The Practice of Statistics for AP - 4th Edition
The Practice of Statistics for AP - 4th Edition
4th Edition
ISBN: 9781429245593
Author: Starnes, Daren S., Yates, Daniel S., Moore, David S.
Publisher: Macmillan Higher Education
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Question
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Chapter 11.2, Problem 38E

(a)

To determine

To summarize these data in a two-way table.

(a)

Expert Solution
Check Mark

Explanation of Solution

In the question, it is given that aspirin prevents blood from clotting and so help prevent strokes. Thus, the data on the strokes for the two-year study is given in the question. Thus, the summary of these data in a two-way table can be represented by the number of no strokes is the number of patients decreased by the number of strokes, that is,

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 38E , additional homework tip  1

(b)

To determine

To make a graph to compare the success rates for the four treatments and describe what you see.

(b)

Expert Solution
Check Mark

Explanation of Solution

In the question, it is given that aspirin prevents blood from clotting and so help prevent strokes. Thus, the data on the strokes for the two-year study is given in the question. Thus, the summary of these data in a two-way table can be represented by the number of no strokes is the number of patients decreased by the number of strokes, that is,

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 38E , additional homework tip  2

So therefore the proportion in each group is the number of strokes/ no strokes divided by the row total given in the last column is calculated as:

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 38E , additional homework tip  3

Thus, the graph to compare the success rates for the four treatments can be constructed as:

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 38E , additional homework tip  4

Thus, by looking at the histogram we note that the proportions of strokes/no strokes do not seems to vary much by group.

(c)

To determine

To explain in words what the null hypothesis H0:p1=p2=p3=p4 says about subjects’ smoking habits.

(c)

Expert Solution
Check Mark

Answer to Problem 38E

It says proportion of successes for each treatment is the same.

Explanation of Solution

In the question, it is given that aspirin prevents blood from clotting and so help prevent strokes. Thus, the data on the strokes for the two-year study is given in the question. Thus, the summary of these data in a two-way table can be represented by the number of no strokes is the number of patients decreased by the number of strokes, that is,

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 38E , additional homework tip  5

Thus, it is given the null hypothesis in the question as:

  H0:p1=p2=p3=p4

And the null hypothesis says about the subjects’ smoking habits that this means that the proportion of successes for each treatment is the same.

(d)

To determine

To find the expected counts if null hypothesis is true and display them in a two-way table of observed counts.

(d)

Expert Solution
Check Mark

Explanation of Solution

In the question, it is given that aspirin prevents blood from clotting and so help prevent strokes. Thus, the data on the strokes for the two-year study is given in the question. Thus, the summary of these data in a two-way table can be represented by the number of no strokes is the number of patients decreased by the number of strokes, that is,

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 38E , additional homework tip  6

Thus, it is given the null hypothesis in the question as:

  H0:p1=p2=p3=p4

Thus, the expected counts are the row total multiplied by the column total, divided by the sample size n=6602 as:

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 38E , additional homework tip  7

Chapter 11 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 11.1 - Prob. 3ECh. 11.1 - Prob. 4ECh. 11.1 - Prob. 5ECh. 11.1 - Prob. 6ECh. 11.1 - Prob. 7ECh. 11.1 - Prob. 8ECh. 11.1 - Prob. 9ECh. 11.1 - Prob. 10ECh. 11.1 - Prob. 11ECh. 11.1 - Prob. 12ECh. 11.1 - Prob. 13ECh. 11.1 - Prob. 14ECh. 11.1 - Prob. 15ECh. 11.1 - Prob. 16ECh. 11.1 - Prob. 17ECh. 11.1 - Prob. 18ECh. 11.1 - Prob. 19ECh. 11.1 - Prob. 20ECh. 11.1 - Prob. 21ECh. 11.1 - Prob. 22ECh. 11.1 - Prob. 23ECh. 11.1 - Prob. 24ECh. 11.1 - Prob. 25ECh. 11.1 - Prob. 26ECh. 11.2 - Prob. 1.1CYUCh. 11.2 - Prob. 1.2CYUCh. 11.2 - Prob. 1.3CYUCh. 11.2 - Prob. 2.1CYUCh. 11.2 - Prob. 2.2CYUCh. 11.2 - Prob. 2.3CYUCh. 11.2 - Prob. 3.1CYUCh. 11.2 - Prob. 3.2CYUCh. 11.2 - Prob. 3.3CYUCh. 11.2 - Prob. 4.1CYUCh. 11.2 - Prob. 4.2CYUCh. 11.2 - Prob. 6.1CYUCh. 11.2 - Prob. 27ECh. 11.2 - Prob. 28ECh. 11.2 - Prob. 29ECh. 11.2 - Prob. 30ECh. 11.2 - Prob. 31ECh. 11.2 - Prob. 32ECh. 11.2 - Prob. 33ECh. 11.2 - Prob. 34ECh. 11.2 - Prob. 35ECh. 11.2 - Prob. 36ECh. 11.2 - Prob. 37ECh. 11.2 - Prob. 38ECh. 11.2 - Prob. 39ECh. 11.2 - Prob. 40ECh. 11.2 - Prob. 41ECh. 11.2 - Prob. 42ECh. 11.2 - Prob. 43ECh. 11.2 - Prob. 44ECh. 11.2 - Prob. 45ECh. 11.2 - Prob. 46ECh. 11.2 - Prob. 47ECh. 11.2 - Prob. 48ECh. 11.2 - Prob. 49ECh. 11.2 - Prob. 50ECh. 11.2 - Prob. 51ECh. 11.2 - Prob. 52ECh. 11.2 - Prob. 53ECh. 11.2 - Prob. 54ECh. 11.2 - Prob. 55ECh. 11.2 - Prob. 56ECh. 11.2 - Prob. 57ECh. 11.2 - Prob. 58ECh. 11.2 - Prob. 59ECh. 11.2 - Prob. 60ECh. 11.2 - Prob. 61ECh. 11.2 - Prob. 62ECh. 11.2 - Prob. 63ECh. 11.2 - Prob. 64ECh. 11 - Prob. 1CRECh. 11 - Prob. 2CRECh. 11 - Prob. 3CRECh. 11 - Prob. 4CRECh. 11 - Prob. 5CRECh. 11 - Prob. 6CRECh. 11 - Prob. 1PTCh. 11 - Prob. 2PTCh. 11 - Prob. 3PTCh. 11 - Prob. 4PTCh. 11 - Prob. 5PTCh. 11 - Prob. 6PTCh. 11 - Prob. 7PTCh. 11 - Prob. 8PTCh. 11 - Prob. 9PTCh. 11 - Prob. 10PTCh. 11 - Prob. 11PTCh. 11 - Prob. 12PTCh. 11 - Prob. 13PT

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