Question

A university claims that the percentage of its graduates who fail to get a job after one year of graduation is less than 10%. A sample of 200 graduates from the university was randomly selected and found that 16 graduates fail to get a job after one year of graduation. Is there enough evidence to support the university's claim? Test at 5% significance level. 

Answer 1

Sample Proportion $\hat{p}=\dfrac{16}{200}=0.08$

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p\geq0.1$

$H_1:p<0.1$

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a left-tailed test is $z_c = -1.6449.$

The rejection region for this two-tailed test is $R = \{z: z<-1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.08-0.1}{\sqrt{\dfrac{0.1(1-0.1)}{200}}}$$

$$\approx-0.9428$$

Since it is observed that $z=-0.9428> -1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=P(Z<-0.9428)=0.1728916,$ and since $p=0.1728916>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is less than $0.1,$ at the $\alpha = 0.05$ significance level.

Therefore, there is not enough evidence to claim that the percentage of the graduates who fail to get a job after one year of graduation is less than 10%, at the $\alpha = 0.05$ significance level.