Bank of America’s Consumer Spending 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 (US Airways Attache,
December 2003). 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, the
sample mean difference was d¯¯¯= $850, and the sample standard deviation was sd= $1123.
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.
b. Use a .05 level of significance. Can you conclude that the population means differ?
What is the p-value?
c. Which category, groceries or dining out, has a higher population mean annual credit
card charge? What is the point estimate of the difference between the population
means? What is the 95% confidence interval estimate of the difference between the
population means?

Given that Bank of America’s Consumer Spending 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 (US Airways Attache, December 2003).

Sample size n =42

Mean difference d = 850

Std dev s = 1125

STd error of sample = [tex]\frac{1125}{\sqrt{42} } =173.59[/tex]

a) [tex]H_0:d=0\\H_a: d\neq 0[/tex]

(Two tailed test at 5% significance level)

b) test statistic t = [tex]\frac{850-0}{173.59} =4.90[/tex]

df = 41

p value = 0.000015

Since p <0.05 our alpha, we reject null hypothesis

c) Point estimate of the difference is 0

95% confidence level would be

[tex]0±t critical * std error\\= ±2.021*173.59\\=(-350.83, 350.83)[/tex]

ogorwyne ogorwyne · Answer 2 · 2022-04-08T17:43:08+02:00

The null hypothesis for the question is mean sample difference = 0, while the alternate hypothesis is that the mean sample difference is not equal to zero.

a. H0: μ = 0

H1: μ ≠ 0

b. The test statistics calculation

xd = 850

ud = 0

sd = 1123

n = 42

[tex]\frac{850-0}{\frac{1123}{\sqrt{42} } }[/tex]

= 4.9053

Next is to find the degree of freedom = 42-1 = 41

critical value tα/2(41) = 2.021

2.021 < 4.905

Therefore the decision would be to reject the null hypothesis.

c. Next we calculate the confidence interval

850 ± 2.021 x 1132/√42

= 850 - 353.06, 850 + 353.06

= 496.93, 1203.06

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