Question

it is claimed that the average monthly income of chemical engineers last year was P27,900.00 A random sample of 36 chemical engineers is selected and it is found out that the average monthly salary is P28,000.00 Using a 0.05 level of significance can it be concluded that there is an increase in the average monthly income of chemical engineers? Assume that the population standard deviation is P250.50

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le40$

$H_1:\mu>40$

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 $\alpha = 0.05,$ 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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{28000-27900}{250.50/\sqrt{36}}\approx2.3952$$

Since it is observed that $z=2.3952>1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>2.3952)= 0.008306,$ and since $p= 0.008306<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than 27900.00, at the $\alpha = 0.05$ significance level.