Question #348902

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


Solution:

step 1: null and alternative hypothesis

H0:__________

H1:__________


step 2: Significance level a =__________


step 3: test statistic


step 4: decision rule


step 5: decision


1
Expert's answer
2022-06-08T14:52:11-0400

1. The following null and alternative hypotheses need to be tested:

H0:μ=18000H_0:\mu=18000

H1:μ18000H_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 α=0.01,\alpha = 0.01, df=n1=18df=n-1=18 and the critical value for a two-tailed test is tc=2.87844.t_c =2.87844.

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

The t-statistic is computed as follows:


t=xˉμs/n=17350180001230/19=2.3035t=\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=tc,|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=18df=18 degrees of freedom, t=2.3035t=-2.3035 is p=0.033391,p=0.033391, and since p=0.033391>0.01=α,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 μ\mu

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


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