The professional organization for private colleges and universities professors reported that more than 17% of professors attended a national convention in the past year. To test this claim, a researcher surveyed 200 professors and found that 45 has attended a national convention in the past year. At 𝛼 = 0.05, test the claim that this figure is correct using p -value method.

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

$H_0:p\le0.17$

$H_1:p>0.17$

This corresponds to a right-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 right-tailed test is $z_c = 1.6449.$

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

The z-statistic is computed as follows:

$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{45}{200}-0.17}{\sqrt{\dfrac{0.17(1-0.17)}{200}}}\approx2.0707$

Using the P-value approach: The p-value is $p = P(Z>2.0707)=0.019193,$ and since $p = 0.019193 < 0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is greater than $0.17,$ at the $\alpha = 0.05$ significance level.