er, the residuals show the pattern of a quadratic curve, indicatin¿

that a quadratic rathcr than a l inear model may have been the co rrect model for this

data.

a a Use a Computer program or the hand calculations show n below.

£ > , = 1 1 6 ¿ * = 1 4 8 0 ¿ > ,* = 3 6 ,1 3 3

= 2 8 18 £ y 2 = 4 6 7 ,6 0 0 n = 5

Then

n

■ S „ = 2 > ,2 - — ^ - = 126.8n2

S = V v2 - ^ —^ 2 - = 29 ,52 0n

1797b = - 2 - = _ _ = 14.17192

S„ 126.8

a = y - b x = 2 9 6 - 1 4 .1 7 1 9 2 ( 2 3 . 2 ) = -3 2 . 7 8 9

and the least squares line is

y = -32 .78 9 + 14.172*.

b T he pro port io n o f th e to ta l vari ation expla ined by re gre ssio n isS~ 17972

r 2 = — 3 - = — ---------- = 0.8627S S m (126.8X29,520)

XX yv

c The diagnostic plots , gene rated by M in i ta b are shown below. The plots do m

show any strong violation of assumptions.

Res iduals Versu s the Fitted Valúes

(response i s Total Yante)

50- •

40

30-

20-

s io - •

-10- •

-20

-30 •

150 200 250 300 350Htted Valué

125

12.43 Re fer to Exe rcise 12.42 and calcúlate

1797 '(SJ SSE = S . - ^ L L . = 2 9 5 2 0 -

an d M S E =

1268SSE 4053.05205

= 4053.05205

= 1351.01735.ti —2 3

a The point est imator for E (y) when x = 21 is

y = -3 2 .78 9 +14 .172(21) = 264 .823

and the 95% confidence interval is

$ ± f 0 2 5 j M S E

2 A

264.823 ± 3.182. / l 351.017351 (2 1- 23 .2 )

5 + 126.8

264 .823± 57 .079

or 207.744 < E ( y ) < 32 1.902 .

b T h e po in t estim ato r fo r y when x = 21 is still y = -32 .78 9 + 14.172(21) = 264.823

and the 95% predict ion interval is

M SE, 1 ( t p - x )1 + - + — ---------

n S „

2 A

264 .823 ±3 .182^135 1 .01735

264.823 ±1 30.143

or 134.680 < > '<39 4 .96 6 .

1 (21 - 2 3 .2 )2 ̂1+ - + - -

5 126.8

126



12.47

12.51

c Th is w ould not be advisable, since you are t rying to est ímate outside the range

of exper imen taron .

a Re fer to f igure below. The sam ple correlat ion coefficient wil l be posi t ive.

Scatterplot of y vs x

4.0- •

3.5-

>. 3.0- •

2 .5 -

2 .0- • •

- 2 - 1 0 1 2X

b C alc úla te

Then r =

S „ = ^ - ^ - 1 0 - y = 10

= « - í f = 4

= 0 .9487 and r 2 = (0.9487)2 = 0.9000 . A pproxim atelyV io

90% of the total sum o f squares of deviat ions wa s reduced by u sing the least squares

equa tion instead o f y as a predictor of y.

W hen the pre-test score x is high, the post-test score y should also be high. The re

should be a posi t ive correlation.

127



Calcúlate

. 70 , 0 0 6 - ^ Z p = 468.42857

s *x = 6 5 , 9 9 3 - - ^ - = 517.42857^ n 7

Syy = Z X2 - ^ y‘ ̂ = 7 4 , 5 8 5 = 733.42857n 7

T , S v 468.42857 n _ Then r = - = . . .. = 0 .760 .

^517 .42857(733 .42857)

The test of hypothesis is

H 0 : p = 0 versus Ha : p > 0


, _ r V ^ 2 _ 0 .7 6 0 ,/5 „í r i — «Z.O I ^

VI - r 2 y l \ - ( 0 . 7 6 0 ) 2

Th e reject ion región for a = 0.05 is t > tM = 2 .01 5 and H0 is rejected. Th ere is

sufficient evidence to indícate posi t ive correlation.

12.55 a Since nei ther of the two var iables , amo unt of sodium or num ber of calor ies , is

control led, thejnethods of correlat ion rather than l inear regression analysis should beused.

b U se a Com pute r program , y ou r scie nti fi c calc u la to r o r th e co m puti ng form ula s

given in the text to calcúlate the correlation coefficient r. The M in i ta b printout for

this data set is show n on the next page.

128

Correlations: Sodium, CaloriesPearson correlation of Sodium and Calories = 0.981P-Value = 0.003

The re is evidence o f a highly signifícant correlat ion, since the p-valu e is so smalThe correlat ion is posi tive.

12.61 An sw ers will vary. The M in i ta b output for this l inear regression p roblem is sho belo w .

Regression Analysis: y versus xThe regression equation isy = 21.9 + 15.0 x

Predictor Coef SE Coef T PConstant 21.867 3.502 6.24 0.000x 14.9667 0.9530 15.70 0.000

S = 3.69098 R-Sq = 96.1% R-Sq(adj) = 95.7%

Analysis of Variance

Source DF SS MS F PRegression 1 3360.0 3360.0 246.64 0.000Residual Error 10 136.2 13.6Total 11 3496.2

Correlations: x, yPearson correlation of x and y = 0.980P-Value = 0.000

a The correlat ion coefficient is r = 0.980.

b The coeffic ie nt o f dete rm in ation is r 2 = 0.961 (or 96.1% ).

c T he least squa res line is y = 2 1.867 + 1 4:9667 x .

d W e wish to es t ím ate the mean percentage of k il l for an appl ication of 4 poi

o f nematicide per acre. Since the percent kil l y is actually a binomial percentage

var iance o f y wil l change depending on the valué of p , the propor t ion o f nematoi

ki lled for a part icular applicat ion rate. Th e residual plot versus the f í t ted valúes

shows this phen om eno n as a “footbal l-shaped ” pat tern. Th e normal probahil i ty

also shows som e deviat ion from no rmali ty in the tai ls of the plot. A transformat

may be neede d to assure that the regression a ssum ptions are sat isfied.

12.65 a Use a Computer program, your scientific calculator or the com puting form i

given in the text to calcúlate the correlat ion co efficient r.

= | , 233. 9 8 7 - 5028i( f 5 6 ) = 3 7 ,3 2 3

= Y x ? - = 2,21'2, i 7 8 - = 1<0 5 , 4 4 6^ n 12

Syy = Z y f ~ ~ ~ ~ = 723’882 ~ = 44 ’154

129

T hen r = ^ = _____ ^ l 3 23 = 0 .5 4 70 .^ V 1'°5 ,446(44,154)

The test of hypothesis isH0 : p = 0 versus H a : p > 0


r y f ñ — 2 0 .5470VÍ0 „l = —■ = .. — = z.Uoo

VI - r y j \ - ( 0 M 1 0 ) 2

with p-value = P (t > 2.066) bounded as

0 .05 < p -va lue <0 .10

If the exp erim enter is willing to tolérate a p-va lue this large, then Ho can be rejected.

Otherwise, you would declare the results not significant; there is insufficient evidenceto indicate that bending stiffness and twisting stiffness are positively correlated.

b r~ = (0.5470 )2 = 0.2992 so that 29.9% o f the total variation in y can be

explained by the independent variable x.

12.69 a Th e plot is shown below. N otice that the relationship is fairly weak.

