Answer to Question #259937 in Statistics and Probability for Jenelle

Question

In 2015, the average duration of long-distance telephone calls from a certain town was 9.4 minutes. A telephone company wants to perform a test to determine whether this average duration of long-distance calls has changed. Twenty calls, originating from the town, was randomly selected and the mean duration was 10.2 minutes with standard deviation 4.8 minutes. a) Using a 1% level of significance, Complete the following: i. Give the null and alternative hypothesis of this test. [2] ii. Determine the critical value(s) of this test. [2] iii. Compute the value of the test statistic. [2] iv. State the decision rule. [1] v. Give your decision based on the available sample evidence. [1] vi. Hence, state your conclusion

Answer 1

A telephone company wants to perform a test to determine whether this average duration of long-distance calls has changed so null and alternate hypotheses can be written as

Null hypothesis H0: $\mu$ = 9.4

Alternate hypothesis Ha: $\mu \not=$  9.4

This is a two-tailed hypothesis test

Number of sample (n) = 20

Sample mean $(\bar{X})$ = 10.2

Sample standard deviation (S) = 4.8

Here we will use one sample mean t distribution as population standard deviation is not known so test statistic value can be calculated as

Test statistic (t) = $\frac{(\bar{X} - \mu )}{(S/\sqrt n)} = \frac{(10.2 - 9.4)}{(4.8/ \sqrt 20) }= 0.745$

Level of significance (α) = 0.01

Degree of freedom (df) = n-1 = 20-1 = 19

from t table we found t critical value = +/-2.861

Decision rule: Reject null hypothesis H0, if test statistic value is less than -2.861 or greater than 2.861 else do not reject the null hypothesis.

Here we can see that the test statistic value is between -2.861 and 2.861 so we are failed to reject the null hypothesis.

So we can not conclude that this average duration of long-distance calls has changed.