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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 11: Q 8. (page 716)

No chi-square The principal in Exercise 7 also asked the random sample of students to record whether they did all of the homework that was assigned on each of the five school days that week. Here are the data:

Short Answer

Expert verified

Because the observations are not independent, a chi-square test for goodness of fit is not appropriate.

Step by step solution

01

Given information

02

Explanation

The observations must be independent in order to use the chi-square test for goodness of fit. Because most kids will have the same teachers, independent observations are not possible. As a result, a chi-square test for goodness of fit would be inappropriate.

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Most popular questions from this chapter

Last python Refer to Exercises 28 and 30.

a. Verify that the conditions for inference are met.

b. Use Table C to find the P-value. Then use your calculator’s χ2 cdf command.

c. Interpret the P-value from the calculator.

d. What conclusion would you draw using α=0.10?

Fewer TVs? The United States Energy Information Administration periodically surveys a random sample of U.S. households to determine how they use energy. One of the variables they track is how many TVs are in a household (None, $1, 2, 3, 4, o r 5$ or more). The computer output compares the distribution of number of TVs for households in $2009 a n d 2015$ .

Cell Contents: Count

Expected count

Contribution to Chi-square

Chi-Squarerole="math" localid="1654195309908" $= 137.137$ , DF $= 5$ , P-Value $= 0.000$

a. Which chi-square test is appropriate to analyze these data? Explain your answer.

b. Show how the numbers $252$ and $14.113$ were obtained for the $2009 / N o n e$ cell.

c. Which $3$ cells contribute most to the chi-square test statistic? How do the observed and expected counts compare for these cells?

Online banking A recent poll conducted by the Pew Research Center asked a random sample of $1846$ Internet users if they do any of their banking online. The table summarizes their responses by age. $23$ Is there convincing evidence of an association between age and use of online banking for Internet users?

Treating ulcers Gastric freezing was once a recommended treatment for ulcers in the upper intestine. Use of gastric freezing stopped after experiments showed it had no effect. One randomized comparative experiment found that $28$ of the $82$ gastric-freezing patients improved, while $30$ of the $78$ patients in the placebo group improved. We can test the hypothesis of “no difference” in the effectiveness of the treatments in two ways: with a two-sample z test or with a chi-square test.

a. State appropriate hypotheses for a chi-square test.

b. Here is Minitab output for a chi-square test. Interpret the P-value. What conclusion would you draw?

c. Here is Minitab output for a two-sample z test. Explain how these results are consistent with the test in part (a).

The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the new school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of $100$ students

and asks them, “Which type of food do you prefer: Ramen, tacos, pizza, or hamburgers?” Here are her data:

An appropriate null hypothesis to test whether the food choices are equally popular is

a. $H_{0} : μ = 25$ where $μ =$ the mean number of students that prefer each type of food.

b. $H_{0} : p = 0.25$ where p = the proportion of all students who prefer ramen.

c. $H_{0} : n R = n T = n P = n H = 25$ where $n R$ is the number of students in the school who would choose ramen, and so on.

d. $H_{0} : p R = p T = p P = p H = 0.25$ where $p R$ is the proportion of students in the school who would choose ramen, and so on.

e. $H_{0} : p^{R} = p^{T} = p^{P} = p^{H} = 0.25$ , where $p^{R}$ is the proportion of students in the sample who chose ramen, and so on.

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