Answer to Question #253339 in Statistics and Probability for help

Question

The percentage of physicians who are women is 28%. In a survey of physicians employed by a large university health system, 60 of 142 randomly selected physicians were women. Is there sufficient evidence at the 0.01 level of significance to conclude that the proportion of women physicians at the university health system exceeds 28%?

a. State the hypothesis and identify the claim of the researcher. b. Find the critical value(s). 

c. Compute the test value. 

d. Make a decision on the null hypothesis. 

e. Make a decision on the claim of the researcher.

Answer 1

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

$H_0:p\leq 0.28$

$H_1:p>0.28$

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 $\alpha = 0.01,$ and the critical value for a right-tailed test is $z_c = 2.3263.$

The rejection region for this right-tailed test is $R = \{z: z > 2.3263\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{60}{142}-0.28}{\sqrt{\dfrac{0.28(1-0.28)}{142}}}=3.7829$$

Since it is observed that $z = 3.7829>2.3263= >z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z>3.7829) = 0.0000775,$ and since $p=0.0000775<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is greater than 0.28, at the $\alpha = 0.01$ significance level.