The average student-loan debt is reported to be $\$ 25,235$. A student believes that student-loan debt is higher in her area. She takes a random sample of 100 college her area and determines the mean student-loan debt is $\$ 27,524$ and the standard d $\$ 6,000$. Is there sufficient evidence to support the student's claim at a $5 \%$ significan
Determine the test statistic. Round to two decimal
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Step 1 ：The student is trying to determine if the mean student-loan debt in her area is significantly higher than the reported average. This is a one-sample, one-tailed t-test. The null hypothesis is that the mean student-loan debt in her area is equal to the reported average, and the alternative hypothesis is that the mean student-loan debt in her area is greater than the reported average.

Step 2 ：The test statistic for a one-sample t-test is calculated as: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] where: \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：In this case, \(\bar{x} = \$ 27,524\), \(\mu = \$ 25,235\), \(s = \$ 6,000\), and \(n = 100\).

Step 4 ：Substituting these values into the formula, we get: \[ t = \frac{27524 - 25235}{6000 / \sqrt{100}} \]

Step 5 ：Solving the above expression, we find that the test statistic is approximately 3.815. This value is used to determine the p-value of the test, which can then be compared to the significance level to decide whether to reject the null hypothesis.

Step 6 ：Final Answer: The test statistic is approximately \(\boxed{3.815}\).