Answer to Question #346252 in Statistics and Probability for hanna

Question #346252

. One of the undersecretaries of the Department of Labor and Employment (DOLE) claims that the average salary of civil engineers is 18,000php. A sample of 19 civil engineers' salaries has a mean of 17,350php and a standard deviation of 1,230php. Is there enough evidence to reject the undersecretary's claim at a 0.01 level of significance?


1
Expert's answer
2022-05-30T23:32:05-0400

The following null and alternative hypotheses need to be tested:

"H_0:\\mu=18000"

"H_1:\\mu\\not=18000"

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is "\\alpha = 0.01," "df=n-1=18" and the critical value for a two-tailed test is "t_c =2.87844."

The rejection region for this two-tailed test is "R = \\{t:|t|>2.87844\\}."


The t-statistic is computed as follows:



"t=\\dfrac{\\bar{x}-\\mu}{s\/\\sqrt{n}}=\\dfrac{17350-18000}{1230\/\\sqrt{19}}=-2.3035"


Since it is observed that "|t|=2.3035<2.87844=t_c," it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, "df=18" degrees of freedom, "t=-2.3035" is "p=0.033391," and since "p=0.033391>0.01=\\alpha," it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean "\\mu"

is different than 18000, at the "\\alpha = 0.01" significance level.


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