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 9.2, Problem 56E
Solution

a)

The parameter of interest.

Given:Let “ p” be the proportion of students who are willing to report cheating by other students.

The Minitab output of the results of a significance test and 95% confidence interval based on survey as shown below

  The Practice of Statistics for AP - 4th Edition, Chapter 9.2, Problem 56E

Explanation:

A parameter is any summary number, which describes for entire population. Here the population is all undergraduates in a large university. The parameter of interest is the proportion of students who are willing to report cheating by other students.

b)

To verify the conditions for performing the significance test are met in this case.

Concept used:The conditions to satisfy to run one sample Z test

1) The sample should be random

2) np = 10

3) n(1-p) = 10

4) There should be at least 10% of people of population in the sample.Calculation:

The sample of 172 undergraduates from a large university are selected randomly.Let p0be the proportion of students who are willing to report cheating by other students.

  p0=0.11

The sample size = n = 172

  np0=172*0.11=18.92np010n(1p0)=172*(10.11)=153n(1p0)10

We know that there will be at least 1720 students who are willing to report cheating by other students. So the 10% condition is satisfied.

Conlcusion:All the conditions for performing the significance test are met in this case.

c)

To interpret the p value.

Explanation:

The p value of the significance test is 0.146. This tell us that at an average 14.6% of students differ from the proportion of students who are willing to report cheating by other students

d)

To check whether there is evidence that the actual population proportion of students who are willing to report cheating by other students differs from 0.15.

Explanation:

The 95% confidence interval is (0.063619, 0.157312) contains the true population proportion of p = 0.15. here null hypothesis cannot be rejected. So at 5% level of significance there is insufficient evidence to conclude that actual population proportion of students who are willing to report cheating by other students differ from 0.15.

Chapter 9 Solutions

The Practice of Statistics for AP - 4th Edition

Ch. 9.1 - Prob. 6ECh. 9.1 - Prob. 7ECh. 9.1 - Prob. 8ECh. 9.1 - Prob. 9ECh. 9.1 - Prob. 10ECh. 9.1 - Prob. 11ECh. 9.1 - Prob. 12ECh. 9.1 - Prob. 13ECh. 9.1 - Prob. 14ECh. 9.1 - Prob. 15ECh. 9.1 - Prob. 16ECh. 9.1 - Prob. 17ECh. 9.1 - Prob. 18ECh. 9.1 - Prob. 19ECh. 9.1 - Prob. 20ECh. 9.1 - Prob. 21ECh. 9.1 - Prob. 22ECh. 9.1 - Prob. 23ECh. 9.1 - Prob. 24ECh. 9.1 - Prob. 25ECh. 9.1 - Prob. 26ECh. 9.1 - Prob. 27ECh. 9.1 - Prob. 28ECh. 9.1 - Prob. 29ECh. 9.1 - Prob. 30ECh. 9.1 - Prob. 31ECh. 9.1 - Prob. 32ECh. 9.2 - Prob. 1.1CYUCh. 9.2 - Prob. 2.1CYUCh. 9.2 - Prob. 3.1CYUCh. 9.2 - Prob. 33ECh. 9.2 - Prob. 34ECh. 9.2 - Prob. 35ECh. 9.2 - Prob. 36ECh. 9.2 - Prob. 37ECh. 9.2 - Prob. 38ECh. 9.2 - Prob. 39ECh. 9.2 - Prob. 40ECh. 9.2 - Prob. 41ECh. 9.2 - Prob. 42ECh. 9.2 - Prob. 43ECh. 9.2 - Prob. 44ECh. 9.2 - Prob. 45ECh. 9.2 - Prob. 46ECh. 9.2 - Prob. 47ECh. 9.2 - Prob. 48ECh. 9.2 - Prob. 49ECh. 9.2 - Prob. 50ECh. 9.2 - Prob. 51ECh. 9.2 - Prob. 52ECh. 9.2 - Prob. 53ECh. 9.2 - Prob. 54ECh. 9.2 - Prob. 55ECh. 9.2 - Prob. 56ECh. 9.2 - Prob. 57ECh. 9.2 - Prob. 58ECh. 9.2 - Prob. 59ECh. 9.2 - Prob. 60ECh. 9.2 - Prob. 61ECh. 9.2 - Prob. 62ECh. 9.3 - Prob. 1.1CYUCh. 9.3 - Prob. 1.2CYUCh. 9.3 - Prob. 1.3CYUCh. 9.3 - Prob. 2.1CYUCh. 9.3 - Prob. 3.1CYUCh. 9.3 - Prob. 3.2CYUCh. 9.3 - Prob. 63ECh. 9.3 - Prob. 64ECh. 9.3 - Prob. 65ECh. 9.3 - Prob. 66ECh. 9.3 - Prob. 67ECh. 9.3 - Prob. 68ECh. 9.3 - Prob. 69ECh. 9.3 - Prob. 70ECh. 9.3 - Prob. 71ECh. 9.3 - Prob. 72ECh. 9.3 - Prob. 73ECh. 9.3 - Prob. 74ECh. 9.3 - Prob. 75ECh. 9.3 - Prob. 76ECh. 9.3 - Prob. 77ECh. 9.3 - Prob. 78ECh. 9.3 - Prob. 79ECh. 9.3 - Prob. 80ECh. 9.3 - Prob. 81ECh. 9.3 - Prob. 82ECh. 9.3 - Prob. 83ECh. 9.3 - Prob. 84ECh. 9.3 - Prob. 85ECh. 9.3 - Prob. 86ECh. 9.3 - Prob. 87ECh. 9.3 - Prob. 88ECh. 9.3 - Prob. 89ECh. 9.3 - Prob. 90ECh. 9.3 - Prob. 91ECh. 9.3 - Prob. 92ECh. 9.3 - Prob. 93ECh. 9.3 - Prob. 94ECh. 9.3 - Prob. 95ECh. 9.3 - Prob. 96ECh. 9.3 - Prob. 97ECh. 9.3 - Prob. 98ECh. 9.3 - Prob. 99ECh. 9.3 - Prob. 100ECh. 9.3 - Prob. 101ECh. 9.3 - Prob. 102ECh. 9.3 - Prob. 103ECh. 9.3 - Prob. 104ECh. 9.3 - Prob. 105ECh. 9.3 - Prob. 106ECh. 9.3 - Prob. 107ECh. 9.3 - Prob. 108ECh. 9 - Prob. 1CRECh. 9 - Prob. 2CRECh. 9 - Prob. 3CRECh. 9 - Prob. 4CRECh. 9 - Prob. 5CRECh. 9 - Prob. 6CRECh. 9 - Prob. 7CRECh. 9 - Prob. 8CRECh. 9 - Prob. 9CRECh. 9 - Prob. 1PTCh. 9 - Prob. 2PTCh. 9 - Prob. 3PTCh. 9 - Prob. 4PTCh. 9 - Prob. 5PTCh. 9 - Prob. 6PTCh. 9 - Prob. 7PTCh. 9 - Prob. 8PTCh. 9 - Prob. 9PTCh. 9 - Prob. 10PTCh. 9 - Prob. 11PTCh. 9 - Prob. 12PTCh. 9 - Prob. 13PT
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