Question 369664
a recent article in vitality magazine reported that the mean amount of leisure time per week for american men is 40.0 hours.
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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. 
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can you conclude that the information in the article is untrue? 
Use the .05 significance level.
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Ho: u = 40
Ha: u < 40 (your belief)
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test stat: t(37.8) = (37.8-40)/[12.2/sqrt(60)] = -1.3968
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determine the p-value and explain its meaning.
p-value = P(t<-1.3968 when df=59) = 0.0839
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Means slightly more than 8% of test results might have produced
stronger evidence for rejecting Ho.
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Conclusion: Since the p-value is greater than 5%, fail to reject Ho
at the 5% significance level.
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Cheers,
Stan H.