Scat terp lo t o f Per-S i te Av ermge v s W eek s i n R e í ca se

7000 ----------------------------------------------------------------------------------------

6000 *

5000

i« 4000<9Í 3000

* 2000 -

1000

o6 8 1 0 12

Weeks h Retease

Regression Analysis. ,The regression equation isv = 3091 - 98 x

Predictor Coef SE Coef T PConstant 3091.4 701.0 4.41 0.000x -97.7 107.2 -0.91 0.374S = 1657.23 R-Sq = 4.4% R-Sq(adj) = 0.0%

Analysis of VarianceSource DF SS MS F PRegression 1 2282896 2282896 0.83 0.374Residual Error 18 49435210 2746401Total 19 51718106

130



12.73

Unusual ObservationsObs X y Fit SE Fit Residual St Resid

1 1.0 6551 2994 613 3557 2. 3 IR

10 16.0 2538 1527 1180 1011 0 .87 X

13 5.06219 2603 375 3616 2 . 24R

b From the printout , r 2 = 0.044 . Only about 4% o fth e overal l variation in y is

explained by using the l inea r model .

c From the printout , the regression equa tion is y = 3091.4 - 97 .7* and the

regression is not significant (t = -0 .91 wi th p -va lue = 0 .374) .

d Since the regression is not significant , i t is not appropriate to use the regressio

line for estimation or prediction.

a Th e calculat ions shown below are done using the com puting formulas. An

appropriate Com puter program wil l provide identical results to within roun ding e rre

X * , = 1 5 ° 2 > , = 91 2 > f = 986

x,2 = 2 7 5 0 £ y? = 1 1 2 0 .0 4 n = 10

Then

1 5 0 ( 91 ) = _ 3 79

* n 10

5 = £ x2 - = 2750 - — = 500“ ' n 10

S = y y ? J 1 * ) = 1 1 2 0 .0 4 - — = 291.94» n 10

- 3 7 9b = — = — — = - .7 58

S „ 5 0 0

a = y - b x = 9 .1 - ( - .758) (15) = 20 .47

and the least squares line is y = a + bx = 2 0 . 4 7 - .7 5 8 x .

b Since Total SS = S = 291.94 andyy

2

SSR = = ( 3?9) - = 287 .282S „ 5 0 0

\2

(SJ T h en S S E = T o ta l S S - S S R = 5 = 4.658

The AN OV A table wi th 1 d f for regression and n - 2 d f for error is show n below.

R em em ber that the mean squares are calculated as M S = SS¡ d f .

Source d f SS M S

Regression 1 287.282 287.282E rror 8 4.658 .58225

Total 9 291.940

131

c T o test H 0 : /? = O, H a : /? * O , the test statistic is

The reject ion región for a = 0 .05 is \t\ > t M5 = 2.306 and we reject H0. Th ere is

sufficien t eviden ce to indícate that x and >' are linea rly related.

d Th e 95% confid enc e interval for the slope (3 is

b ± ía/ 27 M S E ¡ s 2 => - .75 8 ± 2 .896^ .58225 /500 => - .758 ± .099

o r - . 8 5 7 < < - .6 5 9 .

e W hen x = 14, the est ímate o f exp ected freshness E(y) is

y = 20 .47 - .75 8(1 4) = 9 .858 and the 95% con f idence in terval is

The total variat ion has been reduced by 98.4%% by using the l inear model .

12.76-77 Use the H ow a Line W ork s applet . The l ine y = 0.5x + 3 has a slope of 0.5 and a y-

intercept of 3, while the line y = -0 .5 x + 3 has a s lope of -0 .5 and a y- in tercept of 3.

Th e second l ine slopes dow nw ard at the sam e rate as the f irst line slopes upw ard.

They both cross the y axis at the sam e point.

12.80 a Use a Computer, your scientif ic calculator or the com puting form ulas to f ind the

correlat ion between x and y. The M in i ta b correlat ion printout below shows r = 0.231

with p-va lue = 0.549 w hich is not significant at the 5% level of significance. You

cann ot conclude that there is a significant posi tive correlat ion be tween m edian rate

and score for “budg et” hotels.

i , o ,

9 .8 5 8 ± .5 6 2

or 9 .296 < E { y ) < 10.420.

f Calcúlate

_ _ S S R _ _ 287 28 2 _ Q gg4

Total SS 291.94

Correlations: Median Rate, ScorePearson correlation of Median Rate and Score = 0.231

P-Value = 0.549



b-c Use the Correlat ion an d the Scatterplot applet . The re is a rando m ser

o f points, with no outl iers. The studen t’s plot should look similar to the M in i ta b

shown below.

Scatte rplot of Sc ore vs Median Rate~

80- •

75- •

v 70-

•

••

65-

••

•60-

•

40 42 44 46 48 50Median Rate

52 54 56

133



13: Múltiple Regression Analysis

13.1 a W h e n = 2 , £ (> • ) = 3 + * , - 2 ( 2 ) = - 1 .

W he n jc2 = 1,E ( y ) = 3 + * , - 2 ( 1 ) = x, + 1 .

W h e n = 0 , E ( y ) = 3 + x l - 2 ( 0 ) = x l + 3 .

These three straight lines are graphed below.

x1

b N otice th at th e lines are para ll el (th ey have th e sam e slope).

13.5 a The model is quadratic.

b Sin ce R~ = . 8 1 5 , the sum o f squares of deviations is reduced by 81.5% using the

quadratic model rather than y to predict y.

c The hypothesis to be tested is

H 0 : /?, = p : = 0 H a: at least one /?( diffe rs from zero


F = ̂ = 37.37M SE

which has an F distribution with df¡ = k = 2 and d f 2 = n - k - \ = 2 0 - 2 - 1 = 1 7 . T he

p -valu e g iv en in th e prin to u t is P = .0 00 an d H 0 is reje cte d. T here is evid ence th at th e

model contributes information for the prediction o f y.

13.9 a Rate of increase is measure d by the slope o f a line tangent to the curve; this line is

given by an equ ation obtained as d y / d x , the derivative of y w ith respect to x. In part ic ula r,

135

13.15

13.19

^ = ̂ ( A , + / ? , * + A * 2 ) = f l + 2 / ¡ ,*dx dx

which has slope 2/3-,. If /?, is negative , then the rate o f increase is dec reasing .Henee, the hypothesis of interest is

H n : /?2 = 0 , H a: £ < 0

b The in div id ual r- te st is / = -8 .1 1 as in E xerc is e 13 .8b. H ow ever, th e te st is one-

tailed, which m eans that the p-v alue is half of the am ount given in the printout. That

is, p-v alue = -^(.000 ) = .000 . Hen ee, H0 is again rejected. Th ere is evidence to

indícate a decreasing rate of increase.

a The M in itab printou t fitting the m odel to the data is show n on the next page. Theleast squares line is

y = -8 .1 77 + 0.292a:, + 4.434a:2

Regression Analysis: y versus x1, x2The regression equation isy = - 8.18 + 0.292 xl + 4.43 x2

Predictor Coef SE Coef T PConstant -8.177 4.206 -1.94 0.093xl 0.2921 0.1357 2.15 0.068x2 4.4343 0.8002 5.54 0.001

S = 3.30335 R-Sq = 82.3% R-Sq(adj) = 77.2%

Analysis of VarianceSource DF SS MS F PRegression 2 355.22 177.61 16.28 0.002Residual Error 7 76.38 10.91Total 9 431.60

Source DF Seq SSxl 1 20.16x2 1 335.05

b T he F test for the overall u tility o f the m odel is F = 16.28 with P = .002 . The

results are highly significant; the model contributes significant information for the

pre dic tion o f y.c T o test the effect of adve rtising expen diture, the hypo thesis o f interest is

' H 0 : /?, = 0, H a: & * 0

and the test statistic is t = 5.54 with p-value = .001. Since a = .01, H0 is rejected.

W e conclude that advertis ing exp enditure contributes s ignificant information for the

pre dic tion o f y, g iv en th at capit al in vestm ent is alr eady in th e mode l.

d From the M in i tab printout, R -S q = 82 . 3%. which means that 82.3% of the

total variation can be explained by the quadratic m odel. Th e m odel is very effective.

a The variable x 2must be the quantitative variable, since it appears as a quadratic

term in the model. Q ualitative variables appear only with exp onen t 1, although they

136



ma y ap pea r as the coefficient of another quanti tat ive variable w ith expo nen t 2 or

greater.

b W hen x, = 0 , y = 12.6 + 3.9x2 w hile w hen x, = 1,

y = 12.6 + .5 4 ( l ) - 1 .2 x 2 +3 .9x 22 _

= 1 3 . 1 4 - 1 . 2 x2 + 3 . 9 x22

c The graph below shows the two parábolas.

x2

13.23 Th e basic response equation for a specifíc type of bonding comp oun d wo uld be

E ( y ) = A , + P \ X\ + P 2X\

Since the quali tat ive variable “bonding com pou nd” is at tw o levels, one dum m y

variable is needed to incorpórate this variable into the m odel . D efine the dum m y

variable x2 as follow s:

x2 = 1 i f bonding compound 2

= 0 otherwise

The expanded model is now writ ten as

£ (> 0 = A ) + $ x i + M 2 + P i* 2 + P*x \x i + M 2*2

13.25 a From the printout , the predict ion equation is y = 8 . 58 5 + 3 .8 2 0 8 x - 0 .2 1 6 6 3 x 2 .

b R 2 is labeled “R -sq” or R 2 = .94 4. He nee 94.4% o f the total variation is

acco unted for by using x an d x2 in the m odel.

c The hypo thesis o f interest is

H 0 : /?, = p 2 = 0 H a: at least one differs from zero

and the test statistic is F = 33.44 with p-value = .003 . H enee, H 0 is rejec ted, an d we

conc lude that the m odel contributes significant information for the pred ict ion o f y.

d The hypothesis of in terest isH 0 : A = 0 H a:/? 2 * 0

137



13.29

13.31

and the test statistic is t = -4.9 3 w ith p-value = .008 . Henee, H0 is rejected, and we

conclude that the quadratic model provides a better fit to the data than a simple linear

model.

e The pat tern of the diagno st ic plots does not indícate any obvious violation of theregression assum ptions.

a The model is

>’ = fio +M*i+ PiX 2+ M 2+ + A*í*2 + £ and the M initab printout is show n below.

