Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319013387
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 2.6, Problem 121E

(a)

To determine

The explanatory variable and response variable.

(a)

Expert Solution
Check Mark

Answer to Problem 121E

Solution: “Age (years)” of the candidates is the explanatory variable and “Rejected (Yes/No)” is the response variable.

Explanation of Solution

The provided study is conducted to see the healthy teeth. If a person wants a soldier in the Spanish American War, he should have at least 8 teeth. The statement of the requirement is that “As long as an applicant has at least four sound teeth, one above and one below each, from each side of the mouth, and thus aimed at chewing it for the purpose of opposing it, it should be rejected.”

In the study, there are two variables. One is an explanatory variable and the other is a response variable. An explanatory variable is a variable that affects the value of response variable and a response variable measures the result of the study. In the study, “Age (years)” is considered as the explanatory variable and the “Rejected (Yes/No)” is the response variable.

(b)

To determine

To find: The joint distribution.

(b)

Expert Solution
Check Mark

Answer to Problem 121E

Solution: The joint distribution is provided below:

AgeRejectedUnder 202025253030353540Over 40Yes0.00020.00190.00330.00530.00860.0114No0.17610.23330.16630.13160.14230.1196

Explanation of Solution

Calculation: The joint distribution is computed by dividing the cell element by the total observation. The obtained joint distribution for under 20 can be calculated as follows:

Candidate who were rejected (under 20)=68334,321=0.0002

Candidate who were not rejected (Under 20)=58,884334,321=0.1761

Similarly, for other age group, the joint distribution can be calculated by using the above formula.

The joint distribution of the rejection data for the candidates classified by age is provided below:

AgeRejectedUnder 202025253030353540Over 40Yes0.00020.00190.00330.00530.00860.0114No0.17610.23330.16630.13160.14230.1196

From the table above, each cell shows a probability. For each cell, proportion is computed by dividing the cell entry by the total sample size. The collection of these proportions is the joint distribution of the two categorical variables. Because this is a distribution, the sum of the proportion should be 1. For this, the sum is 0.9999 that is approximately equal to 1.

(c)

To determine

To find: The two marginal distributions.

(c)

Expert Solution
Check Mark

Answer to Problem 121E

Solution: The marginal distribution of “Rejected” is given below:

Marginal distribution of “Rejected”

Yes

No

10,300334,321=0.03081

324,021334,321=0.96919

The marginal distribution of “Age” is given below:

Marginal distribution of “Age”

Under 20

20 25

25 30

30 35

35 40

Over 40

=58,952334,321=0.1763

=78,639334,321=0.2352

=56,711334,321=0.1696

=45,777334,321=0.1369

=50,456334,321=0.1509

=43,786334,321=0.1330

Explanation of Solution

Calculation: Now, the marginal distribution is computed by dividing the row or column totals by the overall total. The marginal distribution of “Rejected” is given below:

Marginal distribution of “Rejected”

Yes

No

10,300334,321=0.03081

324,021334,321=0.96919

The marginal distribution of “Age” is given below:

Marginal distribution of “Age”

Under 20

20–25

25–30

30–35

35–40

Over 40

=58,952334,321=0.1763

=78,639334,321=0.2352

=56,711334,321=0.1696

=45,777334,321=0.1369

=50,456334,321=0.1509

=43,786334,321=0.1330

It is clearer to present these distributions as percent of the table total. The marginal distribution of the “Rejection” and “Age” is provided below:

Marginal distribution of “Rejected”

Yes

No

3%

97%

Marginal distribution of “Age”

Under 20

20–25

25–30

30–35

35–40

Over 40

17.63%

23.52%

16.96%

13.69%

15.09%

13.30%

Interpretation: This distribution does not provide any information about the relationship between the variables. The sum of the proportion should be 1.

(d)

To determine

The conditional distribution that will give the better explanation between the variables.

(d)

Expert Solution
Check Mark

Answer to Problem 121E

Solution: The conditional distribution of “Rejection given Age” will give the better explanation between the variables.

Explanation of Solution

In the study, there are two variables, that is, explanatory and response variables. An explanatory variable is “Age in years” and response variable is “Rejection (Yes/No).” The conditional distribution of “Rejection” for provided age is appropriate because “Age” is chosen as explanatory variable.

(e)

To determine

To find: The conditional distribution.

