Answer to Question #318793 in Statistics and Probability for vii

Question

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.

Answer 1

P=17%/100=0.17

$H_0: P\le0.17$

$H_1: P>0.17$

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 α=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:

$z=\frac{p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}=\frac{45/200-0.17}{\sqrt{\frac{0.17(1-0.17)}{200}}}=2.0707$

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

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