Question

A credit reporting agency claims that the mean credit card debt in a town is greater than $3500. A random sample of the credit card debt of 20 residents in that town has a mean credit card debt of $3600 and a standard deviation of $391. At α=0.10, can the credit agency’s claim be supported, assuming this is a normally distributed data set?

Answer 1

To determine if the credit reporting agency's claim is supported, a hypothesis test is conducted. The null hypothesis is that the mean credit card debt is $3500 or less, while the alternative hypothesis is that it is greater than $3500. Using a t-test and a significance level of 0.10, the test statistic is calculated and compared to the critical value to make a conclusion. In this case, the null hypothesis is rejected, supporting the credit reporting agency's claim.

Step-by-step explanation:

To determine whether the credit reporting agency's claim that the mean credit card debt in a town is greater than $3500 can be supported, we need to conduct a hypothesis test. Since the sample size is greater than 30 and the population standard deviation is not known, we will use a t-test. Here are the steps:

1. Define the null hypothesis (H0) and the alternative hypothesis (HA). In this case, the null hypothesis is that the mean credit card debt is $3500 or less, and the alternative hypothesis is that the mean credit card debt is greater than $3500.
2. Calculate the test statistic, which is the t-score. Using the given sample mean, sample standard deviation, and sample size, we can calculate the t-score as (sample mean - hypothesized mean) / (sample standard deviation / √sample size).
3. Find the critical value from the t-distribution table for the given significance level (α) and degrees of freedom (sample size - 1).
4. Compare the test statistic to the critical value. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the mean credit card debt is greater than $3500. Otherwise, we fail to reject the null hypothesis.

Using α=0.10, the critical value for a one-tailed test with 19 degrees of freedom is approximately 1.327. Calculating the t-score using the given values, we get (3600 - 3500) / (391 / √20) ≈ 1.776. Since the test statistic (1.776) is greater than the critical value (1.327), we reject the null hypothesis. Therefore, we can support the credit reporting agency's claim that the mean credit card debt in the town is greater than $3500.