Answer to Question #140423 in Statistics and Probability for Esther Gathenya

Question

A magazine publisher tells potential advertisers that the mean household income of its regular readership is $61,500. An advertising agency wishes to test this claim against the alternative that the mean is smaller. A sample of 40 randomly selected regular readers yields mean income $59,800 with standard deviation $5,850. Perform the relevant test at the 1% level of significance.

Answer 1

$H_0:\mu=61500$

$H_1:\mu<61500$

$tc=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}}$

$=\frac{59800-61500}{\frac{5850}{\sqrt{40}}}=-1.838$

$Cv=t_{{\alpha},n-1}=t_{0.01,39}=2.426$

Since the absolute value of the test statistic tc=1.838 is less than the critical value cv=2.426, we fail to reject the null hypothesis and conclude that there is no sufficient evidence to support the claim that the mean household income is less than $61500.