(e)

Expert Solution
Check Mark

Answer to Problem 121E

Solution: The conditional distribution is provided below:

AgeRejectedUnder 202025253030353540Over 40Yes0.00120.00820.01960.03890.05720.0868No0.99880.99180.98040.96110.94280.9132

Explanation of Solution

Calculation: The conditional distribution is obtained by dividing the row or column elements by the sum of that row or column observation. The conditional distribution of “Rejection” for given age can be calculated as follows:

Candidate who were rejected (Under 20)=6858,952=0.0012

Candidate who were not rejected (Under 20)=58,88458,952=0.9988

Similarly, the conditional distribution for the others can be calculated by using the above formula.

The conditional distribution of “Rejection for given age” is provided below:

AgeRejectedUnder 202025253030353540Over 40Yes0.00120.00820.01960.03890.05720.0868No0.99880.99180.98040.96110.94280.9132

Interpretation: When the value of one variable is conditioned in calculating the distribution of the other variable, we obtain a conditional distribution. From the above table, as the age increases, the number of rejected candidates also increases. This happens due to weak teeth in the older age.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Homework Let X1, X2, Xn be a random sample from f(x;0) where f(x; 0) = (-), 0 < x < ∞,0 € R Using Basu's theorem, show that Y = min{X} and Z =Σ(XY) are indep. -
Homework Let X1, X2, Xn be a random sample from f(x; 0) where f(x; 0) = e−(2-0), 0 < x < ∞,0 € R Using Basu's theorem, show that Y = min{X} and Z =Σ(XY) are indep.
An Arts group holds a raffle.  Each raffle ticket costs $2 and the raffle consists of 2500 tickets.  The prize is a vacation worth $3,000.    a. Determine your expected value if you buy one ticket.     b. Determine your expected value if you buy five tickets.     How much will the Arts group gain or lose if they sell all the tickets?

