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 = $840, and the sample standard deviation was sd = $1,123.

A.Formulate the null abd alternative hypothesis 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 can conclude that the population mean 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?

Answer 1

A. Null and alternative hypothesis:

`H_0: \mu_d=0\\\\H_a:\mu_d\\eq 0`

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

`H_0: \mu_d=0\\\\H_a:\mu_d\\eq 0`

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

`s_M=(s)/(√(n))=(1123)/(√(42))=173.2827`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(840-0)/(173.2827)=(840)/(173.2827)=4.848`

The degrees of freedom for this sample size are:

`df=n-1=42-1=41`

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

`\text{P-value}=2\cdot P(t>4.848)=0.00002`

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

`MOE=t\cdot s_M=2.02 \cdot 173.283=350`

Then, the lower and upper bounds of the confidence interval are:

`LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190`

The 95% confidence interval for the mean difference is (490, 1190).