Answer to Question #350163 in Statistics and Probability for qqqq

Question

A researcher reports that the average salary of College Deans is more than Php 63,000. A sample of 35 College Deans has a mean salary of Php 65,700. At a = 0.01, test the claim that the CollegeDeans earn more than Php 63, 000 a month. The standard deviation of the population is Php 5,250.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 63000$

$H_1:\mu>63000$

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation 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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{65700-63000}{5250/\sqrt{35}}=3.0426$$

Since it is observed that $|z|=3.0426>2.3263=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(z>3.0426)=0.001173,$ and since $p=0.001173<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 63000, at the $\alpha = 0.01$ significance level.