Regression Analysis: y versus x1, x2, xlsq, x1x2, x1sqx2The regression equation isy = 4.5 + 6 .39 xl - 50.9 x2 + 0.132 xlsq ■

Predictor Coef SE Coef T PConstant 4 .51 42 .24 0.11 0.916xl 6.394 5 . 777 1.11 0 .275x2 -50.85 56 .21 -0.90 0.371xlsq 0.1318 0.1687 0.78 0.439xlx2 17.064 7 .101 2.40 0.021xlsqx2 -0.5025 0.1992 -2.52 0.016S = 71.6891 R-Sq = 76.8% R-Sq(adj) = '

Analysis of VarianceSource DF SS MS FRegression 5 664164 132833 25.85Residual Error 39 200434 5139Total 44 864598

P0.000

b The fi tted pre dic tion m odel uses th e coeffic ie nts giv en in th e colu m n m ark ed

“C o e f’ in the printout:

y = 4.51 -t- 6.394jc, - 5 0 .8 5 x , + 17.0 64x,;t2 + .131 8.x,2 - . 5 0 2 5 x f x 2

The F test for the m od el’s utility is F = 25.85 with P = .000 and R~ = .768 . T he

model fits quite well.

c If the dolphin is fema le, x2 = 0 and the predict ion equation becom es

y = 4.51 + 6.394.V, + .1 31 8jc2

d If the dolphin is male, x 2 = 1and the prediction equ ation become s

y = -4 6.3 4 + 23.458.x, - .3707a:2

e The hypo thesis of interest isH „ : / ? 4 = 0 H a : / ? 4 * 0

and the test statistic is t = .78 with p-value = .43 9. H0 is not rejected an d we con clude

that the quadratic term is not important in predicting mercury concentration for

female dolphins.

a- b Th e dató-is plotted on the next page. It app ears to be a curv ilinear relationship,

which cou ld be desc ribed using the quadratic m odel y = /?0 + ¡3xx + f i2 x 2+ £ .

138



X

— -----------------

c The M in i ta b pr intout is shown below.

Regression Analysis: y versus x, x_sqThe regression equation isy = 4114749 - 4113 x + 1.03 x_sq

Predictor Coef SE Coef T PConstant 4114749 343582 11.98 0.001x -4113.4 343.2 -11.99 0.001x_sq 1.02804 0.08568 12.00 0.001S = 0.523521 R-Sq = 99.7% R-Sq(adj) = 99.5%

Analysis of VarianceSource DF SSRegression 2 297.16Residual Error 3 0.82

Total 5 297.98

d The hypo thesis of interest is

H 0 : f l = A = 0

and the test statistic is F = 542.11 with p-value = .00 0. H0 is rejected an d we

conclud e that the model p rovides valuable information for the prediction o f y.

e R 2 = .99 7. Henee, 99.7% o f the total variation is acco unted for by using x and x 2

in the model.

f Th e residual plots are shown below. Th ere is no reason to dou bt the validity of

the regression assumptions. ____

Res iduals V ersus the F itted Valúes(raparse • y)0M -------------------------------------------

| a oo

•0.25

0 5 10 15 20ntted Valué

MS F P148.58 542.11 0.000

0.27

139



13.35 a R 2 = .999 . Henee, 99.9% of the total variation is acco unted for by using x and x 2in the m odel.

b T he hypoth esis o f in te re st is

H 0 : / ? , = A = 0

and the test statistic is F = 1676.61 with p-valu e = .00 0. H0 is rejected and we

conclude that the model provides valuable information for the prediction of y.

c The hypo thesis o f interest is

H 0 : f l = 0

and the test statistic is t = -2.6 5 with p-value = .045 . H0 is rejected and we conclu de

that the linear regression c oeff icient is s ignif icant when x 1 is in the m odel.

d The hypo thesis of interest is

H (, : A = 0

and the test statistic is t = 15.14 with p-valu e = .00 0. H0 is rejected and we conclude

that the quadratic regression coe ff icient is s ignif icant w hen x is in the model.

e W hen the quad ratic term is rem ove d from the model, the valué of R2 decrease s by

9 9 .9 -9 3 .0 = 6.9% . This is the addi tional contribut ion o f the quadrat ic term w hen i t

is included in the model.

f The clear pattem o f a curve in the residual plot indicates that the quadratic term

should be included in the model.

140

14: Analysis of Categorical Data

14.3 Fo r a test of specified cel l probabil i ties, the deg rees o f freedom are k - 1 . U se Ta ble

5, Appendix I:

a d f = 6 ; x l s = 12.59; reje ct H0 if X 2 > 12.59

b d f = 9 ; X m = 21.666; reject H0 if X 2 > 21.666

cd f = 13; x%a = 29.814; reject H0 if X 2 > 29.8194

d d f = 2; Xas = 5-99; reje ct H 0 if X 2 > 5.99

14.7 One thousand cars were each classif ied according to the lañe which they occupied

(one throu gh four). If no lañe is preferred over another, the prob ability that a car will be driven in la ñe i, i = 1,2,3,4 is V a. The nuil hypo thesis is then

H n : Px = P 2 = Pi = P< = j


with £, = npi = 1000(1/4) = 250 for i = 1 ,2 ,3 ,4 . A table of observed and expected

cell counts follows:Lañe 1 2 3 4

o¡ 294 276 238 192

E, 250 250 250 250

Then

v 2 _ ( 2 9 4 - 2 5 0 ) 2 i ( 2 7 6 - 2 5 0 ) 2 _ ( 2 3 8 - 2 5 0 ) 2 _ ( 1 9 2 - 2 5 0 ) 2

250 250 250 250

_ 6 H 0 _ 2 4 4 g

250

The reject ion región with k - 1 = 3 d f is X 2 > x l s = 7.81. Since the observed valuéo f X 2fa lls in the rejection región, we reject H0. Th ere is a differen ce in prefe renc e

for the four lañes.

14.11 Sim ilar to previous exercises. The h ypothesis to be tested is

H 0 : P\ = Pi = " ‘ = Pn ~

ve rsus H a : at least one p¡ is different from the others

with

E. = nPi = 4 0 0 ( 1 /1 2 ) = 3 3 .3 3 3 .The test statistic is

141

, ( 3 8 - 3 3 . 3 3 ) ( 3 5 - 3 3 . 3 3 )X 2 = - ---------------- ¿ - + - + ̂ -------------- — = 13.58

33.33 33.33

Th e up per tai led reject ion región is with a = .05 and k - 1 = 11 d f is

X 2 > Xas - 19.675 . The nuil hypothesis is not rejected and we ca nno t con clude that

the proport ion o f cases varies from m onth to month.

14.15 It is necessary to determ ine whether adm ission rates differ from the previously

reported rates. A table of observed and ex pected cel l counts fol lows:

U nco ndition al T rial Refused T otals

O, 329 43 128 500 E¡ 300 25 175 500

The nuil hypothesis to be tested is

H 0 : P, = -60; p 2 = .05; = .35

ag ains t the alternative that at least one of these prob abilities is incorrect. T he test

statistic is

x 2 = ( 3 2 9 - 3 0 0 ) - + ( « z ^ l + ( 1 2 L - 17 S y

300 25 175

Th e num ber of degrees of f reedom is k - 1= 2 and the r ej ect ion r eg ión2 2

X > X as = ^ -99. The nuil hypothesis is rejected, and we conclude that there has been a departu re fr om previo us adm is sio n ra te s. N otice th a t the percen ta ge o f

unco ndit ional adm issions has r isen sl ightly, the num ber of con dit ional adm issions has

increased, and the percentage refused adm ission has dec reased at the expen se o f the

fist two categories.

