Answer to Question #344897 in Statistics and Probability for Jonathan

Question

A job placement director claims that the average starting salary for nurses is P12,000. A sample of 10 nurses has a mean of P11,750 and a standard deviation of 200. Is there enough evidence to reject the director’s claim at alpha = 0.05

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=12000$

$H_1:\mu\not=12000$

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 $\alpha = 0.05,$ $df=n-1=9$ and the critical value for a two-tailed test is $t_c =2.262156.$

The rejection region for this two-tailed test is $R = \{t:|t|>2.262156\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{11750-12000}{200/\sqrt{10}}=-3.9528$$

Since it is observed that $|t|=3.9528>2.262156=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=9$ degrees of freedom, $t=-3.9528$ is $p= 0.003341,$ and since $p= 0.003341<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 12000, at the $\alpha = 0.05$ significance level.