Question

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 card accounts, 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 = $1,128.

(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 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 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.

(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 or is completely above or is completely below zero. This suggests that the category with higher mean annual expenditure is groceries or dining out.

Answer 1

a)

H₀: μ_d = 0

H_a: μ_d ≠ 0

b)

population mean μ=	0
sample mean 'x̄=	870.000
sample size   n=	42.00
sample std deviation s=	1128.00
std error 'sx=s/√n=	174.0542
test stat t ='(x-μ)*√n/sx=	4.998

p value      =

0.0000

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

5% confidence interval estimate =518 to 1222

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