Answer to Question #348796 in Statistics and Probability for bakekang

Question

a researcher reports that the average salary of a teacher php 27000 a sample of 32 teachers has a mean of php 29500 at a=0.01 test the claim that the co teachers earn more than php 27000 a months the standard deviation of the population is php 2 250

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le27000$

$H_1:\mu>27000$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{29500-27000}{2250/\sqrt{32}}=6.2854$$

Since it is observed that $z=6.2854>2.3263=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>6.2854)= 0,$ and since $p= 0<0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than 27000, at the $\alpha = 0.01$ significance level.