In: Statistics and Probability
We are interested in conducting a study to determine whether a study intervention significantly improves college students’ test scores. The population’s GPA m = 2.64 and s = 0.42. We hope that intervention will improve girls’ body esteem scores by 0.25 points. We plan to recruit a sample of 36 students. We set our alpha level at .05 for two-tailed test. Determine the statistical power that this proposed study would have; report your answer as a percentage.
true mean , µ = 2.89
hypothesis mean, µo = 2.64
significance level, α = 0.05
sample size, n = 36
std dev, σ = 0.4200
δ= µ - µo = 0.25
std error of mean, σx = σ/√n =
0.4200 / √ 36 =
0.07000
Zα/2 = ± 1.960 (two tailed
test)
We will fail to reject the null (commit a Type II error) if we get
a Z statistic between
-1.960
and 1.960
these Z-critical value corresponds to some X critical values ( X
critical), such that
-1.960 ≤(x̄ - µo)/σx≤ 1.960
2.503 ≤ x̄ ≤ 2.777
now, type II error is ,ß = P
( 2.503 ≤ x̄ ≤
2.777 )
Z = (x̄-true
mean)/σx
Z1 = (
2.503 - 2.89 ) /
0.07000 = -5.531
Z2 = (
2.777 - 2.89 ) /
0.07000 = -1.611
so, P( -5.531 ≤ Z
≤ -1.611 ) = P ( Z ≤
-1.611 ) - P ( Z ≤ -5.531
)
= 0.054
- 0.000 = 0.0535 [
Excel function: =NORMSDIST(z) ]
power = 1 - ß = 0.9465 or
94.65%