Question

In: Statistics and Probability

We are interested in conducting a study to determine whether a study intervention significantly improves college students’ test scores.

 

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.

Solutions

Expert Solution

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%   


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