Questions: A credit card company claims that the mean credit card debt for individuals is greater than 4,900. You want to test this claim. You find that a random sample of 37 cardholders has a mean credit card balance of 5,178 and a standard deviation of 625. At α=0.10, can you support the claim? Complete parts (a) through (d) below. Assume the population is normally distributed.
(c) Find the standardized test statistic L.
t= (Round to two decimal places as needed.)
(d) Decide whether to reject or fail to reject the null hypothesis.
A. Reject H₀ because the test statistic is not in the rejection region.
B. Fail to reject H₀ because the test statistic is not in the rejection region.
C. Fail to reject H₀ because the test statistic is in the rejection region.
D. Reject H₀ because the test statistic is in the rejection region.
Transcript text: A credit card company claims that the mean credit card debt for individuals is greater than $\$ 4,900$. You want to test this claim. You find that a random sample of 37 cardholders has a mean credit card balance of $\$ 5,178$ and a standard deviation of $\$ 625$. At $\alpha=0.10$, can you support the claim? Complete parts (a) through (d) below. Assume the population is normally distributed.
(c) Find the standardized test statistic L .
t= $\square$ (Round to two decimal places as needed.)
(d) Decide whether to reject or fail to reject the null hypothesis.
A. Reject $\mathrm{H}_{0}$ because the test statistic is not in the rejection region.
B. Fail to reject $\mathrm{H}_{0}$ because the test statistic is not in the rejection region.
C. Fail to reject $\mathrm{H}_{0}$ because the test statistic is in the rejection region.
D. Reject $\mathrm{H}_{0}$ because the test statistic is in the rejection region.
Solution
Solution Steps
Step 1: Calculate the Standard Error
The standard error \( SE \) is calculated using the formula:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{625}{\sqrt{37}} \approx 102.75
\]
Step 2: Calculate the Test Statistic
The test statistic \( Z_{test} \) is calculated using the formula:
\[
Z_{test} = \frac{\bar{x} - \mu_0}{SE} = \frac{5178 - 4900}{102.75} \approx 2.71
\]
Step 3: Determine the Critical Value
For a right-tailed test at \( \alpha = 0.10 \), the critical value is:
\[
Z_{critical} \approx 1.2816
\]
Step 4: Make a Decision
Since the test statistic \( Z_{test} = 2.71 \) is greater than the critical value \( Z_{critical} = 1.2816 \), we reject the null hypothesis \( H_0 \).
Final Answer
The conclusion is that we can support the claim that the mean credit card debt for individuals is greater than $4,900. Thus, the answer is:
\[
\boxed{D}
\]