Hw2 MS-4373
Thomas Farrell
2/9/2022
Q2
The KNN classifier is mainly used to solve classification problems that have a qualitive output. Whereas the KNN regression method is used to solve problems that have a quantitative response.
Q9
Q9
- Using the name function it displayed all the column headers in the data. Then I had it compute a correlation matrix from the data excluding the name column because that variable is qualitative data.
names(Auto)## [1] "mpg" "cylinders" "displacement" "horsepower" "weight"
## [6] "acceleration" "year" "origin" "name"cor(Auto[1:8])## mpg cylinders displacement horsepower weight
## mpg 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442
## cylinders -0.7776175 1.0000000 0.9508233 0.8429834 0.8975273
## displacement -0.8051269 0.9508233 1.0000000 0.8972570 0.9329944
## horsepower -0.7784268 0.8429834 0.8972570 1.0000000 0.8645377
## weight -0.8322442 0.8975273 0.9329944 0.8645377 1.0000000
## acceleration 0.4233285 -0.5046834 -0.5438005 -0.6891955 -0.4168392
## year 0.5805410 -0.3456474 -0.3698552 -0.4163615 -0.3091199
## origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054
## acceleration year origin
## mpg 0.4233285 0.5805410 0.5652088
## cylinders -0.5046834 -0.3456474 -0.5689316
## displacement -0.5438005 -0.3698552 -0.6145351
## horsepower -0.6891955 -0.4163615 -0.4551715
## weight -0.4168392 -0.3091199 -0.5850054
## acceleration 1.0000000 0.2903161 0.2127458
## year 0.2903161 1.0000000 0.1815277
## origin 0.2127458 0.1815277 1.0000000Q9
fit1 = lm(mpg ~ . - name, data = Auto)
summary(fit1)##
## Call:
## lm(formula = mpg ~ . - name, data = Auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.5903 -2.1565 -0.1169 1.8690 13.0604
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -17.218435 4.644294 -3.707 0.00024 ***
## cylinders -0.493376 0.323282 -1.526 0.12780
## displacement 0.019896 0.007515 2.647 0.00844 **
## horsepower -0.016951 0.013787 -1.230 0.21963
## weight -0.006474 0.000652 -9.929 < 2e-16 ***
## acceleration 0.080576 0.098845 0.815 0.41548
## year 0.750773 0.050973 14.729 < 2e-16 ***
## origin 1.426141 0.278136 5.127 4.67e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.328 on 384 degrees of freedom
## Multiple R-squared: 0.8215, Adjusted R-squared: 0.8182
## F-statistic: 252.4 on 7 and 384 DF, p-value: < 2.2e-161.According to the P-value of the f statistic of 2.2e-16 we can conclude that there is a correlation between the predictors and the response. 2. To figure out which predictors are statistically significant we can check the p-values associated with each predictor. We find from the data that the only predictors not significant are cylinders, horsepower, and acceleration. 3. The coefficient for the variable “Year” says that the average effect of 1 year results in an increase in mpg by .75. Concluding that as newer cars are released they are becoming more fuel efficient.
Q9
par(mfrow = c(2, 2))
plot(fit1)
The plot of residuals vs. fitted data presents a data set that is non-linear in nature. We can also see that in the data of residuals vs. leverage there are a few outliers below -2 and higher than positive 2.
Q9
fit2 = lm(mpg ~ cylinders * displacement+displacement * weight, data = Auto[,1:8])
summary(fit2)##
## Call:
## lm(formula = mpg ~ cylinders * displacement + displacement *
## weight, data = Auto[, 1:8])
##
## Residuals:
## Min 1Q Median 3Q Max
## -13.2934 -2.5184 -0.3476 1.8399 17.7723
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.262e+01 2.237e+00 23.519 < 2e-16 ***
## cylinders 7.606e-01 7.669e-01 0.992 0.322
## displacement -7.351e-02 1.669e-02 -4.403 1.38e-05 ***
## weight -9.888e-03 1.329e-03 -7.438 6.69e-13 ***
## cylinders:displacement -2.986e-03 3.426e-03 -0.872 0.384
## displacement:weight 2.128e-05 5.002e-06 4.254 2.64e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.103 on 386 degrees of freedom
## Multiple R-squared: 0.7272, Adjusted R-squared: 0.7237
## F-statistic: 205.8 on 5 and 386 DF, p-value: < 2.2e-16From the P-values we can see that the relationship between cylinders and displacement is not significant while the relationship between displacement and weight is significant.
