EC315Week4HW

EC315Week4HW - QRM EC 315 Week 4 Homework Chapter 10...

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QRM EC 315 Week 4 Homework Chapter 10, Question 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 is 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. From the data set: Mean number of hours = 47.8 standard deviation = 12.2 number of men is 60 Hypotheses: Critical value: Test value: Critical Value for - one=tailed Test = -1.645 Decision: P-Value for -1.4= 0.5-0.4192 = 0.0.0808 > α Summary: 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

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This note was uploaded on 10/07/2011 for the course EC 315 taught by Professor Barcus during the Spring '10 term at Park.

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EC315Week4HW - QRM EC 315 Week 4 Homework Chapter 10...

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