Chapter 2 Solutions

Introduction to the Practice of Statistics

Ch. 2.2 - Prob. 11UYKCh. 2.2 - Prob. 12UYKCh. 2.2 - Prob. 13UYKCh. 2.2 - Prob. 14UYKCh. 2.2 - Prob. 15UYKCh. 2.2 - Prob. 16ECh. 2.2 - Prob. 17ECh. 2.2 - Prob. 18ECh. 2.2 - Prob. 19ECh. 2.2 - Prob. 20ECh. 2.2 - Prob. 21ECh. 2.2 - Prob. 22ECh. 2.2 - Prob. 23ECh. 2.2 - Prob. 24ECh. 2.2 - Prob. 25ECh. 2.2 - Prob. 26ECh. 2.2 - Prob. 27ECh. 2.2 - Prob. 28ECh. 2.2 - Prob. 29ECh. 2.2 - Prob. 30ECh. 2.2 - Prob. 31ECh. 2.2 - Prob. 32ECh. 2.2 - Prob. 33ECh. 2.2 - Prob. 34ECh. 2.2 - Prob. 35ECh. 2.2 - Prob. 36ECh. 2.2 - Prob. 37ECh. 2.3 - Prob. 38UYKCh. 2.3 - Prob. 39UYKCh. 2.3 - Prob. 40ECh. 2.3 - Prob. 41ECh. 2.3 - Prob. 42ECh. 2.3 - Prob. 43ECh. 2.3 - Prob. 44ECh. 2.3 - Prob. 45ECh. 2.3 - Prob. 46ECh. 2.3 - Prob. 47ECh. 2.3 - Prob. 48ECh. 2.3 - Prob. 49ECh. 2.3 - Prob. 50ECh. 2.3 - Prob. 51ECh. 2.3 - Prob. 52ECh. 2.3 - Prob. 53ECh. 2.3 - Prob. 54ECh. 2.3 - Prob. 55ECh. 2.3 - Prob. 56ECh. 2.3 - Prob. 57ECh. 2.3 - Prob. 58ECh. 2.3 - Prob. 59ECh. 2.3 - Prob. 60ECh. 2.4 - Prob. 61UYKCh. 2.4 - Prob. 62UYKCh. 2.4 - Prob. 63UYKCh. 2.4 - Prob. 64UYKCh. 2.4 - Prob. 65ECh. 2.4 - Prob. 66ECh. 2.4 - Prob. 67ECh. 2.4 - Prob. 68ECh. 2.4 - Prob. 69ECh. 2.4 - Prob. 70ECh. 2.4 - Prob. 71ECh. 2.4 - Prob. 72ECh. 2.4 - Prob. 73ECh. 2.4 - Prob. 74ECh. 2.4 - Prob. 75ECh. 2.4 - Prob. 76ECh. 2.4 - Prob. 77ECh. 2.4 - Prob. 78ECh. 2.4 - Prob. 79ECh. 2.4 - Prob. 80ECh. 2.4 - Prob. 81ECh. 2.4 - Prob. 82ECh. 2.4 - Prob. 83ECh. 2.4 - Prob. 84ECh. 2.4 - Prob. 85ECh. 2.4 - Prob. 86ECh. 2.4 - Prob. 87ECh. 2.4 - Prob. 88ECh. 2.4 - Prob. 89ECh. 2.4 - Prob. 90ECh. 2.4 - Prob. 91ECh. 2.5 - Prob. 92UYKCh. 2.5 - Prob. 93UYKCh. 2.5 - Prob. 94ECh. 2.5 - Prob. 95ECh. 2.5 - Prob. 96ECh. 2.5 - Prob. 97ECh. 2.5 - Prob. 98ECh. 2.5 - Prob. 99ECh. 2.5 - Prob. 100ECh. 2.5 - Prob. 101ECh. 2.5 - Prob. 102ECh. 2.5 - Prob. 103ECh. 2.5 - Prob. 104ECh. 2.5 - Prob. 105ECh. 2.5 - Prob. 106ECh. 2.5 - Prob. 107ECh. 2.5 - Prob. 108ECh. 2.5 - Prob. 110ECh. 2.5 - Prob. 111ECh. 2.5 - Prob. 112ECh. 2.6 - Prob. 113UYKCh. 2.6 - Prob. 114UYKCh. 2.6 - Prob. 115UYKCh. 2.6 - Prob. 116UYKCh. 2.6 - Prob. 117UYKCh. 2.6 - Prob. 118UYKCh. 2.6 - Prob. 119ECh. 2.6 - Prob. 120ECh. 2.6 - Prob. 121ECh. 2.6 - Prob. 122ECh. 2.6 - Prob. 123ECh. 2.6 - Prob. 124ECh. 2.6 - Prob. 125ECh. 2.6 - Prob. 126ECh. 2.6 - Prob. 127ECh. 2.6 - Prob. 128ECh. 2.6 - Prob. 129ECh. 2.6 - Prob. 130ECh. 2.7 - Prob. 131ECh. 2.7 - Prob. 132ECh. 2.7 - Prob. 133ECh. 2.7 - Prob. 134ECh. 2.7 - Prob. 135ECh. 2.7 - Prob. 136ECh. 2.7 - Prob. 137ECh. 2.7 - Prob. 138ECh. 2.7 - Prob. 139ECh. 2.7 - Prob. 140ECh. 2.7 - Prob. 141ECh. 2.7 - Prob. 142ECh. 2.7 - Prob. 143ECh. 2 - Prob. 144ECh. 2 - Prob. 145ECh. 2 - Prob. 146ECh. 2 - Prob. 147ECh. 2 - Prob. 148ECh. 2 - Prob. 149ECh. 2 - Prob. 150ECh. 2 - Prob. 151ECh. 2 - Prob. 152ECh. 2 - Prob. 153ECh. 2 - Prob. 154ECh. 2 - Prob. 155ECh. 2 - Prob. 156ECh. 2 - Prob. 157ECh. 2 - Prob. 158ECh. 2 - Prob. 159ECh. 2 - Prob. 160ECh. 2 - Prob. 161ECh. 2 - Prob. 162ECh. 2 - Prob. 163ECh. 2 - Prob. 164ECh. 2 - Prob. 165ECh. 2 - Prob. 166ECh. 2 - Prob. 167ECh. 2 - Prob. 168ECh. 2 - Prob. 169ECh. 2 - Prob. 170ECh. 2 - Prob. 171ECh. 2 - Prob. 172ECh. 2 - Prob. 173ECh. 2 - Prob. 174ECh. 2 - Prob. 175ECh. 2 - Prob. 176E
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
Propositional Logic, Propositional Variables & Compound Propositions; Author: Neso Academy;https://www.youtube.com/watch?v=Ib5njCwNMdk;License: Standard YouTube License, CC-BY
Propositional Logic - Discrete math; Author: Charles Edeki - Math Computer Science Programming;https://www.youtube.com/watch?v=rL_8y2v1Guw;License: Standard YouTube License, CC-BY
DM-12-Propositional Logic-Basics; Author: GATEBOOK VIDEO LECTURES;https://www.youtube.com/watch?v=pzUBrJLIESU;License: Standard Youtube License
Lecture 1 - Propositional Logic; Author: nptelhrd;https://www.youtube.com/watch?v=xlUFkMKSB3Y;License: Standard YouTube License, CC-BY
MFCS unit-1 || Part:1 || JNTU || Well formed formula || propositional calculus || truth tables; Author: Learn with Smily;https://www.youtube.com/watch?v=XV15Q4mCcHc;License: Standard YouTube License, CC-BY