Question

Open the Undergrad Debt data in the Excel file. Nationally, on average, a college student last year graduated with $27,200 in debt with a historical standard deviation of $5,000. The file gives sample date from Arkansas.
Does Arkansas students have higher debt than the national average at alpha equal to 10?

Answer 1

Answer:

Arkansas students have higher debt than the national average, $27,200.

Explanation:

The provided data is:

S = {24040, 19153, 26762, 31923, 31533, 34207, 14623, 24370, 31016}

In this case we need to test whether Arkansas students have higher debt than the national average at alpha equal to 0.10.

The hypothesis can be defined as follows:

H₀: Arkansas students does not have higher debt than the national average, i.e. μ ≤ $27,200.

Hₐ: Arkansas students have higher debt than the national average, i.e. μ > $27,200.

Compute the sample mean:

$\bar x=\frac{1}{n}\sum X=\frac{1}{10}\times 213537=21353.70$

As the population standard deviation is provided, we will use a z-test for single mean.

Compute the test statistic value as follows:

$z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{21353.70-27200}{5000/\sqrt{10}}=-3.70$

The test statistic value is -3.70.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the test as follows:

 $p-value=P(Z>-3.70)=P(Z$

*Use a z-table for the probability.

The p-value of the test is 0.00011.

p-value = 0.00011 < α = 0.10

The null hypothesis will be rejected.

Thus, it can be concluded that Arkansas students have higher debt than the national average, $27,200.