Answer to Question #200209 in Statistics and Probability for Lesley

Question

USE ONE SAMPLE T TEST

One of the undersecretary of the Department of Labor and Employment (DOLE) claims that the average salary of civil engineer is P18,000. A sample of 19 civil engineers salary has a mean of 

Given:

STEP 1: STATE THE HYPOTHESIS AND IDENTIFY THE CLAIM.

STEP 2: THE LEVEL OF SIGNIFICANCE 

STEP 3: THE Z CRITICAL VALUE

STEP 4: COMPUTE THE ONE SAMPLE Z TEST VALUE 

STEP 5: DECISION RULE

STEP 6: CONCLUSION

Answer 1

Hypothesized Population Mean $\mu=18000$

Sample Standard Deviation $s=1230$

Sample Size $n=19$

Sample Mean $\bar{x}=17350$

Significance Level $\alpha=0.01$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu=18000$

$H_1: \mu\not=18000$

This corresponds to 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 $\alpha=0.01,$

$df=n-1=18$ degrees of fredom, and the critical value for two-tailed test is $t_c=2.87844.$ 

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

The $t$ - statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{17350-18000}{1230/\sqrt{19}}\approx-2.30348$$

Since it is observed that $|t|=2.30348<2.87844=t_c,$ it is then 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.

Using the P-value approach: The p-value for two-tailed, the significance level $\alpha=0.01, df=18, t=-2.30348,$ is $p=0.033425,$ and since $p=0.033425>0.01=\alpha,$ 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.