14.17 Re fer to Section 14.4 o f the text. For a 3 x 5 contingency table with r = 3 and c = 5 ,

th er e are ( r - l ) ( c - l ) = ( 2 ) ( 4 ) = 8 d eg re es o f fr ee do m .

14.21 a Th e hypothesis o f independ ence between at tachm ent pat tern and child care t ime is

tested using the chi-square stat ist ic. The c ontingency table, including colum n and

row totals and the est imated ex pected cel l counts, fol lows. ________

C h i l d C a r e

A t t a c h m e n t Low M odérate H igh T o t a l

Secure 24

(24.09)

35

(30.97)

5

(8.95)

64

A nxious 11

(10.91)

10

(14.03)

8

(4.05)

29

T o ta l 111 51 297 459

The test statistic isv 2 _ ( 2 4 - 2 4 .0 9 ) 2 | ( 3 5 -3 0 . 9 7 )2 <

24.09 30.97

( 8 - 4 . 0 5 )+ - ------------ — = 7.2 67

4.05

142

and the rejection región is X 2 > X m = $- 99 w ' t h 2 d f . is rejected . Th ere is

evidence of a dependence between at tachm ent pat tern and c hi ld care t ime,

b The valu é X 2 = 7.2 67 is betw een x l ¡ and Xm s so l^at < p-v alue < .05 .

resu lts are significant.

14.25 a The hypothesis of independence between salary and numb er of workdays at he

is tested using the chi-square stat ist ic. The co ntingency table, including colum n a

row totals and the est im ated exp ected cel l counts, generated by M in i ta b follows.

Chi-Square Test: Less than one, At least one, not all, All at homeExpected counts are printed below observed countsChi-Square contributions are printed below expected counts

Atleast

1

Lessthanone38

36.270.083

one,notall16

21.081.224

Allat

home14

10.651.051

Total68

2 5449.070.496

2628.520 .223

1214 .410.404

92

3 3535.200 . 001

2220 .460.116

910 . 340 . 174

66

4 33

39.471. 060

29

22 . 941.601

12

11.590.014

74

Total 160 93 47 300

Chi-Sq = 6.447, DF = 6, P-Value = 0.375

Th e test statistic is

v 2 _ ( 3 8 - 3 6 .2 7 ) 2 | ( 1 6 - 2 1 .0 8 ) 2 ( | ( 1 2 - 1 1 .5 9) 2 _ ^

36.27 21.08 11.59

and the reject ion región with a = .05 and d f = 3(2) = 6 is X 2 > Xa s = 12.59 and

nuil hyp othes is is not rejected . Th ere is insufficien t evid enc e to indícate that salai

dependent on the num ber of w orkdays spent at home.

b T he observed valu é o f th e te st sta ti stic , X 2 = 6.447 , is less than Xa o = 10.64¿

that the p-value is more than .10. This w ould confirm the non-reject ion of the nul

hypothesis from part a.

14.29 Be cause a set num ber of A m ericans in each sub-populat ion were each fíxed at 20'

we have a contingen cy table with fixed rows. The table, with estima ted expected

counts appea ring in parentheses, is show n on the next page.

143



Yes No T otal

W hite-A m erican 40

(62)

160

(138)

200

A frican-A merican 56

(62)

144

(138)

200

H ispanic-A merican 68

(62)

132

(138)

200

A sian-A m erican 84

(62)

116

(138)

200

Total 248 552 800


v 2 _ ( 4 0 - 6 2 ) 2 t ( 5 6 - 6 2 ) 2 | , ( 1 1 6 - 1 3 8 ) 2 21

62 62 138

and the rejection región w ith 3 d f '\s X 2 >1 1.34 49 . H0 is rejected and we conclud e

that the incidence o f parental support is depend ent on the sub-po pulation o f

Amer icans .

14.33 Th e number of observations per column were se lected pr ior to the exper imen t . The

test proce dure is identical to that used for an r x c con tingenc y table. The

contingency table, including column and row totals and the estimated expected cell

counts, follows.

Type

FamilyMembers

Apar tment D úplex SingleResidence

Total

1 8 20 1 29

(9.67) (9.67) (9 .67)

2 16 8 9 33

(11) (11) (11)3 10 10 14 34

(11.33) (11.33) (11.33)

4 or more 6 2 16 24

(8 ) (8) (8)

Total 40 40 40 120The test statistic is

. (8 —9.67 )2 (2 0 - 9 .6 7 )2 ( 1 6 - 8 )2X - ------------ ’ - + ±----------------- + + ± — = 36.499

9.67 9.67 8

u sin g C om p ute r a cc ura cy . W i th ( r - l ) ( c - l ) = 6 d fa n d a = .01 , the rejection región

is X 2 > 16.8449 . The nuil hypo thesis is rejected. There is suff icient eviden ce to

indicate that fam ily size is depende nt on type of family residence. I t appears that as

the family size increases, it is more likely that people will live in single residences.

14.35 If the hou sekee per actually has no preference, he or she has an equ al chan ce of

p ic k ing any o f th e five floor poli shes. H enee, the nui l hypoth esis to be te ste d is

144



14.39

H, P\ = Pi = Py = P\ = Ps = ^

The valúes of O, are the actual counts observed in the experim ent, and

E. = / i p , = 1 0 0(1 /5 ) = 2 0 . -

Polish A B C D E

o¡ 27 17 15 22 19

E¡ 20 20 20 20 20

Then X =( 2 7 - 2 0 ) ' (1 7 -2 0 )" ( 1 9 - 2 0 ) 2

= 4.4020 20 20

The p-value with d f = k -1 = 4 is greater than .10 and Ho is not rejected. W e cannot

con clude that there is a difference in the preferenc e for the five floor polishes. Even

if this hyp othesis h ad been rejected, the conclusión wo uld be that at least one o f thevalué of the p, was signif icantly different from 1/6. H ow ever, this does not imply that

p¡ is necessarily greater than 1/6. Hen ee, we cou ld not con clude that polish A is

superior.

If the objective of the exp erimen t is to show that polish A is superior , a better

p ro cedure would be to te st an hypoth es is as fo llow s:

H 0 '■P\ = 1/6 H a : p, >1 /6

From a sample of n = 100 hou sewives, x = 27 are found to prefer polish A. A s- test

can be performed on the single binomial parameter p¡.

a To test for hom ogen eity of the five binom ial pop ulations, we use chi-square

s ta tist ic and the 5x 2c on t ing en cy table shown below T he nuil hypothes is is that voter

choice and chu rch attenda nce are independent, w ith p be the proportion o f voters who

intend to vote for G.W. Bush in the 20 04 election for a particular church attendance

group. The contingency table generated by M in i ta b is shown below

Chi-Square Test: G.W. Bush, DemocratExpected counts are printed below observed countsChi-Square contributions are printed below expected counts

Total142

155

178

248

G.W. Bush Democrat

i 89 5373 .64 68.36

3 .205 3 .453

2 87 6880.38 74 . 620.546 0.588

3 93 8592.30 85.700.005 0.006

4 114 134128.60 119.401.658 1.786

145



5 22 36 5830.08 27.922.169 2.336

Total 405 376 781

Chi-Sq = 15.752, DF = 4, P-Value = 0.003

The observed valué of the test statistic is X 2 = 15.752 with p-v alue = .003 and the

nuil hy poth esis is rejected at the 5% level o f signifícance. Th ere is sufficient

evide nce to indicate that the proportion o f adults wh o intend to vote for G .W . Bush in

the 2004 election is depend ent on church attendance.

