In: Statistics and Probability
Question: For each of the following scenarios, conduct the complete hypothesis testing process as appropriate give that scenario.
A regional manager implements a policy change for stores in his region to begin greeting customers whenever they are standing within a 3 metre distance in the store. He compares mean daily purchase numbers over one month from pre-change to post-change for each store to see if the change makes a difference in sales.
Mean daily purchases for Store 1 through Store 10 before policy: 82, 103, 91, 91, 83, 76, 90, 114, 88, 92
Mean daily purchases for Store 1 through Store 10 after policy: 102, 83, 113, 87, 94, 78, 91, 117, 101, 107
Ho :   µd=   0
Ha :   µd ╪   0
| Sample #1 | Sample #2 | difference , Di =sample1-sample2 | (Di - Dbar)² | 
| 82 | 102 | -20.000 | 187.69 | 
| 103 | 83 | 20.000 | 691.69 | 
| 91 | 113 | -22.000 | 246.49 | 
| 91 | 87 | 4.000 | 106.09 | 
| 83 | 94 | -11.000 | 22.09 | 
| 76 | 78 | -2.000 | 18.49 | 
| 90 | 91 | -1.000 | 28.09 | 
| 114 | 117 | -3.000 | 10.89 | 
| 88 | 101 | -13.000 | 44.89 | 
| 92 | 107 | -15.000 | 75.69 | 
| sample 1 | sample 2 | Di | (Di - Dbar)² | |
| sum = | 910 | 973 | -63 | 1432.1 | 
mean of difference ,    D̅ =ΣDi / n =  
-6.300          
       
          
           
   
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =
   12.6144      
           
          
           
   
std error , SE = Sd / √n =    12.6144   /
√   10   =   3.9890  
   
          
           
   
t-statistic = (D̅ - µd)/SE = (   -6.3  
-   0   ) /    3.9890  
=   -1.579
          
           
   
Degree of freedom, DF=   n - 1 =   
9          
       
p-value =        0.14872  
[excel function: =t.dist.2t(t-stat,df) ]   
           
Conclusion:     p-value>α=0.05 , Do not reject
null hypothesis     
There is not enough evidence to conclude that the change makes a difference in sales at α=0.05