The average student-loan debt is reported to be $\$ 25,235$. A student believes that the student-loan debt is higher in her area. She takes a random sample of 100 college students in her area and determines the mean student-loan debt is $\$ 27,524$ and the standard deviation is $\$ 6,000$. Is there sufficient evidence to support the student's claim at a 5\% significance level? Find the ptvalue. Round to 4 decimals

Step 1 ：The student believes that the average student-loan debt in her area is higher than the reported average. This is a one-tailed hypothesis test. The null hypothesis is that the average student-loan debt in her area is equal to the reported average, and the alternative hypothesis is that the average student-loan debt in her area is higher than the reported average.

Step 2 ：To test this hypothesis, we can use a t-test. The t-value is calculated as follows: \(t = \frac{{\text{{sample mean}} - \text{{population mean}}}}{{\text{{sample standard deviation}} / \sqrt{{\text{{sample size}}}}}}\)

Step 3 ：The p-value is the probability of observing a t-value as extreme as the one calculated, assuming the null hypothesis is true. If the p-value is less than the significance level (0.05 in this case), we reject the null hypothesis and conclude that there is sufficient evidence to support the student's claim.

Step 4 ：Given that the sample mean is $27,524, the population mean is $25,235, the sample standard deviation is $6,000, and the sample size is 100, we can calculate the t-value as \(t = \frac{{27524 - 25235}}{{6000 / \sqrt{100}}} = 3.815\)

Step 5 ：With a t-value of 3.815 and degrees of freedom of 99, we can find the p-value to be 0.0001185669089480701

Step 6 ：The p-value is less than the significance level (0.05), so there is sufficient evidence to reject the null hypothesis and support the student's claim that the average student-loan debt in her area is higher than the reported average.

Step 7 ：Final Answer: The p-value is \(\boxed{0.0001}\).