14.43 The flower fall into one of four classifications, with theoretical ratio 9:3:3:1.

Co nverting these ratios to probabil it ies ,

Pl = 9 /1 6 = .5625 p2

=3 /16 = . 1875

p , = 3 /1 6 = .1 875 p 4 = 1 /1 6 = .0625

We will test the nuil hypothesis that the probabilities are as above against the

alternative that they differ. Th e table of obse rved and expected cell cou nts follows:

AB Ab aB aa

o¡ 95 30 28 7

E, 90 30 30 10

T he tes t statistic is

v 2 _ ( 9 5 - 9 0 ) 2 ^ 3 0 - 3 0 ) * t ( 2 8 - 3 0 ) 2 , ( 7 - 1 0 ) 2 _ ^ ^ ]

90 30 30 10

The number o f degrees of f reedom is k - 1= 3 and the re jec t ion región w i th a = .01 is

X 2 > x \ \ = 1 1-3449 . Since the observed v alué of X 2 does no t fall in the rejection

región, we do not reject H0. W e do not have enough information to contradict the

theoretical model for the classification of flower color and shape.

14.48 In order to perform a chi-square “goodness of fit” test on the given data, it is

necessa ry that the va lúes O, and E¡ are know n for each o f the fíve cells. T he O, (the

num ber of measurements falling in the i-th cell) are given. H ow ever, £. = np( must

be calc ula te d . R em em ber th at p , is th e pro babil ity th at a m easurem ent fa ll s in th e j'-th

cell. Th e hypothesis to be tested is

H0 : the experim ent is binom ial versus Ha : the ex perim ent is not binom ial

Let x be the n um ber o f successes and p be the probabil i ty o f success on a s ingle tr ial.

Then, assuming the nuil hypothesis to be true,

^ = p ( x = 0 ) = Coy (1 - P ) 4 p l = p ( x = l ) = c t 4 p ' ( l - p )3

p 2 = /> ( * = 2 ) = c 24p 2( l - p ) 2 p 3 = p ( x = 3 ) = C * p 3( l - p ) ‘

p 4 = p ( x = 4 ) = C ; p 4( l - p ) °

146

H enee, once an est ím ate for p is obtained, the expec ted cel l frequen cies can be

calculated using the above proba bil i ties. No te that eac h o f the 100 experim ents

consists of four t r iáis and henee the comp lete expe rime nt involves a total o f 400 triáis.

14.51

The best est imator of p is p = x / n (as in Ch apter 9). Then,

„ _ x _ num ber of successes _ 0(1 l ) + l ( l7 ) + 2( 42 ) + 3(12)-t -'4‘(9) 1

n num ber of t riá is 400 2

The ex perimen t (consisting of four tr iáis) was repeated 100 t imes. Th ere are a total of

40 0 tr iáis in which the result “no succes ses in four tr iáis” was obse rved 11 t imes, the

result “one succes s in four t r iáis” was observed 17 t ime s, and so on. Then

Po = c * ( 1 / 2 ) 0 ( l / 2 ) 4 = 1/16 p , = c ; (1 /2 )' (1 /2 )3 =4 /16

p 2 = < T Í ( l / 2 ) ' ( l /2 ) ’ = 6 /1 6 p , = C, ( l /2 )? (1 /2) ' =4 /16

P ,= C 44( l /2 )4 (1 /2 ) ° =1 /16

The observed and expected cel l frequencies are shown in the fol lowing table.

X 0 1 2 3 4

o , 11 17 42 21 9

E, 6.25 25.00 37.50 25.00 6.25

and the statistic is

X =( 1 1 - 6 .2 5 )2 ( 1 7 - 2 5 .0 0 ) 2 ( 9 - 6 .2 5 )

= 8.566.25 25.0 0 6.25

In orde r to bound the p-value or set up a rejection región, i t is necessary to determ inethe appropriate degrees of freedom associated w ith the test stat ist ic. Tw o degrees of

freedom are lost because:

1 The cel l probab il it ies are restr icted by the fact that Z p, = 1.

2 The binomial param eter p i s unknown and m ust be es timated before calculat ing

the expected cel l counts. The num ber of degrees o f freedom is equa l to

k - \ - \ = k - 2 = 3 . Wi th d f = 3 , the p-value for X 2 = 8.56 is betwe en .025 and

.05 and the nuil hypothesis can be rejected at the 5% level of significance. We

conclude that the exp erimen t in quest ion do es not fulfi ll the requireme nts for a

b in om ia l experi m ent.

The nuil hypothesis to be tested is

u 1H 0 : p , = p 2 = p 3 = -


x 2 = I

with E j = np¡ = 200(1/3) = 66.67 for / = 1 ,2 ,3 . A table of observed and expected cell

counts follows:

147

E ntrance 1 2 3

o¡ 83 61 56

E¡ 66.67 66.67 66.67Then

v 2 _ ( 8 4 - 6 6 . 6 7 ) 2 | ( 6 1 - 6 6 . 6 7 ) 2 | ( 5 6 -6 6 .Ó 7 ) 2 _ 6 ^

66.67 66.67 66.67

Wi th d f = k - 1= 2 , the p-v alue is between .025 and .05 ánd we can reject H 0 at the

5% level of significance . Th ere is a difference in preferen ce for the three doors. A

95% confidence interval for p \ is given as

± Z o 2 5 \ r ^ - => — ± l - 9 6 j 4 1 5^'5 85 ^ => .4 15 ±.0 68n 025V n 200 V 200

or .347 < p, < .483.

14.55 Since the cards for each of the three holidays will be either “hum orous” or “not

hum orou s”, the table actually consis ts of two rows and three colum ns, and is shown

with estimated expected and observed cell counts in the M in itab printout below.

Chi-Square Test: Fathers, Mothers, ValentinesExpected counts are printed below observed countsChi-Square contributions are printed below expected counts

Fathers Mothers Valentines Total

i 100 125 120 345

115.00 115.00 115.001. 957 0.870 0 .2172 400 375 380 1155

385.00 385.00 385.000 .584 0 .260 0 . 065

Total 500 500 500 1500Chi-Sq = 3.953, DF = 2, P-Value = 0. 139

The test statistic for the equality of the three population proportions is

X 2 = 3.953 with p-value = . 139 and H0 is not rejected. There is insufficient evidence

to indícate a difference in the proportion o f hum orous cards for the three holidays.

14.59 a The 2 x 3 cont ingency table i s ana lyzed as in previous exerc ises . The Min i tab

p rin to ut be lo w sh ow s th e observed and estim ate d expecte d cell counts , th e te ststatistic and its associated p-value .

Chi-Square Test: 3 or fewer, 4 or 5, 6 or moreExpected counts are printed below observed countsChi-Square contributions are printed below expected counts

3 orfewer 4 or 5 6 or more Total

1 49 43 34 12637.89 42 .63 45.47

T . 254 0.003 2 .8952 31 47 62 140

42.11 47.37 50.532 . 929 0.003 2 .605

Total 80 90 96 266

Chi-Sq = 11.690, DF = 2, P-Value = 0.003

148



14.65

14.69

The results are highly significant ( p-value = .003 ) and we conclude that there is a

difference in the suscep tibili ty to colds dep ending on the num ber of relationships you

have.

b The proportion o f people with colds is calculated cond itionally for each of the

T h ree o r few er F ou r o r fiv e Six or more

Coid i » = .6180

« = . 4 890

— = .3596

No coid — = .3980

— = .5290

— = .6596

Total 1.00 100 1.00

As the researche r suspects, the susceptibility to a co id seem s to decrease as the

num ber o f relationships increases!

U se the first Goodness-of-Fit applet. Enter the observed valúes into the three cells in

the first row. and the expected cell counts will automatically appear in the second

row. The valué of X : = 18.5 with p-value = .0001 provide suff icient evidence to

reject H0 and conclude that custom ers have a preference for one o f the three brands

(in this case, Brand II).

Oreen R e d B iue To«*l

« 1 1 5 120 65 300

..............

Uoo o

■ i

100.0 100 0 30 0

Obse ivéd Fr«Qu«nel*s

ChiSq(2) = 18.5, p-value = 1 OE-4

The data is analyzed as a 2 x 3 contingency table with estimated expected cell counts

shown in parentheses. Use the Chi-Square Tes t o f Independence applet. Your

Opinión

Party 1 2 3 T otal

R epublican 114

(120.86)

53

(48.10)

17

(15.03)

184

Dem ocrat 87

(80.14)

27

(31.89)

8

(9.97)

122

Total 201 80 25 306

149




y 2 _ ( 1 1 4 - 1 2 0 . 8 6 ) 2 t ( 5 3 - 4 8 . 1 0 ) 2 [ ( ( B - 9 . 9 7 ) 2 = 2

120.86 48.1 0 9.97

With d f = 2 , the p-valu e is greater than .10 ( the applet reports p-valu e = .237 8 ) and

H0 is not rejected . Th ere is no eviden ce that party affiliation has any effect onopinión.

