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The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset Vaia Explanations$ Math

6th Edition · 837 pages

ISBN: 9781319113339

Chapter 10: Q. 72. (page 672)

Based on the P-value in Exercise 71, which of the following must be true?
a. A $90 %$ confidence interval for μM−μF $\frac{305}{1526} = 0.200 = 20.0 % μ_{M} - μ_{F}$ will contain $0$
b. A $95 %$ confidence interval for μM−μF $\frac{305}{1526} = 0.200 = 20.0 % μ_{M} - μ_{F}$ will contain $0$
c. A $99 %$ confidence interval for μM−μF $\frac{305}{1526} = 0.200 = 20.0 % μ_{M} - μ_{F}$ will contain $0$
d. A $99.9 %$ confidence interval for μM−μF $\frac{305}{1526} = 0.200 = 20.0 % μ_{M} - μ_{F}$ will contain $0$

Short Answer

Expert verified

The required correct option is $(d)$

Step by step solution

01

Given information

Given,

$P = 0.002$

02

Explanation

If the P-value is less than the significance level, the null hypothesis is rejected. As a result, we see that we reject the null hypothesis at the significance level $0.001$ , but not at the significance level $0.01, 0.05, 0.10$

If we reject the null hypothesis at the corresponding significance level, the confidence interval will contain zero. The $99.9 %$ confidence interval will contain zero because we reject the null hypothesis at the corresponding significance level of $0.001$

Therefore, option $(d)$ is the correct option.

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Most popular questions from this chapter

I want red! Refer to Exercise 1.

a. Find the probability that the proportion of red jelly beans in the Child sample is less than or equal to the proportion of red jelly beans in the Adult sample, assuming that the company’s claim is true.

b. Suppose that the Child and Adult samples contain an equal proportion of red jelly beans. Based on your result in part (a), would this give you reason to doubt the

company’s claim? Explain your reasoning.

Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select $10$ Variety A and $10$ Variety B tomato plants. Then the researchers divide in half each of $10$ small 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. The $10$ differences (Variety A − Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and single-peaked with no outliers. The mean difference is $\bar{x} = 0.34$ $\frac{305}{1526} = 0.200 = 20 % {\bar{x}}_{A - B} = 0.34$ and the standard deviation of the differences is s A-B $= 0.83$ $\frac{305}{1526} = 0.200 = 20 % = s_{A - B} = 0.83$ .Let $μ_{A - B} = \frac{305}{1526} = 0.200 = 20 %$ μA−B = the true mean difference (Variety A − Variety B) in yield for tomato plants of these two varieties.

The P-value for a test of H0: μA−B=0 $\frac{305}{1526} = 0.200 = 20 %$ versus Ha: μA−B≠0 is 0.227. Which of the following is the

correct interpretation of this P-value?

a. The probability that μA−B is $0.227$ .

b. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is $0$ , the probability of getting a sample mean difference of $0.34$ is $0.227$ .

c. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is $0$ , the probability of getting a sample mean difference of $0.34$ or greater is $0.227$ .

d. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is $0$ , the probability of getting a sample mean difference greater than or equal to $0.34$ or less than or equal to − $0.34$ is $0.227$ .

e. Given that the true mean difference (Variety A – 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 $0.34$ or less than or equal to − $0.34$ is $0.227$ .

Friday the $13 t h$ Refer to Exercise $88$ .

a. Construct and interpret a $90 %$ confidence interval for the true mean difference. If you already defined parameters and checked conditions in Exercise $88$ , you don’t need to do them again here.

b. Explain how the confidence interval provides more information than the test in Exercise $88$ .

A beef rancher randomly sampled $42$ cattle from her large herd to obtain a $95 %$ confidence interval for the mean weight (in pounds) of the cattle in the herd. The interval obtained was ( $1010, 1321$ ). If the rancher had used a $98 %$ confidence interval instead, the interval would have been

a. wider with less precision than the original estimate.

b. wider with more precision than the original estimate.

c. wider with the same precision as the original estimate.

d. narrower with less precision than the original estimate.

e. narrower with more precision than the original estimate.

Quit smoking Nicotine patches are often used to help smokers quit. Does giving medicine to fight depression help? A randomized double-blind experiment assigned $244$ smokers to receive nicotine patches and another $245$ to receive both a patch and the antidepressant drug bupropion. After a year, $40$ subjects in the nicotine patch group had abstained from smoking, as had $87$ in 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.

See all solutions

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