Answer to Question #276796 in Statistics and Probability for ros

Question

An ambulance service claims that it takes on the average 8.9 minutes to reach its destination in

emergency calls. To check this claim, the agency which licenses ambulance services has them

timed on 50 emergency calls, getting a mean of 9.3 minutes

with a standard deviation of 1.6

minutes. What can they conclude at the level of significance of 0.05?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=8.9$

$H_1:\mu\not=8.9$

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 $\alpha = 0.05,$ $df=n-1=50-1=49$ degrees of freedom, and the critical value for a two-tailed test is $t_c = 2.009575.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.009575\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{9.3-8.9}{1.6/\sqrt{50}}\approx1.767767$$

Since it is observed that $|t| = 1.767767 \le 2.009575=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for two-tailed, $df=49$ degrees of freedom, $t= 1.767767$ is significance level $p=0.083326,$ and since $p = 0.083326 \ge 0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$  is different than 8.9, at the $\alpha = 0.05$ significance level.