Q9
par(mfrow = c(2, 2))
plot(log(Auto$horsepower), Auto$mpg)
plot(sqrt(Auto$horsepower), Auto$mpg)
plot((Auto$horsepower)^2, Auto$mpg)
We can see in the top left that is the plot given by the log function and it appears to be the most linear of the three. Then on the top right we have the sqrt function plot and on the bottom left we have the squared function plot.
Q10
data("Carseats")
cr1 = lm(Sales ~ Price + Urban + US, data = Carseats)
summary(cr1)##
## Call:
## lm(formula = Sales ~ Price + Urban + US, data = Carseats)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.9206 -1.6220 -0.0564 1.5786 7.0581
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 13.043469 0.651012 20.036 < 2e-16 ***
## Price -0.054459 0.005242 -10.389 < 2e-16 ***
## UrbanYes -0.021916 0.271650 -0.081 0.936
## USYes 1.200573 0.259042 4.635 4.86e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.472 on 396 degrees of freedom
## Multiple R-squared: 0.2393, Adjusted R-squared: 0.2335
## F-statistic: 41.52 on 3 and 396 DF, p-value: < 2.2e-16B. To interpret each variable we must first look at “Price” variable which can be interpreted as for every 1 dollar increase there is a decrease of 54.459. The coefficient of “Urban” can be interpreted as on average unit sales in urban locations are 21.91 units less than rural locations. The coefficient of “US” can be interpreted as unit sales in us stores are 1200,57 more than in non-U.S stores.
C. Sales = 13.043 + (-.054459) X Price + (-.021916) X Urban + (1.200573) X US
D. We reject the null hypothesis for the variables of “Price” and “US”.
Q10
f4 = lm(Sales ~ Price + US, data = Carseats)
summary(f4)##
## Call:
## lm(formula = Sales ~ Price + US, data = Carseats)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.9269 -1.6286 -0.0574 1.5766 7.0515
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 13.03079 0.63098 20.652 < 2e-16 ***
## Price -0.05448 0.00523 -10.416 < 2e-16 ***
## USYes 1.19964 0.25846 4.641 4.71e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.469 on 397 degrees of freedom
## Multiple R-squared: 0.2393, Adjusted R-squared: 0.2354
## F-statistic: 62.43 on 2 and 397 DF, p-value: < 2.2e-16F. When looking at the two models we can see that the new smaller model constructed fits better as it has a better r-squared.
Q10
confint(f4)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632Computing of a 95% confidence interval for the smaller data model.
Q10
par(mfrow = c(2,2))
plot(f4)
This code shows that the plot of residuals vs leverage shows outliers between higher than 2 and lower than -2.
Q12
A.The coefficients estimate will be the same when the sum of x^2 is equal to the sum of y^2.
x = 1:100
sum(x^2)## [1] 338350y = 2*x + rnorm(100, sd = .1)
sum(y^2)## [1] 1353448ft.y = lm(y ~ x + 0)
ft.x = lm(x ~ y + 0)
summary(ft.y)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.183244 -0.065632 -0.000408 0.064997 0.219022
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 2.00003 0.00016 12500 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.09307 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.563e+08 on 1 and 99 DF, p-value: < 2.2e-16summary(ft.x)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.10949 -0.03248 0.00025 0.03283 0.09165
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 5e-01 4e-05 12500 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.04653 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 1.563e+08 on 1 and 99 DF, p-value: < 2.2e-16These lines of code will make an observation X with 100 observations. We will then create a y-variable that will create a normal distribution of 100 observations but to make the coefficient estimates different we will include 2 * X in the equation.
x = 1:100
sum(x^2)## [1] 338350y = 100:1
sum(y^2)## [1] 338350ft.y = lm(y ~ x + 0)
ft.x = lm(x ~ y + 0)
summary(ft.y)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -49.75 -12.44 24.87 62.18 99.49
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 0.5075 0.0866 5.86 6.09e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 50.37 on 99 degrees of freedom
## Multiple R-squared: 0.2575, Adjusted R-squared: 0.25
## F-statistic: 34.34 on 1 and 99 DF, p-value: 6.094e-08summary(ft.x)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -49.75 -12.44 24.87 62.18 99.49
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 0.5075 0.0866 5.86 6.09e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 50.37 on 99 degrees of freedom
## Multiple R-squared: 0.2575, Adjusted R-squared: 0.25
## F-statistic: 34.34 on 1 and 99 DF, p-value: 6.094e-08