15: Nonparametric Statistics

15.1 a If distr ibution 1 is shifted to the r ight of distr ibution 2, the rank sum for sam ple 1

(T¡) will tend to be large. T he test statistic w ill be T ¡ , the rank sum for sample 1 if

the obse rvations had been ranked from large to small . Th e nuil hyp othesis wil l be

rejected if 7j’ is unusually small.

b F rom T ab le 7 a with n, = 6 , n 2 = 8 and a = .05 , H0 will be rejected if 7'* <31.

c From Table 7c wi th n, = 6 , n2 = 8 and a = .0 1, H0 will be reje cte d if 7”, < 2 1 .

15.5 If Ha is t rue and populat ion 1 l ies to the r ight o f popu lat ion 2, then 7, w ill be large and

7* will be sm all. H enee, the test statistic w ill be 7* and the large sa m ple

approximation can be used. Calcúlate

7* = n, (rtj + n2 + 1) - 7 , = 1 2( 27 ) —193 = 131

_ n , ( t t i + n 2 + l ) _ 1 2 (2 6 + 1) ^

Mt 2 2

= nin2( n ] + n 2+ 1) = 12 (14 )(27 ) = 3?g

r 12 12T he tes t statistic is

7 , - A r _ 13 1 -1 6 2 _ Z — / v.jyGr V378

The reject ion región with a = .05 is z < -1 .6 4 5 and Hq is not rejected. Th ere is

insufficient evidence to indicate a difference in the two po pulat ion distr ibutions.

15.9 Sim ilar to previous exercises. The data, with correspo nding ranks, are shown in the

fol lowing table. ___________ ______________

D eaf (1) H ea r i n g (2 )

2 . 7 5 ( 1 5 ) 0 .8 9 ( 1 )

2 .1 4(1 1) 1.43 (7)

3 .2 3 ( 1 8 ) 1.06 (4)

2 .0 7(1 0) 1.01 (3)

2 . 4 9 ( 1 4 ) 0.94 (2)

2 .1 8(1 2) 1.79 (8)

3 .1 6(1 7 ) 1 .1 2 (5 .5 )

2 .9 3(1 6) 2.01 (9)

2 .2 0(1 3 ) 1 .1 2 (5 .5 )

7, =126

Calcúlate

151

T{ = 126

T¡ = « , ( « , + n 2 + l ) - r , = 9 (1 9 ) —126 = 4 5


T = m in( 7J ,7’1‘ ) = 45

With n, = n 2 = 9 , the tw o-tailed rejection re gión with ar = .05 is found in Tab le 7b to

be 7j’ < 62 . T he observed valu é, T = 45, falls in the rejection región and H0 is

rejected . W e conclude that thé d ea f chi ldren do di f fer from the hear ing children in

eye-mov emen t rate .

15.13 a If a paired difference expe riment has been used and the sign test is one-tai led

(H a : p > .5) , then the experimenter would l ike to show that one populat ion of

m easurem ents lies above the other populat ion. An exac t pract ical statem ent of theal ternat ive hyp othesis would depen d on the expe rimental si tuat ion.

b I t is necessary th at a ( the probab ili ty of reject ing the nuil hyp othesis w hen i t is

true) take valúes less than a = .15 . Assum ing the nuil hypothesis to be true, the two

popu la tions are id entical and consequen tly ,

p = P { A exceeds B for a given pair of observations) is 1/2. The binom ial probabil i ty

wa s discussed in Cha pter 5. In particular , i t was noted that the distr ibution o f the

random variable * is symmetrical about the mean np when p = 1/2 . For exam ple,

with n = 25, P ( x = 0 ) = P ( x = 2 5 ) . Simi larly , P { x = 1) = P ( x = 24) and so on.

He nee, the lower tailed probabil i ties tabulated in Tab le 1, Ap pendix I wil l be identicalto their upper tailed equivalent probab ili t ies. The valúes of a avai lable for this upper

tai led test and the correspon ding reject ion regions are show n below.

Reject ion Región a

* > 2 0 .002

* > 1 9 .007

* > 1 8 .022

* > 1 7 .054

* > 1 6 .115

15.17 a If assessors A and B are equal in their property assessm ents, then p , the

p robab il it y th at A ’s assessm en t exceeds B ’s assessm en t fo r a g iv en property , should

equal 1/2. If one of the assessors tends to be m ore conservative than the other, then

either p > 1/2 or p < 1/2 . Henee, we can test the equivalence o f the two assessors by

testing the hypothesis

H n : p = l / 2 versus H a : p í ¿ l / 2

using the test stat ist ic *, the nu m ber of t im es that assessor A exce eds a ssessor B for a

parti cu la r property assessm ent. To find a tw o-t ail ed re je cti on re g ió n w ith a cióse to.05, use Table 1 wi th n = 8 and p = .5 . Fo r the rejec tion región {* = 0, * = 8) the

152

valué of a is .004 + .004 = .008 , while for the rejec tion región { x = 0,1,7,8} the

valué of a is .03 5+ .035 = .070 w hich is closer to .05. H enee, using the reject ion

región {jc < 1 or * > 7 } , the nuil hypoth esis is not rejected, since x = number o f

p roperties fo r w hic h A exceeds B = 6 . T he p -v alu e fo r th is tw o-t ailed te s t is

p -va lu e = 2 P ( x > 6 ) = 2 (1 - .8 5 5 ) = .290

Since the p-valu e is greater than .10, the results are no t significant; H0 is not rejected

(as with the critical valué approach).

b T he t statistic use d in Exe rcise 10.45 allows the exp erim en ter to rejec t Ho, wh ile

the sign test fails to rejec t H0. Th is is beca use the sign test usedless inform ation and

makes few er assumpt ions than does the t test. If al l norm ali ty assum ptions are met,

the t test is the m ore pow erful test and can reject when the sign test cannot .

15.21 a Ho'. po pula tion distributions 1 and 2 are identical

H a: the distribu tions d iffer in location

b S in ce T ab le 8, A ppendix I giv es cri ti cal valú es fo r re je ction in th e lo w er ta il o f the

distribution, we use the smaller of T +and T~ as the test statistic.

c From Table 8 wi th n = 3 0 , a = .05 and a two-tai led test, the reject ion región is

T < 137 .

d Since T + = 24 9, we can calcúlate

n ( n + 1) . 3 0( 3 1)T~ = — - - T * = — - 2 4 9 = 2 1 6 .

2 2The test statistic is the sm aller of T + and T~ or T = 216 and H0 is not rejected. The re

is no evidenc e o f a difference b etween the two distr ibutions.

15.25 a Th e hypo thesis to be tested is

H 0: po pu lation dis tribu tions 1 and 2 are iden tical

H a: the d istributions diffe r in location

and the test statistic is T, the rank sum o f the posit ive (or negative) differences. The

ranks are obtained by ordering the differences according to their absolute valué.Def ine d¡ to be the d i fference between a pai r in populations 1 and 2(i.e., x u - x 2¡).

The differences, along with their ranks (according to absolute magnitude), are shown

in the following table. _______

di .1 .7 .3 -.1 .5 .2 .5

R a n k | d¡ \ 1.5 7 4 1.5 5.5 3 5.5

The rank sum for posi t ive differences is T + = 26.5 and the rank sum for negative

differences is T~ =1.5 wi th n = 7 . Con sider the sm aller rank sum an d determine the

appropriate low er portion of the two-tai led reject ion región. Indexing n = l and

a = .05 in Table 8, the rejection región is T < 2 and Hq is rejected. Th ere is adifference in the two p opu lat ion locations.

153

b The results do not agree with those obtained in Exercise 15.16. W e are able

to reject H0 with the more p ow erful W ilcoxon test.

15.29

15.31

a The paired data are given in the exercise. The d ifferences, along w ith their ranks

el i 2 -1 1 3 1 -1 3 -2 3 1 0

Rank

K l

3.5 7.5 3.5 3.5 10 3.5 3.5 10 7.5 10 2.5 -

Let p = P ( A exceeds B for a given intersec tion) and x = num ber of intersect ions a t

w hich A exceeds B. The hyp othesis to be tested is

H 0 : p = 1/2 versu s H a : p * 1/2

using the sign test with x as the test statistic.

