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Q.2.2

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The Practice of Statistics for AP
Found in: Page 703
The Practice of Statistics for AP

The Practice of Statistics for AP

Book edition 4th
Author(s) David Moore,Daren Starnes,Dan Yates
Pages 809 pages
ISBN 9781319113339

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Illustration

Short Answer

Calculate the expected counts. Show your work

The expected count is,

Use Facebook Main Campus Common wealth Several times a month or less 77.5653.44 At least once a week 220.25151.75 At least once a day 612.19421.81

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given Information

The table is

Step 2: Explanation

The formula to compute the degree of freedom is:

Expected count = Row total × Coloumn total Table total

The expected count is Use Facebook Main Campus Common wealth Several times a month or less 1820×1313074=77.561254×1313074=53.44 At least once a week 1820×3723074=220.251254×3723074=151.75 At least once a day 1820×10343074=612.191254×10343074=421.81

Most popular questions for Math Textbooks

Refer to Exercise 28. Do the data provide convincing evidence of a difference in the distributions of opinions about how high schools are doing among black, Hispanic, and white parents?

(a) State appropriate null and alternative hypotheses for a significance test to help answer this question.

(b) Calculate the expected counts. Show your work.

(c) Calculate the chi-square statistic. Show your work.

Gregor Mendel (1822–1884), an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. In one experiment, he observed 423 smooth and 133 wrinkled peas. The data were produced in such a way that the Random and Independent conditions are met. Carry out a chi-square goodness-of-fit test based on Mendel’s prediction. What do you conclude?

Canada has universal health care. The United States does not but often offers more elaborate treatment to patients with access. How do the two systems compare in treating heart attacks? Researchers compared random samples of 2600 U.S. and 400 Canadian heart attack patients. One key outcome was the patients’ own assessment of their quality of life relative to what it had been before the heart attack. Here are the data for the patients who survived a year:

Quality of life Canada United States Much better 75541 Somewhat better 71498 About the same 96779 Somewhat worse 50282 Much worse 1965 Total 3112165

Construct an appropriate graph to compare the distributions of opinions about the quality of life among heart attack patients in Canada and the United States.

An appropriate null hypothesis to test whether the trees in the forest are randomly distributed is

(a) H0:μ=25, where μ= the mean number of trees in each quadrant.

(b) H0:p=0.25, where p= the proportion of all trees in the forest that are in Quadrant 1 .

(c) H0:n1=n2=n3=n4=25, where ni is the number of trees from the sample in Quadrant i.

(d) H0:p1=p2=p3=p4=0.25, where pi is the actual proportion of trees in the forest that are in Quadrant i.

(e) H0:p^1=p^2=p^3=p^4=0.25, where p^i is the proportion of trees in the sample that are in Quadrant i.

A chi-square goodness-of-fit test is used to test whether a 0 to 9 spinner is "fair" (that is, the outcomes are all equally likely). The spinner is spun 100 times, and the results are recorded. The degrees of freedom for the test will be

(a) 8 .

(c) 10 .

(e) None of these.

(b) 9 .

(d) 99 .
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