The Analysis of Biological Data
2nd Edition
ISBN: 9781936221486
Author: Michael C. Whitlock, Dolph Schluter
Publisher: W. H. Freeman
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Chapter 2, Problem 1PP
(a)
To determine
To find: To estimate the relative frequency of the shaded area by eye.
(a)
Expert Solution
Answer to Problem 1PP
The relative frequency of the shaded area is
Explanation of Solution
Given:
Calculation:.The area of the rectangular bar of a histogram is proportional to the relative frequency of the considering numerical variable.
Interpretation:As the area of the rectangular bar of the histogram proportional to the relative frequency, the area of the shaded bar should be estimated by eye, comparing to the total area of the graph. The width of each bar of the histogram is equal to one another. Therefore the area of the shaded part is about one-fourth of the total area of the graph. Therefore the relative frequency of the shaded area is
(b)
To determine
To find: To estimate the relative frequency of the shaded area by eye.
(b)
Expert Solution
Answer to Problem 1PP
The relative frequency of the shaded area is
Explanation of Solution
Given:
Calculation:.The area of the rectangular bar of a histogram is proportional to the relative frequency of the considering numerical variable.
Interpretation: As the area of the rectangular bar of the histogram proportional to the relative frequency, the area of the shaded bar should be estimated by eye, comparing to the total area of the graph. The width of each bar of the histogram is equal to one another and height of last two bars is exactly half of first two bars. Which makes the total area of last two bars equal to one complete bar. Therefore the area of the shaded part is about one-third of the total area of the graph. Therefore the relative frequency of the shaded area is
(c)
To determine
To find: To estimate the relative frequency of the shaded area by eye.
(c)
Expert Solution
Answer to Problem 1PP
The relative frequency of the shaded area is
Explanation of Solution
Given:
Calculation:. The area of the rectangular bar of a histogram is proportional to the relative frequency of the considering numerical variable.
Interpretation: As the area of the rectangular bar of the histogram proportional to the relative frequency, the area of the shaded bar should be estimated by eye, comparing to the total area of the graph. The width of each bar of the histogram is equal to one another. Therefore the area of the shaded part is about one-third of the total area of the graph. Therefore the relative frequency of the shaded area is
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(c) Because logistic regression predicts probabilities of outcomes, observations used to build a logistic regression model need not be independent.
A. false: all observations must be independent
B. true
C. false: only observations with the same outcome need to be independent
I ANSWERED: A. false: all observations must be independent.
(This was marked wrong but I have no idea why. Isn't this a basic assumption of logistic regression)
Business discuss
Spam filters are built on principles similar to those used in logistic regression. We fit a probability that each message is spam or not spam. We have several variables for each email. Here are a few: to_multiple=1 if there are multiple recipients, winner=1 if the word 'winner' appears in the subject line, format=1 if the email is poorly formatted, re_subj=1 if "re" appears in the subject line. A logistic model was fit to a dataset with the following output:
Estimate
SE
Z
Pr(>|Z|)
(Intercept)
-0.8161
0.086
-9.4895
0
to_multiple
-2.5651
0.3052
-8.4047
0
winner
1.5801
0.3156
5.0067
0
format
-0.1528
0.1136
-1.3451
0.1786
re_subj
-2.8401
0.363
-7.824
0
(a) Write down the model using the coefficients from the model fit.log_odds(spam) = -0.8161 + -2.5651 + to_multiple + 1.5801 winner + -0.1528 format + -2.8401 re_subj(b) Suppose we have an observation where to_multiple=0, winner=1, format=0, and re_subj=0. What is the predicted probability that this message is spam?…
Chapter 2 Solutions
The Analysis of Biological Data
Ch. 2 - Prob. 1PPCh. 2 - Prob. 2PPCh. 2 - Prob. 3PPCh. 2 - Prob. 4PPCh. 2 - Prob. 5PPCh. 2 - Prob. 6PPCh. 2 - Prob. 7PPCh. 2 - Prob. 8PPCh. 2 - Prob. 9PPCh. 2 - Prob. 10PP
Ch. 2 - Prob. 11PPCh. 2 - Prob. 12PPCh. 2 - Prob. 13PPCh. 2 - Prob. 14PPCh. 2 - Prob. 15PPCh. 2 - Prob. 16PPCh. 2 - Prob. 17PPCh. 2 - Prob. 18PPCh. 2 - Prob. 19APCh. 2 - Prob. 20APCh. 2 - Prob. 21APCh. 2 - Prob. 22APCh. 2 - Prob. 23APCh. 2 - Prob. 24APCh. 2 - Prob. 25APCh. 2 - Prob. 26APCh. 2 - Prob. 27APCh. 2 - Prob. 28APCh. 2 - Prob. 29APCh. 2 - Prob. 30APCh. 2 - Prob. 31APCh. 2 - Prob. 32APCh. 2 - Prob. 33APCh. 2 - Prob. 34APCh. 2 - Prob. 35APCh. 2 - Prob. 36APCh. 2 - Prob. 37AP
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