A coach is trying to determine whether the shooting ability

A coach is trying to determine whether the shooting ability of the kids on his basketball team has improved after a month of training with him. He had each kid attempt 15 baskets in a row on the first day of practice and recorded how many baskets each of the 9 kids made. He had them attempt 15 more baskets now one month in and interested in knowing if they have increased the number of baskets. Below are the pre and post baskets for each of the 9 kids. and the gained basket score of each kid (G=Post-Pre).

Solution

a)

Getting the mean and standard deviation of the \"G\" column,

X = sample mean =    3.333333333          

s = standard deviation =    2.449489743   [ANSWERS]

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b)  

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   0  
Ha:    u   =/   0  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    8          
tcrit =    +/-   2.306004135      
              
Getting the test statistic, as              
              
X = sample mean =    3.333333333          
uo = hypothesized mean =    0          
n = sample size =    9          
s = standard deviation =    2.449489743          
              
Thus, t = (X - uo) * sqrt(n) / s =    4.082482905          
              
Also, the p value is              
              
p =    0.003522021          
              
As t > 2.306, and P < 0.05, we   REJECT THE NULL HYPOTHESIS.          
              
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c)

Thus, there is significant evidence at 0.05 level that the gain in scores is significantly different from 0 at 0.05 level. [CONCLUSION]              
For the   0.95   confidence level, then      
df =    8          
alpha/2 = (1 - confidence level)/2 =    0.025          
t(alpha/2) =    2.306004135          
              
lower bound = X - z(alpha/2) * s / sqrt(n) =    1.450488841          
upper bound = X + z(alpha/2) * s / sqrt(n) =    5.216177825          
              
Thus, the confidence interval is              
              
(   1.450488841   ,   5.216177825   )