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?

Preliminary:
a. Is it safe to assume that $n \leq 5 \%$ of all college students in the local area?
No
Yes
of
b. Is $n \geq 30$ ?
No
Yes
$\infty$
Test the claim:
a. Determine the null and alternative hypotheses. Enter correct symbol and value.
\[
\begin{array}{l}
H_{0}: \mu= \\
H_{a}: \mu \text { ? } 0
\end{array}
\]
b. Determine the test statistic. Round to two decimals.
\[
t=
\]
c. Find the $p$-value. Round to 4 decimals.
\[
p \text {-value }=
\]

Step 1 ：The null hypothesis (H0) is \(\mu = \$25,235\), which is the reported average student-loan debt. The alternative hypothesis (Ha) is \(\mu > \$25,235\), as the student believes that the student-loan debt is higher in her area.

Step 2 ：The test statistic is calculated using the formula: \(t = \frac{{\text{{sample mean}} - \text{{population mean}}}}{{\text{{standard deviation}} / \sqrt{n}}}\)

Step 3 ：Substituting the given values into the formula, we get \(t = \frac{{\$27,524 - \$25,235}}{{\$6,000 / \sqrt{100}}}\)

Step 4 ：Solving the above expression, we get \(t = \frac{{\$2,289}}{{\$600}}\)

Step 5 ：The calculated t-value is \(t = 3.82\) (rounded to two decimal places)

Step 6 ：The p-value is the probability that the observed data would occur if the null hypothesis were true. Since we are dealing with a one-tailed test, and the t-value is 3.82, we can say that the p-value will be less than 0.0001.

Step 7 ：Since the p-value is less than the significance level of 0.05, we reject the null hypothesis. \(\boxed{\text{{There is sufficient evidence to support the student's claim that the student-loan debt is higher in her area.}}}\)