Question

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A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment. Using data from a sample of 42 credit cardaccounts, assume that each account was used to identify the annual credit card charges for groceries (population 1) and the annual credit card charges for dining out (population 2). Using the difference data, with population 1 − population 2, the sample mean difference was

d = $870,

and the sample standard deviation was

s_d = $1,121.

(a)

Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

H₀: μ_d ≠ 0

H_a: μ_d = 0

H₀: μ_d < 0

H_a: μ_d = 0

    

H₀: μ_d ≤ 0

H_a: μ_d > 0

H₀: μ_d ≥ 0

H_a: μ_d < 0

H₀: μ_d = 0

H_a: μ_d ≠ 0

(b)

Calculate the test statistic. (Round your answer to three decimal places.)

What is the p-value? (Round your answer to four decimal places.)

Can you conclude that the population means differ? Use a 0.05 level of significance.

The p ≤ 0.05, therefore we can conclude that there is a difference between the annual population mean expenditures for groceries and for dining out.The p > 0.05, therefore we can conclude that there is a difference between the annual population mean expenditures for groceries and for dining out.    The p > 0.05, therefore we cannot conclude that there is a difference between the annual population mean expenditures for groceries and for dining out.The p ≤ 0.05, therefore we cannot conclude that there is a difference between the annual population mean expenditures for groceries and for dining out.

(c)

What is the point estimate (in dollars) of the difference between the population means?

$

What is the 95% confidence interval estimate (in dollars) of the difference between the population means? (Round your answers to the nearest dollar.)

$  to $

Which category, groceries or dining out, has a higher population mean annual credit card charge?

The 95% confidence interval  ---Select--- contains is completely below is completely above zero. This suggests that the category with higher mean annual expenditure is  ---Select--- groceries dining out .

Answer 1

a)

Ho :   µd=   0
Ha :   µd ╪   0

b)

mean of difference ,    D̅ =    870.000                  
                          
std dev of difference , Sd =        1121.0000                  
                          
std error , SE = Sd / √n =    1121.0000   / √   42   =   172.9741      
                          
t-statistic = (D̅ - µd)/SE = (   870   -   0   ) /    172.9741   =   5.030

                          
Degree of freedom, DF=   n - 1 =    41                  
  
p-value =        0.0000 [excel function: =t.dist.2t(t-stat,df) ]               

The p ≤ 0.05, therefore we can conclude that there is a difference between the annual population mean expenditures for groceries and for dining out

c)

point estimate (in dollars) of the difference between the population means=870

sample size ,    n =    42          
Degree of freedom, DF=   n - 1 =    41   and α =    0.05  
t-critical value =    t α/2,df =    2.0195   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =        1121.0000          
                  
std error , SE = Sd / √n =    1121.0000   / √   42   =   172.9741
margin of error, E = t*SE =    2.0195   *   172.9741   =   349.3282
                  
mean of difference ,    D̅ =   870.000          
confidence interval is                   
Interval Lower Limit= D̅ - E =   870.000   -   349.3282   =   520.6718
Interval Upper Limit= D̅ + E =   870.000   +   349.3282   =   1219.3282
                  
so, confidence interval is ( $521 < µd < $1219 )  

The 95% confidence interval contains is ( completely above zero). This suggests that the category with higher mean annual expenditure is (groceries ) .