Answer to Question #348958 in Statistics and Probability for Cecel

Question

The leader of the association of jeepney drivers claims that the average daily take home pay of all jeepney drivers in a particular city is php 450. A random sample of 100 jeepney drivers in the city was interviewed and the average daily take home pay of these drivers is found to be php 500. Use a 0.05 significance level to find out if the average daily take home pay of all jeepney drivers in the city is greater than php 450. Assume that the population variance is php 8644.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le450$

$H_1:\mu>450$

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:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{500-450}{\sqrt{8644}/\sqrt{100}}=5.3779$$

Since it is observed that $z=5.3779>1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>5.3779)=0,$ 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 greater than 450, at the $\alpha = 0.05$ significance level.