Answer to Question #278684 in Statistics and Probability for zing

Question

The average hospitality girls working in a certain area is 19 years old. A civic oriented group made a recent survey on the age of the hospitality girls working in the same locality. A random sample of 20 respondents showed an average of 15.4 years old with a standard deviation of 2.14 years old. Using a 0.05 level of significance, has the hospitality girls gone lower?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq 19$

$H_1:\mu<19$

This corresponds to a left-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 $\alpha = 0.05,$ $df=n-1=20-1=19$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -1.729133.$

The rejection region for this left-tailed test is $R = \{t: t < -1.729133\}.$

The t-statistic is computed as follows:

$$t=\dfrac{X-\mu}{s/\sqrt{n}}=\dfrac{157.4-19}{2.14/\sqrt{20}}\approx-7.52322$$

Since it is observed that $t = -7.52322<-1.729133=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for left-tailed, $df=19$ degrees of fredom, $t=-7.52322$ is $p \approx 0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$  is less than $19,$ at the $\alpha = 0.05$ significance level.