Answer to Question #335786 in Statistics and Probability for Ben

Question

It is claimed that the mean annual salary of call center customer service representatives

is P188,584.00. A researcher randomly selected 50 call center customer service representatives. He computed the mean of their annual salaries and obtained a mean of P188,600.00.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le188584$

$H_a:\mu>188584$

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$

and the critical value for a right-tailed test is $z_c = 1.6449.$

The rejection region for this right-tailed test is $R = \{z: z > 1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{188600-188584}{39.5/\sqrt{50}}\approx2.86423$$

Since it is observed that $z = 2.86423 >1.6449= z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z>2.86423)=0.002090,$ and since $p = 0.002090<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is greater than $188584,$ at the $\alpha = 0.05$ significance level.