# Week 4 Ec315 Homework

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Week 4 EC315 Homework

Chapter 10

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 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.

H0: u = 40
H1: u < 40 (my belief)

test stat: t(37.8) = (37.8-40)/[12.2/sqrt(60)] = -1.3968 determine the p-value and explain its meaning. p-value = P(t25

Critical value:
Statistic= t=(26.067-25)/(1.534/√15)=2.694
Critical value= inv-t(14,0.01) =2.624

Test value:
Reject region: {t/t>2.624}
Decision: statistic=2.694>2.624=critical value
We reject H0, At the .01 significance level we conclude that the mean number of patients per day is more than 25.

Decision:
P-value= t(2.694,14)=0.0087
So p-value < 0.01 (we reject H0)

Summary:
We can interpret the p-value as: p-value= probability of obtaining a statistic value greater than 2.694 if μ =25

61. Refer to the Baseball 2010 data, which report information on the 30 Major League Baseball teams for the 2010 season.
a. Conduct a test of hypothesis to determine whether the mean payroll of the teams was different form \$80.0 million. Use the .05 significance level.
b. Conduct a test of hypothesis to determine whether the mean attendance was more than 2,000,000 per team.

a. H0: mu = 80, H1: Mu is not equal to 80. This is a 2 tail test. The test statistic is (91.01667 - 80)*30^.5/ 38.25954= 1.577144

The critical value will be 1.96, we use a z test since 1.577 < 1.96 we accept the null hypothesis the mean payroll is not different from 80 million

b. H0: mu less…...

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