Question

A marketing consultant was hired to visit a random sample of five sporting goods stores across the state of California. Each store was part of a large franchise of sporting goods stores. The consultant taught the managers of each store better ways to advertise and display their goods. The net sales for 1 month before and 1 month after the consultant’s visit were recorded as follows for each store (in thousands of dollars):

Before visit: 57.1 94.6 49.2 77.4 43.2
After visit: 63.5 101.8 57.8 81.2 41.9

use a 1% level of significance

a. State the null and alternative hypotheses ?0: ?1:

b. What calculator test will you use? List the requirements that must be met to use this test, and indicate whether the conditions are met in this problem.

c. Run the calculator test and obtain the P-value.

d. Based on your P-value, will you reject or fail to reject the null hypothesis?

e. Interpret your conclusion from part d in the context of this problem

Answer 1

a)

µd = µbefore - µ after

Ho :   µd=   0
Ha :   µd <   0

b)

paired t test will be used.

we assume that d has normal distribution

c)

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
57.1	63.5	-6.400	2.1316
94.6	101.8	-7.200	5.1076
49.2	57.8	-8.600	13.3956
77.4	81.2	-3.800	1.2996
43.2	41.9	1.300	38.9376

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	321.5	346.2	-24.7	60.9

mean of difference ,    D̅ =ΣDi / n =   -4.940                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    3.9010                  
                          
std error , SE = Sd / √n =    3.9010   / √   5   =   1.7446      
                          
t-statistic = (D̅ - µd)/SE = (   -4.94   -   0   ) /    1.7446   =   -2.83

d)

Degree of freedom, DF=   n - 1 =    4      
  
p-value =        0.0236 [excel function: =t.dist(t-stat,df) ]  
Conclusion:     p-value>α , Do not reject null hypothesis         

e)

there is not enough evidence to conclude that  average net sales improved at α=0.01