Question

A magazine uses a survey of readers to obtain customer satisfaction ratings for the nation's largest retailers. Each survey respondent is asked to rate a specified retailer in terms of six factors: quality of products, selection, value, checkout efficiency, service, and store layout. An overall satisfaction score summarizes the rating for each respondent with 100 meaning the respondent is completely satisfied in terms of all six factors. Sample data representative of independent samples of Retailer A and Retailer B customers are shown below.

Retailer A	Retailer B
n1 = 25	n2 = 30
x1 = 80	x2 = 71

Formulate the null and alternative hypotheses to test whether there is a difference between the population mean customer satisfaction scores for the two retailers. (Let μ₁ = population mean satisfaction score for Retailer A customers and μ₂ = population mean satisfaction score for Retailer B customers.)

(b) Assume that experience with the satisfaction rating scale of the magazine indicates that a population standard deviation of 11 is a reasonable assumption for both retailers. Conduct the hypothesis test.

Find the value of the test statistic. (Round your answer to two decimal places.) =

Find the p-value. (Round your answer to four decimal places.)

p-value = .

(c)

Provide a 95% confidence interval for the difference between the population mean customer satisfaction scores for the two retailers. (Round your answers to two decimal places.)

_______ to ______

Which retailer, if either, appears to have the greater customer satisfaction?

Answer 1

SOLUTION

a)

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 ╪   0
      

b)

Level of Significance ,    α =    0.05          
                  
sample #1   ------->   A          
mean of sample 1,    x̅1=   80          
population std dev of sample 1,   σ1 =    11          
size of sample 1,    n1=   25          
                  
sample #2   --------->   B          
mean of sample 2,    x̅2=   72          
population std dev of sample 2,   σ2 =    11          
size of sample 2,    n2=   30          
                  
difference in sample means = x̅1 - x̅2 =    80   -   72   =   8
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    2.9788          
                  
Z-statistic = ((x̅1 - x̅2)-µd)/SE =    8   /   2.9788   =   2.69   
p-value =        0.0072   [excel formula =2*NORMSDIST(z)]      
Decison:   p-value <α , Reject null hypothesis  

      Reject H0. There is sufficient evidence to conclude that the population mean satisfaction scores differ for the two retailers.

c)

Level of Significance ,    α =    0.05          
z-critical value =    Z α/2 =    1.9600   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    2.9788          
margin of error, E = Z*SE =    1.9600   *   2.9788   =   5.8384
                  
difference of means = x̅1 - x̅2 =    80   -   72   =   8.000
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    8.000   -   5.8384   =   2.16
Interval Upper Limit= (x̅1 - x̅2) + E =    8.000   +   5.8384   =   13.84

The 95% confidence interval  is completely above zero. This suggests that the Retailer A has a higher population mean customer satisfaction score than Retailer B.