MATLAB: A Practical Introduction to Programming and Problem Solving
5th Edition
ISBN: 9780128163450
Author: ATTAWAY
Publisher: ELSEVIER
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Chapter 7, Problem 7.5P
To determine
To explain:
The results of the following expressions that are given below, and then verify the answers of the expressions using MATLAB.
dept = "Electrical";
strfind(dept,'e')
strfind(lower(dept),'e')
phone_no = '703-987-1234';
[area_code, rest] = strtok(phone_no,'-')
rest = rest(2:end)
strcmpi('Hi','HI')
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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 7 Solutions
MATLAB: A Practical Introduction to Programming and Problem Solving
Ch. 7 - Prob. 7.1PCh. 7 - Prob. 7.2PCh. 7 - Prob. 7.3PCh. 7 - Prob. 7.4PCh. 7 - Prob. 7.5PCh. 7 - Prob. 7.6PCh. 7 - Prob. 1ECh. 7 - Prob. 2ECh. 7 - Prob. 3ECh. 7 - Prob. 4E
Ch. 7 - Prob. 5ECh. 7 - Prob. 6ECh. 7 - Prob. 7ECh. 7 - Prob. 8ECh. 7 - Prob. 9ECh. 7 - Prob. 10ECh. 7 - Prob. 11ECh. 7 - Prob. 12ECh. 7 - Prob. 13ECh. 7 - Prob. 14ECh. 7 - Prob. 15ECh. 7 - Prob. 16ECh. 7 - Prob. 17ECh. 7 - Prob. 18ECh. 7 - Prob. 19ECh. 7 - Prob. 20ECh. 7 - Prob. 21ECh. 7 - Prob. 22ECh. 7 - Prob. 23ECh. 7 - Prob. 24ECh. 7 - Prob. 25ECh. 7 - Prob. 26ECh. 7 - Prob. 27ECh. 7 - Prob. 28ECh. 7 - Prob. 29ECh. 7 - Prob. 30ECh. 7 - Prob. 31ECh. 7 - Prob. 32ECh. 7 - Prob. 33ECh. 7 - Prob. 34ECh. 7 - Prob. 35ECh. 7 - Prob. 36ECh. 7 - Prob. 37E
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