Question

The marketing manager of a branch office of a local telephone operating company wants to study the characteristics of residential customers served by her office. In particular, she wants to estimate the mean monthly cost of calls within the local calling region. In order to determine the sample size necessary, she needs an estimate of the standard deviation. On the basis of her past experience and judgment, she estimates that the standard deviation is equal to $12. Suppose a small scale study of 15 residential customers a sample standard deviation of $9.25. At the 0.10 level of significance, is there evidence that the population standard deviation is different from $12?

Answer 1

Answer:

No. There is not enough evidence to support the claim that the population standard deviation is different from $12.

Step-by-step explanation:

The null hypothesis is that the true standard deviation is 12.

The alternative hypothesis is that the true standard deviation differs from 12.

We can state:

$H_0: \sigma=12\\\\H_a: \sigma\neq12$

The significance level is 0.10.

The sample size is n=15, so the degrees of freedom are:

$df=n-1=15-1=14$

The sample standard deviation is 9.25.

The test statistic is

$T=(n-1)(s/\sigma_0)^2=14*(9.25/12)^2=14*0.77^2=14*0.59=8.32$

The critical values for rejecting the null hypothesis are:

$\chi_{0.025,13}=5.00875\\\\\chi_{0.975,13}=24.7356$

As T=8.32 is within the acceptance region (5.01, 24.74), the null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the population standard deviation is different from $12.