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. \(\boxed{\text{{There is sufficient evidence to support the student's claim that the student-loan debt is higher in her area.}}}\)