Question

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: Is it safe to assume that n ≤ 5 % of all college students in the local area? No Yes Is n ≥ 30 ? Yes No Test the claim: Determine the null and alternative hypotheses. Enter correct symbol and value. H 0 : μ = H a : μ Determine the test statistic. Round to two decimals. t = Find the p -value. Round to 4 decimals. p -value = Make a decision. Fail to reject the null hypothesis. Reject the null hypothesis. Write the conclusion. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area. There is not sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.

Answer 1

Explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 25235

For the alternative hypothesis,

µ > 25235

This is a right tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 100,

Degrees of freedom, df = n - 1 = 100 - 1 = 99

t = (x - µ)/(s/√n)

Where

x = sample mean = 27524

µ = population mean = 25235

s = samples standard deviation = 6000

t = (27524 - 25235)/(6000/√100) = 3.815

We would determine the p value using the t test calculator. It becomes

p = 0.000119

Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.