Confiabilidad

4
Sujeto Item Item Item Item Item Item Item Item Item Item Item Item 1 2 3 4 5 6 7 8 9 10 11 12 Total (X) 1 4 4 3 2 1 1 2 3 4 5 1 2 32 2 3 1 3 3 1 2 3 4 5 1 2 3 31 3 4 5 4 4 4 4 3 5 4 2 3 1 43 4 4 3 2 3 1 1 4 5 3 2 1 2 31 5 5 2 4 4 3 1 3 4 5 1 2 3 37 6 5 4 3 2 4 2 2 4 3 1 2 3 35 7 4 4 2 3 2 3 4 3 2 1 5 3 36 8 4 4 2 5 2 4 4 3 5 2 1 3 39 9 4 4 3 3 2 4 3 4 2 5 3 1 38 10 5 4 4 4 4 4 3 4 4 5 1 3 45 Total 43 37 33 37 29 32 38 47 46 35 32 36 367 X2 180 135 96 117 72 84 101 157 149 91 59 64 0.4 1.38 0.66 0.9 1.6 1.82 0.54 0.54 1.34 3.16 1.65 0.91 Vi2 Calculo de la varianza X 2 – ( X) VF= n______ n-1 13675 – (367 ) 2 10 Vf= _____________________ = 23 10-1 Calculo de la varianza individual X 2 – ( X) VF= n______ n-1 180-(42) 2 10 V1 = _________________ = 0,4 10-1 135-(35) 2 10 V2 = _________________ = 1,38 10-1 96-(30) 2 10 V3 = _________________ = 0,66 10-1 117-(33) 2 10 V4 = _________________ = 0,9 10-1 72-(24) 2 10 V5 = _________________ = 1,6 10-1 84-(26) 2 10 V6 = _________________ = 1,82 10-1 101-(31) 2 10 V7 = _________________ = 0,54 10-1 157-(39) 2 10 V8 = _________________ = 0,54 10-1 149-(37) 2 10 V9 = _________________ = 1,34 10-1 91-(25) 2 10 V10 = _________________ = 3,16 10-1 59-(21) 2 10 V11 = _________________ = 10-1 641-(24) 2 10 V12 = _________________ = 10-1 Calculo del coeficiente de confiabilidad form ula alfa de cronbach (n) * (1- Vi) Rt= n-1 vt Rt= (12 ) * (1-15) = -0,66 12-1 23

description

ciencia

Transcript of Confiabilidad

Page 1: Confiabilidad

Sujeto Item Item Item Item Item Item Item Item Item Item Item Item1 2 3 4 5 6 7 8 9 10 11 12 Total (X) ∑X2

1 4 4 3 2 1 1 2 3 4 5 1 2 32 10242 3 1 3 3 1 2 3 4 5 1 2 3 31 9613 4 5 4 4 4 4 3 5 4 2 3 1 43 18494 4 3 2 3 1 1 4 5 3 2 1 2 31 9615 5 2 4 4 3 1 3 4 5 1 2 3 37 13696 5 4 3 2 4 2 2 4 3 1 2 3 35 12257 4 4 2 3 2 3 4 3 2 1 5 3 36 12968 4 4 2 5 2 4 4 3 5 2 1 3 39 15219 4 4 3 3 2 4 3 4 2 5 3 1 38 1444

10 5 4 4 4 4 4 3 4 4 5 1 3 45 2025Total 43 37 33 37 29 32 38 47 46 35 32 36 367 13675

∑X2 180 135 96 117 72 84 101 157 149 91 59 640.4 1.38 0.66 0.9 1.6 1.82 0.54 0.54 1.34 3.16 1.65 0.91Vi2

Calculo de la varianza

∑X2 – (∑X) VF= n______

n-1

13675 – ( 367)2 10 Vf = _____________________ = 23 10-1 Calculo de la varianza individual

∑X2 – (∑X) VF= n______

n-1 180- ( 42)2 10 V1 = _________________ = 0,4 10-1 135- ( 35)2 10 V2 = _________________ = 1,38 10-1 96- ( 30)2 10 V3 = _________________ = 0,66 10-1 117- ( 33)2 10 V4 = _________________ = 0,9 10-1

72- ( 24)2 10 V5 = _________________ = 1,6 10-1 84- ( 26)2 10 V6 = _________________ = 1,82 10-1 101- ( 31)2 10 V7 = _________________ = 0,54 10-1 157- ( 39)2 10 V8 = _________________ = 0,54 10-1

149- ( 37)2 10 V9 = _________________ = 1,34 10-1 91- ( 25)2 10 V10 = _________________ = 3,16 10-1

59- ( 21)2 10 V11 = _________________ = 1,65 10-1 641- ( 24)2 10 V12 = _________________ = 0,91 10-1

Calculo del coeficiente de confiabilidad formula alfa de cronbach (n) * (1- ∑Vi) Rt = n-1 vt Rt = (12) * ( 1-15) = -0,66 12-1 23

Page 2: Confiabilidad

149- ( 37)2 10 V9 = _________________ = 1,34 10-1 91- ( 25)2 10 V10 = _________________ = 3,16 10-1

59- ( 21)2 10 V11 = _________________ = 1,65 10-1 641- ( 24)2 10 V12 = _________________ = 0,91 10-1

Calculo del coeficiente de confiabilidad formula alfa de cronbach (n) * (1- ∑Vi) Rt = n-1 vt Rt = (12) * ( 1-15) = -0,66 12-1 23

Page 3: Confiabilidad

Sujeto Item Item Item Item Item Item Item Item Item Item Item Item1 2 3 4 5 6 7 8 9 10 11 12 Total (X) ∑X2

1 1 1 1 1 1 0 0 1 1 1 1 1 10 1002 0 1 1 1 1 1 1 1 1 0 0 1 9 813 1 0 0 0 0 0 1 0 0 1 1 0 4 164 1 1 1 1 1 1 0 1 1 1 1 1 11 1215 0 0 0 0 1 1 1 1 1 1 1 1 8 646 1 1 1 1 0 0 0 0 0 0 0 1 5 257 1 1 0 0 1 1 1 1 1 1 1 1 10 1008 1 0 0 0 1 1 1 0 1 1 1 1 8 649 0 1 1 1 0 0 0 1 0 0 0 0 4 16

10 0 0 0 0 0 0 1 0 0 0 0 1 69 4Total de frecuencia 6 6 5 5 6 5 6 6 6 6 6 8 591P 0.6 0.6 0.5 0.5 0.6 0.5 0.6 0.6 0.6 0.6 0.6 0.8q 0.4 0.4 0.5 0.5 0.4 0.5 0.4 0.4 0.4 0.4 0.4 0.2p*q 0.24 0.24 0.25 0.25 0.24 0.25 0.24 0.24 0.24 0.24 0.24 0.16 2.83

Calculo de la varianza

∑(X – Ẋ)2

VF= ______ n-1

591 – ( 63)2 Vf = _____________________ = 21,56 10-1

Calculo del coeficiente de confiabilidad formula alfa de KR20 (n) * (1- ∑p.q) Rt = n-1 vt Rt = (12) * ( 2,83) = 0,94 12-1 21,56