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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. T10.4 (page 698)

A quiz question gives random samples of $n = 10$ observations from each of two Normally distributed populations. Tom uses a table of t distribution critical values and $9$ degrees of freedom to calculate a $95 %$ confidence interval for the difference in the two population means. Janelle uses her calculator's two-sample t Interval with $16.87$ degrees of freedom to compute the $95 %$ confidence interval. Assume that both students calculate the intervals correctly. Which of the following is true?
(a) Tom's confidence interval is wider.
(b) Janelle's confidence Interval is wider.
(c) Both confidence Intervals are the same.
(d) There is insufficient information to determine which confidence interval is wider.
(e) Janelle made a mistake, degrees of freedom has to be a whole number.

Short Answer

Expert verified

The correct answer is:

(a) Tom's confidence interval is wider.

Step by step solution

01

Given information

We are given that Janelle uses a higher degree of freedom than Tom.

02

Explanation

A higher degree of freedom results in a lower $t^{*} - v a l u e .$

A lower $t^{*} - v a l u e$ results in a lower margin of error and thus also a narrower confidence interval.

Then we know that Janelle's confidence interval is narrower than Tom's confidence interval, or Tom's confidence interval is wider than Janelle's confidence level.

Thus answer (a) is correct.

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Most popular questions from this chapter

I want red!A candy maker offers Child and Adult bags of jelly beans with

different color mixes. The company claims that the Child mix has $30 %$ red jelly beans, while the Adult mix contains $15 %$ red jelly beans. Assume that the candy maker’s claim is true. Suppose we take a random sample of $50$ jelly beans from the Child mix and a separate random sample of $100$ jelly beans from the Adult mix. Let ${\hat{p}}_{C}$ and ${\hat{p}}_{A}$ be the sample proportions of red jelly beans from the Child and

Adult mixes, respectively.

a. What is the shape of the sampling distribution of ${\hat{p}}_{C} - {\hat{p}}_{A}$ ? Why?

b. Find the mean of the sampling distribution.

c. Calculate and interpret the standard deviation of the sampling distribution.

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 $0.227$ .

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 $0$

The correlation between the heights of fathers and the heights of their grownup sons, both measured in inches, is $r = 0.52$ . If fathers’ heights were measured in feet instead, the correlation between heights of fathers and heights of sons would be

a. much smaller than $0.52$ .

b. slightly smaller than $0.52$ .

c. unchanged; equal to $0.52$ .

d. slightly larger than $0.52$ .

e. much larger than $0.52$ .

Each day I am getting better in math A "subliminal" message is below our threshold of awareness but may nonetheless influence us. Can subliminal messages help students learn math? A group of $18$ students who had failed the mathematics part of the City University of New York Skills Assessment Test agreed to participate in a study to find out. All received a daily subliminal message, flashed on a screen too rapidly to be consciously read. The treatment group of $10$ students (assigned at random) was exposed to "Each day I am getting better in math." The control group of $8$ students was exposed to a neutral message, "People are walking on the street." All $18$ students participated in a summer program designed to improve their math skills, and all took the assessment test again at the end of the program. The following table gives data on the subjects' scores before and after the program.

a. Explain why a two-sample t-test and not a paired t-test is the appropriate inference procedure in this setting.

b. The following boxplots display the differences in pretest and post-test scores for the students in the control (C) and treatment (T) groups. Write a few sentences comparing the performance of these two groups.

c. Do the data provide convincing evidence at the $α = 0.01, \frac{305}{1526} = 0.200 = 20 %$ significance level that subliminal messages help students like the ones in this study learn math, on average?

d. Can we generalize these results to the population of all students who failed the mathematics part of the City University of New York Skills Assessment Test? Why or why not?

See all solutions

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