Question

30 pts.

4. Management of a major company has made the following assumptions regarding their consumer profile.

• Over 60% of all shoppers are women.
• Less than 40% of our customer get information on our products through online resources
• The average age of a purchasers is below 45 years.
• The average annual expenditure by the shoppers exceed $45.

The company hired market researchers to determine the validity of their consumers’ demographic assumptions. The researchers conducted survey and focus groups to evaluate the purchasing pattern of the company’s current consumers. In a survey of 800 consumers, 510 respondents were women and 300 indicated they primarily collect information on the company’s products online. The average age of the respondents was 43 years with standard deviation of 8 years and the average reported annual expenditure was $48 with standard deviation of $10.5

Determine which of the management’s demographic assumption is supported by the data. Use 0.05 level of significance. Provide conclusion in non-technical language.

Answer 1

• Over 60% of all shoppers are women.

Ho :   p =    0.6                  
H1 :   p >   0.6       (Right tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   510                  
Sample Size,   n =    800                  
                          
Sample Proportion ,    p̂ = x/n =    0.6375                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0173                  
Z Test Statistic = ( p̂-p)/SE = (   0.6375   -   0.6   ) /   0.0173   =   2.1651
                          

                          
p-Value   =   0.0152   [Excel function =NORMSDIST(-z)              
Decision:   p-value<α , reject null hypothesis                       
                      

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Less than 40% of our customer get information on our products through online resources

Ho :   p =    0.4                  
H1 :   p <   0.4       (Left tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   300                  
Sample Size,   n =    800                  
                          
Sample Proportion ,    p̂ = x/n =    0.3750                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0173                  
Z Test Statistic = ( p̂-p)/SE = (   0.3750   -   0.4   ) /   0.0173   =   -1.4434
                          
critical z value =        -1.645   [excel function =NORMSINV(α)]              
                          
p-Value   =   0.0745   [excel function =NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis                       

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The average age of a purchasers is below 45 years.

Ho :   µ =   45                  
Ha :   µ <   45       (Left tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s =    8.0000                  
Sample Size ,   n =    800                  
Sample Mean,    x̅ =   43.0000                  
                          
degree of freedom=   DF=n-1=   799                  
                          
Standard Error , SE = s/√n =   8.0000   / √    800   =   0.2828      
t-test statistic= (x̅ - µ )/SE = (   43.000   -   45   ) /    0.2828   =   -7.07
                             
                          
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       

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• The average annual expenditure by the shoppers exceed $45.

Ho :   µ =   45                  
Ha :   µ >   45       (Right tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s =    10.5000                  
Sample Size ,   n =    800                  
Sample Mean,    x̅ =   48.0000                  
                          
degree of freedom=   DF=n-1=   799                  
                          
Standard Error , SE = s/√n =   10.5000   / √    800   =   0.3712      
t-test statistic= (x̅ - µ )/SE = (   48.000   -   45   ) /    0.3712   =   8.08
                             
                          
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       

There is sufficient evidence that Over 60% of all shoppers are women.

There is not sufficient evidence that Less than 40% of our customer get information on our products through online resources

There is sufficient evidence that The average age of a purchasers is below 45 years.

There is sufficient evidence thatThe average annual expenditure by the shoppers exceed $45.

Please revert back in case of any doubt.

Please upvote. Thanks in advance.