Answer to Question #337096 in Statistics and Probability for Paowii

Question #337096

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.


1
Expert's answer
2022-05-04T17:31:10-0400

The following null and alternative hypotheses need to be tested:

H0:σ2=1600H_0:\sigma^2=1600

Ha:σ21600H_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 α=0.05,\alpha = 0.05, df=n1=9df=n-1=9 degrees of freedom and the rejection region for this two-tailed test is R={χ2:χ2<2.7004 or χ2>19.0228}.R = \{\chi^2: \chi^2 < 2.7004 \text{ or } \chi^2 > 19.0228\}.

The Chi-Squared statistic is computed as follows:


χ2=(n1)s2σ2=(101)381600=0.21375\chi^2=\dfrac{(n-1)s^2}{\sigma^2}=\dfrac{(10-1)38}{1600}=0.21375

Since it is observed that χ2=0.21375<2.7004=χL2,\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 σ2\sigma^2

is different than 1600, at the 0.05 significance level.

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



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