MIDTERM – SOLUTION KEY
CODING QNS
Q21) Please estimate a linear regression model (using the lm function) with Personal as the
dependent variable and Room.Board as the independent variable. What are the model’s R-
squared and adjusted R-squared values?
a) 0.00549, 0.048
b) 0.0143, 0.022
c) 0.0398, 0.0385
d) 0.0325, 0.0336
Answer: C (Week 1 Lesson 4)
library("ISLR")
data("College")
summary(lm(College$Personal~College$Room.Board))
##
## Call:
## lm(formula = College$Personal ~ College$Room.Board)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1153.1 -444.6 -92.3 316.0 5505.2
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1877.14827 97.64374 19.224 < 2e-16 ***
## College$Room.Board -0.12312 0.02173 -5.666 2.06e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 663.9 on 775 degrees of freedom
## Multiple R-squared: 0.03977, Adjusted R-squared: 0.03853
, ## F-statistic: 32.1 on 1 and 775 DF, p-value: 2.065e-08
Q22)Based on the linear-linear regression model in the previous question (with Personal as the
dependent variable and Room.Board as the independent variable), fit three nonlinear models
using those two variables. Based on their adjusted R-squared values, which one of the four
models is most appropriate to use?
a) Log-Linear
b) Log-Log
c) Linear-Linear
d) Linear-Log
Answer: B (Week 3 lesson 4)
summary(lm(log(College$Personal)~College$Room.Board))
## Call:
## lm(formula = log(College$Personal) ~ College$Room.Board)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.61024 -0.31235 0.03383 0.31037 1.77383
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.485e+00 6.992e-02 107.057 < 2e-16 ***
## College$Room.Board -9.187e-05 1.556e-05 -5.904 5.3e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4754 on 775 degrees of freedom
## Multiple R-squared: 0.04304, Adjusted R-squared: 0.04181
## F-statistic: 34.86 on 1 and 775 DF, p-value: 5.303e-09
summary(lm(log(College$Personal)~log(College$Room.Board)))
CODING QNS
Q21) Please estimate a linear regression model (using the lm function) with Personal as the
dependent variable and Room.Board as the independent variable. What are the model’s R-
squared and adjusted R-squared values?
a) 0.00549, 0.048
b) 0.0143, 0.022
c) 0.0398, 0.0385
d) 0.0325, 0.0336
Answer: C (Week 1 Lesson 4)
library("ISLR")
data("College")
summary(lm(College$Personal~College$Room.Board))
##
## Call:
## lm(formula = College$Personal ~ College$Room.Board)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1153.1 -444.6 -92.3 316.0 5505.2
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1877.14827 97.64374 19.224 < 2e-16 ***
## College$Room.Board -0.12312 0.02173 -5.666 2.06e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 663.9 on 775 degrees of freedom
## Multiple R-squared: 0.03977, Adjusted R-squared: 0.03853
, ## F-statistic: 32.1 on 1 and 775 DF, p-value: 2.065e-08
Q22)Based on the linear-linear regression model in the previous question (with Personal as the
dependent variable and Room.Board as the independent variable), fit three nonlinear models
using those two variables. Based on their adjusted R-squared values, which one of the four
models is most appropriate to use?
a) Log-Linear
b) Log-Log
c) Linear-Linear
d) Linear-Log
Answer: B (Week 3 lesson 4)
summary(lm(log(College$Personal)~College$Room.Board))
## Call:
## lm(formula = log(College$Personal) ~ College$Room.Board)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.61024 -0.31235 0.03383 0.31037 1.77383
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.485e+00 6.992e-02 107.057 < 2e-16 ***
## College$Room.Board -9.187e-05 1.556e-05 -5.904 5.3e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.4754 on 775 degrees of freedom
## Multiple R-squared: 0.04304, Adjusted R-squared: 0.04181
## F-statistic: 34.86 on 1 and 775 DF, p-value: 5.303e-09
summary(lm(log(College$Personal)~log(College$Room.Board)))
