Answer to Question #337691 in Statistics and Probability for mikay

Question

 A report in LTO stated that the average age of taxis in the Philippines is 12 years. An operations manager of a large taxi company selects a sample of 40 taxis and finds the average age of the taxis is 11.2 years. The standard deviation of the population is 2.3 years. At 0.05 level of significance, can it be concluded that the average age of the taxis in his company is less than the national average?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\ge12$

$H_a:\mu<12$

This corresponds to a left-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 left-tailed test is $z_c = -1.6449.$

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

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{11.2-12}{1.2/\sqrt{40}}=-2.1998$$

Since it is observed that $z=-2.1998<-1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z<-2.1998)= 0.013911,$and since $p= 0.013911<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

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

is less than 12, at the $\alpha = 0.05$ significance level.