EC315_HW4_JamieBoyer

EC315_HW4_JamieBoyer - Jamie Boyer EC 315 Week 4 Homework...

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Unformatted text preview: Jamie Boyer EC 315 Week 4 Homework 34. 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 I 47.8 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 the meaning. Critical Value for - one=tailed Test = -1.645 P-Value for -1.4= 0.5-0.4192 = 0.0.0808 > The P-value is the smallest level of significance for which the sample data tells us to reject the null hypothesis. A P-value of 0.0808 (which is greater than alpha), therefore, tells us that we should not reject the null hypothesis. Therefore, at the 5% level of significance, the data does not provide sufficient evidence that the mean amount of leisure time per week for American men...
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EC315_HW4_JamieBoyer - Jamie Boyer EC 315 Week 4 Homework...

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