Critical valué approach: Various two tailed rejection regions are tr ied in order to

f ind a región w ith a ~ .05 . T he se are show n in the follow ing table.Rejection Región a

jc < 1;jc > 10 .012

x < 2 , x > 9 .066

* < 3 ;jc > 8 .226

W e ch oo se to reject H o i f x < 2 o r ; t > 9 w ith a = .066 . Since x = 8 , H qis not

rejected. There is insufficient evidence to indícate a difference betw een the two

methods.

p-value approach: For the observed valué x = 8 , calcúlate the tw o-tailed p-value:

p -v alu e = 2 P ( x > 8 ) = 2(1 - .88 7 ) = .226Since the p-value is greater than .10, H0 is not rejected.

b To use the W ilcoxon signed rank test, we use the ranks o f the absolute differences

show n in the table above. The n T + = 5 1 . 5 a n d T~ =14.5 with n = 11 . I ndexing

« = 11 and a = .05 in Table 8, the low er portion of the tw o-tailed rejection región is

T < 11 and H0 is not re jecte d, a s in pa rt a.

a Since the experimen t has been design ed as a paired exp eriment, there are three

tests available for testing the differences in the distributions with and without imagery

- (1) the paired difference t test; (2) the sign test or (3) the Wilcoxon signed rank test.

In order to use the paired difference t test, the scores must be ap proxim ately normal;since the num ber of words recalled has a binomial d istr ibution with n = 25 and

unknown recall probahility, this distr ibution may not be approximately normal,

b Using the sign test, the hypothesis to be tested is

H 0 : p = 1/2 versus H a : p > 1/2

For the observed valué x = 0 we calcúlate the tw o-tailed p-value:

- p -v alu e = 2 P ( x < 0 ) = 2 (.000 ) = .000

Th e results are highly signif icant; Ho is rejected and we co nclud e there is a difference

in the recall scores with and without imagery.

Using the W ilcoxon s igned-rank test , the differences w ill all be positive ( x = 0 for

the sign test), so that and

154

15.35

15.39

. n ( n + 1) 20(21)T + = — -------- = — -— - = 210 and T ~ = 2 1 0 - 2 1 0 = 0

2 2

Indexing n = 20 and a = .01 in Table 8, the lower po rtion o f the two -tailed rejection

región is T < 37 and H0 is rejected.

The d ata w ith corresponding ranks in parentheses are shown b elow.

A ge

1 0 - 1 9 20 - 39 4 0 - 5 9 6 0 - 6 9

29 (21) 24 (8) 3 7 (3 9 ) 28 (18)

33 (29.5) 2 7 (1 5 ) 2 5 (1 0 .5 ) 29 (21)

2 6 (1 2 .5 ) 33 (29.5) 22 (5.5) 34 (34)

27 (15) 31 (24) 33 (29.5) 36 (37.5)

39 (40) 21 (3) 2 8 ( 1 8 ) 2 1 ( 3 )

35 (36) 2 8 (1 8 ) 2 6 (1 2 .5 ) 2 0 ( 1 )33 (29.5) 24 (8) 30 (23) 2 5 (1 0 .5 )

29 (21) 34 (34) 34 (34) 2 4 (8 )

36 (37.5) 2 1 ( 3 ) 2 7 (1 5 ) 33 (29.5)

22 (5.5) 32 (25.5) 33 (29.5) 32 (25.5)

7j = 247.5 T2 = 168 T 3 = 2 1 6 .5 7; = 188

n, = 1 0 n 2 = 10 n 3 = 1 0 n4 = 10

Th e test statistic, based on the ran k sums, is

12 H = 3 ( „ + l)

n ( n + 1) /i,

12

4 0 ( 4 1 )

(2 47 .5 )2 (1 68 )2 (2 16 .5 )2 (1 88 )2

10 10 10 10- 3 ( 4 1 ) = 2.6 3

Th e rejection región w ith a = .01 and k -1 = 3 d f is based on the chi-square

dis tribution, o r H > = 11.35 . The nuil hyp othesis is not rejected. Th ere is no

evidence of a difference in location.

b S in ce th e obse rved valu é H = 2.63 is less than %2l0 = 6.25 , the p-v alue is greater

than .10.

c-d From Ex ercise 11.60, F = .87 w ith 3 and 36 df . Again, the p-v alue is greaterthan .10 and the results are the same.

Th e ranks o f the data are shown on the next page.

155

Treatment

Block 1 2 3 4

1 4 1 2 3

2 4 1.5 1.5 3

3 4 1 3 2

4 4 1 2 3

5 4 1 2.5 2.5

6 1 2 3

7 4 3 2

8 4 1 2 3

7; =32 T 2 = 8.5 7 ; = 1 8 7 ; =21 .5

a Th e test statistic is

Fr = ,12 , y T - - ? , b { k + \)

r bk(lc + 1 ) ^ ' V ’

= 8( ^ 5 )[ (32)2 +(8*5)2+182 +(21-5)2] - 3(8)(5) = 2119

and the rejection región is Fr > ^ 205 = 7.81. Hen ee, Ho is rejected and we conclude

that there is a difference am ong the four treatments .

b The ob served valué, Fr = 2 1 . 1 9 , e x c e e d s x 2m , p-valu e < .005 .

c-e The analysis of variance is perform ed as in Ch apter 11. The A NO VA table is

shown below.

Source d f SS M S F

T reatm ents 3 198.34375 66 .114583 75.43B locks 7 220.46875 31.495536

E rro r 21 18.40625 0.876488

Total 31 437 .40625

The analysis of variance F test for treatm ents is F = 75.43 and the appro xim ate p-

value with 3 and 21 d f is p-v alue < .005 . Th e result is identical to the param etric

result.

15.43 Ta ble 9, A ppen dix I gives critical valúes r0 such that P ( r s >r0 ) = a . Hen ee, for an

upper-tailed test, the critical valué for rejection can be read directly from the table.a rs > . 4 2 5 b a; >.601

15.47 a The two variables (rating and dis tance) are ranked from low to high, and the

results are shown in the following table.

Voter X v Voter X y

1 7.5 3 7 6 4

2 4 7 8 11 2

3 3 12 9 1 10

4 12 1 10 5 9

5 10 8 11 9 5.5

6 7.5 11 12 2 5.5

156

Calcúlate I j c ¡y, = 4 4 2 .5

n = \2

I x f = 6 4 9 .5 X y f = 649.5

I x , = 7 8 X y, = 78

Then

S n, = 42 2.12

7 8 :S a = 6 4 9 .5 ---------= 142.5

12

S yy = 6 4 9 .5 -12

and

b T he h yp oth esis o f in tere st is H 0: no co rr e la ti on vers us H a: n egati ve corr e la ti on.

Co nsul t ing Table 9 for a = .05 , the critical valué of rs, deno ted by r0 is - .49 7. Since

the valué of the test statistic is less than the

critical valué, the nuil hyp othesis is rejected. Th ere is eviden ce o f a significant

negative correlation between rating and distance.

15.51 Re fer to Ex ercise 15.50. To test for positive correlation with a = .05 , index .05 in

Table 9 and the rejection región is rs > .600 . W e reject the nuil hypothe sis of no

association and conclude that a positive correlation exists between the teacher’s ranks

and the ran ks o f the IQs.

15.55 a Def ine p = P (res po ns e for stimulus 1 exceed s that for stimulus 2 ) and x = n u m b e r

of t im es the respon se for stimulus 1 exceed s that for stimulus 2. Th e hyp othesis to be

tested is

using the sign test with x as the test s tatistic. N otice that for this exercise n = 9 , and

the ob served valué o f the test s tatistic is x = 2 . Various two tailed rejection regions

are tr ied in order to f ind a región w ith a ~ .05 . Th ese are show n in the following

table.

