Answer to Question #348902 in Statistics and Probability for ...

Question

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

Answer 1

1. 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.