a recent article in Vitality magazine reported that the mean amount of leisure t

Question

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 this article is untrue? Use the .05 significance level. Determine the p-value and explain its meaning

Explanation / Answer

We have a mean of 37.8 with a standard deviation of 12.2 hours. We need to find the p-value for 40.0 the probability distrubution function for a normal distribution is f= 1/(s*sqrt(2pi)) exp( -1/2 * (x-mu)/s^2) s = standard deviation (12.2) mu = our mean (37.8) x = mean of the original study (40) So we can solve for f p = 1-f gives us the p-value if p > 0.05 we can conclude that the article is untrue with a 0.05 significance level.

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