Question

The Department of Education would like to check if the average debt load of graduating students with a bachelor’s degree is different from $17,000. A random sample of 34 students had an average debt load of $18,200. It is believed that the population standard deviation for student debt load is $4,200. The significance level α is set to 0.02 for the hypothesis test.

State your conclusion for the test.

A. $LaTeX: -z_{\alpha/2}<z_{\bar x}<z_{\alpha/2} \; \Rightarrow$ we reject H_o. Therefore, there is enough evidence to conclude that the average debt load of graduating students with a bachelor’s degree is different from $17,000.

B. $LaTeX: z_{\bar x}>-z_{\alpha/2} \; \Rightarrow$ we reject H_o. Therefore, there is enough evidence to conclude that the average debt load of graduating students with a bachelor’s degree is different from $17,000.

C. $LaTeX: z_{\bar x}<z_{\alpha/2} \; \Rightarrow$ we do not reject H_o. Therefore, there is not enough evidence to conclude that the average debt load of graduating students with a bachelor’s degree is different from $17,000.

D. $LaTeX: -z_{\alpha/2}<z_{\bar x}<z_{\alpha/2} \; \Rightarrow$ we do not reject H_o. Therefore, there is not enough evidence to conclude that the average debt load of graduating students with a bachelor’s degree is different from $17,000.

Answer 1

Here population standard deviation is known so we will use z statistics

$z=\frac{\overline{x}-\mu}{\frac{\sigma }{\sqrt{n}}}=\frac{18200-17000}{\frac{4200}{\sqrt{34}}}=1.67$

The z-critical values for a two-tailed test, for a significance level of α=0.02

zc​=−2.33 and zc​=2.33

Graphically

$normaldistributiongrapher.php?mean=0&sig$

So answer here is

D. $LaTeX: -z_{\alpha/2}<z_{\bar x}<z_{\alpha/2} \; \Rightarrow$ we do not reject H_o. Therefore, there is not enough evidence to conclude that the average debt load of graduating students with a bachelor’s degree is different from $17,000.