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Chapter 10: Q 12. (page 640)

Quit smoking Nicotine patches are often used to help smokers quit. Does giving medicine to fight depression help? A randomized double-blind experiment assigned 244smokers to receive nicotine patches and another 245to receive both a patch and the antidepressant drug bupropion. After a year, 40subjects in the nicotine patch group had abstained from smoking, as had 87in the patch-plus-drug group. Construct and interpret a 99%confidence interval for the difference in the true proportion of smokers like these who would abstain when using bupropion and a nicotine patch and the proportion who would abstain when using only a patch.

Short Answer

Expert verified

Confidence Interval is(-0.2908,-0.0916).

Step by step solution

01

Given Information

It is given that x1=40

x2=87

n1=244

n2=245

c=99%=0.99

02

Checking the conditions

Random: Samples are independent random samples.

Independent: 224<10%of all smokers.

Normal: The success are 40,87and failures are 204,158which are greater than ten. So, is satisfied.

03

Calculations

Sample Proportion: p^1=x1n1=40244=0.1639

and p^2=x2n2=87245=0.3551

For 1-伪=0.99, value of role="math" localid="1654639742329" z伪/2=z0.005using table is

za/2=2.575

Hence, confidence interval is p^1-p^2-za/2脳p^11-p^1n1+p^21-p^2n2

=(0.1639-0.3551)-2.575脳0.1639(1-0.1639)244+0.3551(1-0.3551)245

鈮�-0.2908

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

=(0.1639-0.3551)+2.575脳0.1639(1-0.1639)244+0.3551(1-0.3551)245

鈮�-0.0916

Hence, confidence interval is(-0.2908,-0.0916)

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Most popular questions from this chapter

Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select10Variety A and10Variety B tomato plants. Then the researchers divide in half each of10small plots of land in different locations. For each plot, a coin toss determines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The10differences (Variety A 鈭� Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and single-peaked with no outliers. The mean difference is x-=0.343051526=0.200=20%x-A-B=0.34and the standard deviation of the differences is s A-B=0.833051526=0.200=20%=sA-B=0.83.Let渭A-B=3051526=0.200=20%渭A鈭払 = the true mean difference (Variety A 鈭� Variety B) in yield for tomato plants of these two varieties.

The P-value for a test of H0: 渭A鈭払=03051526=0.200=20%versus Ha: 渭A鈭払鈮�0 is 0.227. Which of the following is the

correct interpretation of this P-value?

a. The probability that 渭A鈭払 is0.227.

b. Given that the true mean difference (Variety A 鈥� Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference of0.34is0.227.

c. Given that the true mean difference (Variety A 鈥� Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference of0.34or greater is0.227.

d. Given that the true mean difference (Variety A 鈥� Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference greater than or equal to0.34or less than or equal to 鈭�0.34is0.227.

e. Given that the true mean difference (Variety A 鈥� Variety B) in yield for these two varieties of tomato plants is not 0, the probability of getting a sample mean difference greater than or equal to 0.34or less than or equal to 鈭�0.34is0.227.

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.

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worked for pay last summer is different from the proportion of female college students in the study who worked for pay last summer.

b. Because the P-value is less than 1%, fail to reject H0. There is not

convincing evidence that the proportion of all male college students who worked for pay last summer is different from the proportion of all female college students who worked for pay last summer.

c. Because the P-value is less than 1%, reject H0. There is convincing

evidence that the proportion of all male college students who worked for pay last

summer is the same as the proportion of all female college students who worked for pay last summer.

d. Because the P-value is less than 1%, reject H0. There is convincing

evidence that the proportion of all male college students in the study who worked for pay last summer is different from the proportion of all female college students in the study who worked for pay last summer.

e. Because the P-value is less than 1%, reject H0 . There is convincing

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I. Confirm graphically that the scores of the caffeine drinkers could have come from a Normal distribution.

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