A recent article in Vitality magazine reported that the mean

A recent article in Vitality magazine reported that the mean amount of leisure time per week for American men is 40.0 hours. You believe this figure is too large and decide to conduct your own test. In a random sample of 60 men, you find that the mean is 37.8 hours of leisure per week and that the standard deviation of the sample is 12.2 hours. Can you conclude that the information in the article is untrue? Use the .05 significance level. Determine the p-value and explain its meaning.

Solution

The test hypothesis is Ho: u = 40 Ha: u < 40 The test statistic: t(37.8) = (37.8-40)/[12.2/sqrt(60)] = -1.3968 The p-value is p-value = P(t<-1.3968 when df=59) = 0.0839 Means slightly more than 8% of test results might have produced stronger evidence for rejecting Ho. Conclusion: Since the p-value is greater than 5%, fail to reject Ho at the 5% significance level.