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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Chapter 11.2, Problem 37E

(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 different ways so that a person can quit smoking and the table with different ways and with their success. Thus, the summary of these data in a two-way table can be represented as:

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 37E , 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 different ways so that a person can quit smoking and the table with different ways and with their success. And the proportion in each group is the number of successes/failures divided by the row total given in the last column is as:

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 37E , additional homework tip  2

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 37E , additional homework tip  3

Thus, by looking at the histogram we can say that there appears to be no different in proportion between Nicotine Patch and Placebo. And the group Drug or Both seem to have a better success rate.

(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 37E

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

Explanation of Solution

In the question, it is given different ways so that a person can quit smoking and the table with different ways and with their success. 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 different ways so that a person can quit smoking and the table with different ways and with their success. Thus, it is given the null hypothesis in the question as:

  H0:p1=p2=p3=p4

And the summary of these data in a two-way table can be represented as:

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 37E , additional homework tip  4

Thus, the expected counts are the row total multiplied by the column total, divided by the sample size n=893 as:

  The Practice of Statistics for AP - 4th Edition, Chapter 11.2, Problem 37E , additional homework tip  5

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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