Answer to Question #342338 in Statistics and Probability for Fox

Question #342338

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


1
Expert's answer
2022-05-19T09:47:45-0400

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.



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