W e choose to reject Hq if x < 1or x > 8 w ith a = .04 0 . S ince x = 2 , H0 is not

rejected. T here is insuff icient eviden ce to indícate a difference between the two

stimuli.

b T h e ex perim en t has been d esig ned in a pair ed m anner, and th e pair ed dif fe re nce

test is used. The differences are show n below.

d, - . 9 - 1 .1 1.5 - 2 . 6 - 1 . 8 - 2 . 9 - 2 . 5 2 .5 - 1 . 4

The hypothesis to be tested isH 0 - / / 2 = 0 H a : / / , - / / 2 * 0

Calcúlate

H 0 : p = 1/2 ve rsu s H a : p 1/2

Reject ion Región a_

x = 0 , x = 9 .004

jc < 1;jc > 8 .040

x < 2 - , x > 7 .180

157

15.59

- = I < = - 9 2 = _ 1

n 'd = ^ L = — = - \ m i

n 9

, > „ ' 2

5: = Z d ? -

n - 1


—— = 3 7 -1 4 -9 .4 0 4 = 3 467

The re jection región with a = .05 and 8 d f is |/| > 2.3 06 and H 0 is not rejec ted.

The data, with corresponding ranks, are shown in the following table.

A (1) B (2)

6.1 (1) 9.1 (16)

9 . 2 ( 1 7 ) 8.2 (8)

8 .7 (1 2 ) 8 .6 (1 1 )

8 .9 (1 3.5 ) 6 .9 (2)

7.6 (5) 7.5 (4)

7.1 (3) 7.9 (7)

9 . 5 ( 1 8 ) 8.3 (9.5)8.3 (9.5) 7.8 (6)

9.0 (1.5) 8.9(13.5)

7¡ =94

The d ifference in the brightness levels using the two p rocesse s can be tested using the

nonparametr ic W ilcoxon rank sum test , or the parametr ic tw o-samp le t test.

1 To test the nuil hypo thesis that the two pop ulation distr ibution s are identical,

calcúlate

7¡ =1 + 1 7 h------h 1.5 = 94

7 T = * , ( n , + « 2 + l ) - 7 ¡ = 9 ( 18 + l ) - 9 4 = 77


T = m in(7¡ ,r ,* ) = 77

With n x = n 2 = 9, the two-tailed rejection región with a = .05 is found in Ta ble 7 b to

be Tx < 62 . The obse rved valué, T = 77 , doe s not fall in the rejection región and H 0

is not rejected. We canno t conclud e that the distribution s of br ightne ss measu reme nts

is different for the two processes.

2 To test the nuil hypo thesis that the two pop ulation m eans are identical, calcúlate

_ I * , ; 74.4= 8.2667

_ 1 * 2 , 73.2 jc, = ------ - ---------

n, 9= 8.1333

158

15.61

15.65

2 _ (w, -1 )J ,2 + (n 2 - l ) s 22 _

n\ + n 2 - 2 and the test statistic is

*1 ~*2

6 2 5 . 0 6 -( 7 4 . 4 ) 2

+ 5 9 9 . 2 2 -(73 .2)2

t =8 . 2 7 - 8 . 1 3

1 1 — +n, n2

- i - J . 8 6 75

16

= .304

- — = .8675

The rejection región with a = .05 and 16 deg rees o f freedo m is |f| > 1.746 and H0 is

not rejected. Th ere is insuff icient eviden ce to indicate a difference in the average

bri gh tn ess m easure m ents fo r th e tw o pro cesses.N oti ce th at th e nonpara m etr ic and

para m etr ic te sts re ach th e sam e conclu sio ns.

Since this is a paired exp erimen t, you can cho ose either the sign test, the Wilcoxon

signed rank test, or the param etr ic paired t test. Since the tenderizers have been

score d on a scale of 1 to 10, the param etric test is not applicab le. Sta rt by using the

easiest of the two nonp aram etr ic tests - the sign test.

Def ine p = ^( te nd er iz er A exceeds B for a given cut) and x = num ber of times that A

excee ds B. The hy pothes is to be tested is

H 0 : p = l / 2 v e rs us H a : p * 1/2

using the sign test w ith x as the test statistic. N otice that for this exe rcise n = 8 (there

are tw o ties), and the observe d valué o f the test s tatistic is x = 2 .

p -v a lu e a p p ro a c h : For th e observ ed valu é x = 2, calcúlate

p -v alu e = 2 P ( x < 2 ) = 2 ( .145 ) = .290

Since the p-value is greater than .10, H0 is not rejected. The re is insuff icient eviden ce

to indicate a difference betw een the two tenderizers.

If you use the W ilcoxon signed rank test, you will f ind T* = 7 and T~ = 29 which w ill

not allow rejection o f H0 at the 5% level of signif icance. The results are the same.

The hypothesis to be tested isH0: pop ulation distr ibutions 1 and 2 are identical

Ha: the distributions differ in location

and the test statistic is T, the rank sum o f the positive (or negative) differences. The

ranks are o btained by o rdering the differences according to their absolute valué.

Def ine d¡ to be the difference b etween a p air in populations 1 and 2 ( i .e. , x u - x 2i).

The differences, along with their ranks (according to absolute magnitude) , are shown

di -31 -3 1 - 6 -1 1 - 9 - 7 7

R a n k | d¡ | 14.5 14.5 4.5 12.5 10.5 7 7

di -1 1 7 - 9 - 2 - 8 -1 - 6 - 3

R a n k | d¡ \ 12.5 7 10.5 2 9 1 4.5 3

159

15.69

15.73

15.77

The rank sum for positive differences is T * = 14 and the rank sum for negative

differences is T~ = 106 with n = 15 . Co nsider the smaller rank sum and determ ine

the appropriate lower portion of the two-tailed rejection región. Indexing n = 15 and

a = .05 in Table 8, the rejection región is T < 25 and H 0 is rejected. W e concludethat there is a difference betw een m ath and ar t scores.

a-b Since the exper iment is a complete ly random ized des ign, the K ruskal W all is H

test is used. The com bined ranks are show n below.

P la n t R a n k s

A 9 12 5 1 7 34

B 11 15 4 19 14 63

C 3 13 2 9 6 33

D 20 17 9 16 18 80

The test statistic, based on the rank sums, is

H =12

/ t ( n + l )

12

20(21)

Z ^ ~ 3 ( « + l)

(34)2 , (63 )2 , (33 )2 _ (80)- 3 ( 2 1 ) = 9.0 8

Wi t h d f = k -1 = 3 , the observed valué H = 9.08 is between %m5 and X 05so l^at

.025 < p - \alu e < .05 . The n uil hyp othesis is rejected and we con clude that there is a

difference am ong the fou r plants.

c From Exercise 11.66, F = 5.2 0 , and H0 is rejected. The results are the same.

Th e data are already in rank form. The “substantial experien ce” sample is design ated

as sample 1, and n, = 5 ,n2 = 7 .C alcúla te

7, =19

T¡ = « , ( / ! , + n 2+ \ ) - T { = 5 (13 ) —19 = 46


T = m in(7 ^,7 'l*) = 19

With n, = n 2 = 12, the one-tailed rejection región w ith a = .05 is found in Table 7a to

be Tx < 21 . The ob served valué, 7 = 19, falls in the rejection región and H 0 is

rejected. There is suff icient eviden ce to indicate that the review boa rd considers

expe rience a pr ime factor in the selection of the best cand idates.

The data with corresponding ranks in parentheses are shown on the next page.

160



Training Per iods (hours)

.5 1.0 1.5 -2:0

8 (9.5) 9( 1 1 . 5 ) 4 ( 1 . 5 ) 4( 1 . 5 )

14( 14) 7 ( 7 ) 6 ( 5 ) 7 ( 7 )

9 ( 1 1 . 5 ) 5 (3.5) 7 ( 7 ) 5 (3.5)

1 2( 1 3) 8 (9.5)

7 , =48 T 2= 2 2 T, =23 7, =12

« , = 4 n2 = 3 * , = 4 n4 = 3

The test statistic, based on the rank sums, is

12 H =

n ( n + \

12

14(15)

' (48 ) : _ (22 ) : 1 (2 3) : , (12)2

- 3 ( 1 5 ) = 7 .4 3 33

The re jection región with a = .01 and k - 1= 3 d f is based on the chi-square

distribution, or H > %20l = 11.34 . The nuil hyp othesis is not rejected and we

conclude that there is insufficient evidence to indícate a difference in the distribution

of times for the four groups.

161

SOLU Introducción a La Probabilidad y Estadística, 12ma Edición - W. Mendenhall, R. Beaver

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Transcript of SOLU Introducción a La Probabilidad y Estadística, 12ma Edición - W. Mendenhall, R. Beaver