Answer to Question #337096 in Statistics and Probability for Paowii

Question

The standard deviation of the number of minutes of calls of a certain subscriber is 40 min per day. A sample of 10 days gave a variance of 38 min. Test the hypothesis with α = 0.05 that the standard deviation of the call usage is not equal to 40.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\sigma^2=1600$

$H_a:\sigma^2\not=1600$

This corresponds to a two-tailed test, for which a Chi-Square test for one population variance will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=9$ degrees of freedom and the rejection region for this two-tailed test is $R = \{\chi^2: \chi^2 < 2.7004 \text{ or } \chi^2 > 19.0228\}.$

The Chi-Squared statistic is computed as follows:

$$\chi^2=\dfrac{(n-1)s^2}{\sigma^2}=\dfrac{(10-1)38}{1600}=0.21375$$

Since it is observed that $\chi^2 = 0.21375<2.7004= \chi_L^2,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population variance $\sigma^2$

is different than 1600, at the 0.05 significance level.

Therefore, there is enough evidence to claim that the population standard deviation $\sigma$ is different than 40, at the 0.05 significance level.