Question

A new program of imagery training is used to improve the performance of basketball players shooting free-throw shots. The first group did an hour imagery practice, and then shot 30 free throw basket shots with the number of shots made recorded. A second group received no special practice, and also shot 30 free throw basket shots. The data is below. Did the imagery training make a difference? Set alpha = .05.
Group 1: 15, 17, 20, 25 26, 27
Group 2: 5, 6, 10, 15, 18, 20

1. You must use all five steps in hypothesis testing:
A. Restate the question as a research hypothesis and a null hypothesis about the populations.
B. Determine the characteristics of the comparison distribution.
C. Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected.
D. Determine your sample’s score on the comparison distribution.
E. Decide whether to reject the null hypothesis.
2. Solve for and evaluate the effect size of this study using Cohen's D.
3. Create a 90% confidence interval for this problem.

Answer 1

1)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ----> With Imagery training
mean of sample 1,    x̅1=   21.67                  
standard deviation of sample 1,   s1 =    5.05                  
size of sample 1,    n1=   6                  
                          
Sample #2   ---->   Without Imagery training   
mean of sample 2,    x̅2=   12.33                  
standard deviation of sample 2,   s2 =    6.28                  
size of sample 2,    n2=   6                  
                          
difference in sample means =    x̅1-x̅2 =    21.6667   -   12.3   =   9.33  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.6980                  
std error , SE =    Sp*√(1/n1+1/n2) =    3.2897                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   9.3333   -   0   ) /    3.29   =   2.837
                          
Degree of freedom, DF=   n1+n2-2 =    10                  
t-critical value , t* =        1.8125   (excel function: =t.inv(α,df)              
Decision:   | t-stat | > | critical value |, so, Reject Ho                      
p-value =        0.008818   [excel function: =T.DIST.RT(t stat,df) ]              
Conclusion:     p-value <α , Reject null hypothesis         

There is sufficient evidence to prove that A new program of imagery training improve the performance of basketball players shooting free-throw shots.           

2)

effect size,      
cohen's d =    |( x̅1-x̅2 )/Sp | =    1.638

3)

Degree of freedom, DF=   n1+n2-2 =    10              
t-critical value =    t α/2 =    1.8125   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    5.6980              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    3.2897              
margin of error, E = t*SE =    1.8125   *   3.2897   =   5.9625  
                      
difference of means =    x̅1-x̅2 =    21.6667   -   12.333   =   9.3333
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    9.3333   -   5.9625   =   3.3709
Interval Upper Limit=   (x̅1-x̅2) + E =    9.3333   +   5.9625   =   15.2958

Please revert back in case of any doubt.

Please upvote. Thanks in advance.