Question

Question 2: Testing a variance A call center is concerned that call durations for a customer service representative are too erratic: high variations is call durations can lead to customer dissatisfaction who have to wait longer for a resolution. The company has collected data from 24 randomly selected phone calls from that specific customer representative and found that s = 4.15 minutes. Answer the following questions. (a) Is there enough evidence to suggest that o = 4 minutes? Use a = 0.05. (b) Assume that we do not care about the standard deviation being lower than 4 minutes; instead, we are only interested if the standard deviation is higher than that. Is there enough ev- idence to suggest that o = 4 minutes or is it higher than that? Again, you may use that a = 0.05.

Answer 1

a) $H_{0}: \sigma = 4$

    $H_{1}: \sigma \neq 4$

The test statistic is $\chi^2 = \frac{(n - 1)s^2}{\sigma^2}$

                                      $= \frac{(24 - 1)(4.15)^2}{4^2}$

                                      

At alpha = 0.05, the critical values are

$\chi^2_{L} = \chi^2_{0.975, 23} = 11.6886$

$\chi^2_{R} = \chi^2_{0.025, 23} = 38.0756$

Since the test statistic value lies between the critical values (11.6886 < 24.757 < 38.0756), so we should not reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that $\sigma = 4$ minutes.

b) $H_{0}: \sigma = 4$

    $H_{1}: \sigma > 4$

The test statistic is $\chi^2 = \frac{(n - 1)s^2}{\sigma^2}$

                                      $= \frac{(24 - 1)(4.15)^2}{4^2}$

                                      

At alpha = 0.05, the critical value is $\chi^2_{0.05, 23}$ = 35.1725

Since the test statistic value is not greater than the critical value (24.757 < 35.1725), so we should not reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that $\sigma = 4$ minutes.