Question

the average credit card debt for a recent year was $9205. Five years earlier it was $6618. Assume sample sizes of 35 were used and the population standard deviation for both was $1928. Construct a 90% confidence interval for the true difference of means, is there sufficeint evidence to support the claim that the average has increased, why or why not?

Answer 1

The provided sample means are shown below:

X1 = 6618 X2 = 9205

Also, the provided population standard deviations are:

01 = 1928 02 = 1928

and the sample sizes are

ni = 35 and n2 = 35

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ1​=μ2​

Ha:μ1​<μ2​

This corresponds to a left-tailed test, for which a z-test for two population means, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.10, and the critical value for a left-tailed test is

z = -1.28

The rejection region for this left-tailed test is R = { z : z < − 1.28}

(3) Test Statistics

The z-statistic is computed as follows:

X1 - X2 ani+ož/n 6618 - 9205 = = -5.613 19282/35 + 19282/35

(4) The decision about the null hypothesis

Since it is observed that Zc = −5.613 < z∗=−1.28, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0001, and since p = 0.00001 <0.10, it is concluded that the null hypothesis is rejected.

There is sufficient evidence to support the claim that the average has increased. As the p-value is less than alpha, we reject the null hypothesis and conclude that average has increased over 5 years.