Question

1. Utilizing the sample size chart, what would be the minimum sample size for the following situations?

a. one sample test              ES = .8,*a = .05/2, 1- B = .90

b. two sample test (independent)   ES = .8, *a = .01, 1- B = .80

c. two sample test (independent)    ES = .2, *a = .05/2, 1- B = .95

d. one sample test      ES = .5, *a = .01, 1- B = .80

e. two sample test       ES = .8, *a = .05/2, 1- B = .95

f. one sample test        ES = .5, *a = .01, 1- B = .90

Answer 1

Solution:

From given data,

a. one sample test              ES = .8,*a = .05/2, 1- B = .90

n= Z(1-0.05/2)+Z(0.9)/(0.8)^2

= ((1.96+1.28)/0.8)*(1.96+1.28)/0.8))

n =16.4 (next whole number 17)

b. two sample test (independent)   ES = .8, *a = .01, 1- B = .80

n= Z(1-0.01/2)+Z(0.8)/(0.8)^2

= ((2.58+0.84)/0.8)^2

n =18.25 (next whole number 19)

c. two sample test (independent)    ES = .2, *a = .05/2, 1- B = .95

n= Z(1-0.05/2)+Z(0.95)/(0.2)^2

= ((1.96+1.64)/0.2)^2

n = 324

d. one sample test      ES = .5, *a = .01, 1- B = .80

n= Z(1-0.01/2)+Z(0.8)/(0.5)^2

= ((2.58+0.84)/0.5)^2

n =46.78 (next whole number 47)

e. two sample test       ES = .8, *a = .05/2, 1- B = .95

n= Z(1-(0.05/2)/2)+Z(0.95)/(0.8)^2

= Z(1-0.025/2)+Z(0.95)/(0.8)^2

  = ((2.22+1.64)/0.5)^2

n =59.59 (next whole number 60)

f. one sample test        ES = .5, *a = .01, 1- B = .90

n= Z(1-0.01/2)+Z(0.90)/(0.5)^2

= ((2.58+1.28)/0.5)^2

n =59.59 (next whole number 60)