Young adults living at home A surprising number of young adults (ages 19

to 25) still live in their parents’ homes. The National Institutes of Health surveyed

independent random samples of 2253men and 2629women in this age group. The survey found that 986 of the men and 923 of the women lived with their parents.

a. Construct and interpret a 99%confidence interval for the difference in the true

proportions of men and women aged 19-25who live in their parents’ homes.

b. Does your interval from part (a) give convincing evidence of a difference between the population proportions? Justify your answer.

Short Answer

Expert verified

a. Confidence Interval is between 0.051and0.123.

b. Yes, it gives convincing evidence of difference between population proportions.

Step by step solution

01

Given Information

It is given that x1=986

x2=923

n1=2253

n2=2629

c=99%=0.99

02

Calculating Confidence Interval

The conditions are:

Random: Samples are independent random samples.

Independent: 2252men are <10%of all mean and it applies to women also.

Normal: In first sample, there are 983success and 2253-983=1270failures.

In second sample, there are 923success and 2629-923=1706failures. All are greater than ten.

All conditions are satisfied.

Sample proportion is p^1=x1n1=9862253=0.438

p^2=x2n2=9232629=0.351

For 1-α=0.99,using table, za/2=2.575

Confidence interval is p^1-p^2-za/2×p^11-p^1n1+p^21-p^2n2

=(0.438-0.351)-2.575×0.438(1-0.438)2253+0.351(1-0.351)26290.051

and p^1-p^2+za/2×p^11-p^1n1+p^21-p^2n2

=(0.438-0.351)+2.575×0.438(1-0.438)2253+0.351(1-0.351)26290.123

Hence, confidence interval is0.051,0.123

03

To check if it gives evidence of difference between population proportions

As Confidence Interval is (0.051,0.123). it does not contain zero. It is unlikely that population proportion are equal.

Hence, it gives convincing evidence for difference between population proportions.

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Most popular questions from this chapter

Two samples or paired data? In each of the following settings, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice.

a. To compare the average weight gain of pigs fed two different diets, nine pairs of pigs were used. The pigs in each pair were littermates. A coin toss was used to decide which pig in each pair got Diet A and which got Diet B.

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sample of 68 workers from Company B admitted that they had used sick leave when

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in the proportions of workers at the two companies who would admit to using sick

leave when they weren’t ill?

(a) 0.03±(0.28)(0.72)125+(0.25)(0.75)68

(b) 0.03±1.96(0.28)(0.72)125+(0.25)(0.75)68

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(d) 0.03±1.96(0.269)(0.731)125+(0.269)(0.731)68

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The P-value for the stated hypotheses is 0.002Interpret this value in the context of this study.

a. Assuming that the true mean road rage score is the same for males and females, there is a 0.002probability of getting a difference in sample means equal to the one observed in this study.

b. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study.

c. Assuming that the true mean road rage score is different for males and females, there is a 0.002 probability of getting a difference in sample means at least as large in either direction as the one observed in this study.

d. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the null hypothesis is true.

e. Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the alternative hypothesis is true.

Suppose the null and alternative hypothesis for a significance test are defined as

H0: μ=403051526=0.200=20.0%H0 : μ=40

Ha: μ<403051526=0.200=20.0%Ha : μ<40

Which of the following specific values for Ha will give the highest power? a. μ=383051526=0.200=20.0%μ=38

b. μ=393051526=0.200=20.0%μ=39

c. μ=413051526=0.200=20.0%μ=41

d. μ=423051526=0.200=20.0%μ=42

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a. gives z=2.25,P<0.02 .

b. gives z=2.60,P<0.005 .

c. gives z=2.25,P<0.04 but not<0.02

d. should not be used because the Random condition is violated.

e. should not be used because the Large Counts condition is violated.

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