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Study more! A student group claims that first-year students at a university study 2.5hours per night during the school week. A skeptic suspects that they study less than that on average. He takes a random sample of 30first-year students and finds that x=137minutes and sx=45minutes. A graph of the data shows no outliers but some skewness. Carry out an appropriate significance test at the 5%significance level. What conclusion do you draw?

Short Answer

Expert verified

We are in the acceptance region we need not to reject H.

Step by step solution

01

Given information

A student group claims that first-year students at a university study 2.5hours per night during the school week.

A skeptic suspects that they study less than that on average.

He takes a random sample of 30first-year students and finds that x=137minutes andsx=45minutes.

02

Explanation

Normal distribution:

mean μ=150

Sample:

Sample size n=30

Sample mean x=137

Sample standard deviation sx=45

The standard error of the sample mean

localid="1650891685864" SE=sx/nSE=45/30SE=8.22

Test Hypothesis:

Null hypothesis Hx=μ

Alternative hypothesis Hx<μ

z(s) test statistics is:

z(s)=(x-μ)/s/nz(s)=-13/8.22z(s)=-1.58

p-value for that z(s) localid="1650892596487" p-value=0.062

Then for α=0.05p-value>0.05

We are in the acceptance region we need not to rejectH.

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Most popular questions from this chapter

A government report says that the average amount of money spent per U.S. household per week on food is about \(158. A random sample of 50households in a small city is selected, and their weekly spending on food is recorded. The Minitab output below shows the results of requesting a confidence interval for the population mean M. An examination of the data reveals no outliers.

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