Answer to Question #305494 in Statistics and Probability for marky boy

Question

A survey of Davao Region finds the average commute time of employees on one way is 30

minutes. The Digos Chamber of Commerce feels that in their city is greater and want to

publicize this. They randomly select 28 commuters and find the average is 35 minutes with

a standard deviation of 6 minutes. At 𝛼 = 0.05, are they correct?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le30$

$H_1:\mu>30$

This corresponds to a right-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 $\alpha = 0.05, df=n-1=27$ degrees of freedom and the critical value for a right-tailed test is $t_c = 1.703288.$

The rejection region for this right-tailed test is $R = \{t: t > 1.703288\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{X}-\mu}{s/\sqrt{n}}=\dfrac{35-30}{6/\sqrt{28}}\approx4.409586$$

Since it is observed that $t = 4.409586 > 1.703288=t_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed $df=27$ degrees of freedom, $t=4.409586,$ is $p = 0.000074,$ and since $p= 0.000074<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is greater than 30, at the $\alpha = 0.05$ significance level.