STA 6543 : Assignment 2
Mihreteab Teklehaimanot
Exercise 2. Carefully explain the differences between the KNN classifier and KNN regression methods.
- KNN regression is similar to the KNN classifiers. While KNN classifier results in the classification of Y, which is qualitative, the KNN Regression method predict quantitative value of y for a given value of x. KNN regression considers k closest points to x in training the data and take the average of the responses.
Exercise 9. auto
auto = read.csv("auto.csv")auto$horsepower = as.integer(auto$horsepower)
auto = na.omit(auto)
str(auto)## 'data.frame': 392 obs. of 9 variables:
## $ mpg : num 18 15 18 16 17 15 14 14 14 15 ...
## $ cylinders : int 8 8 8 8 8 8 8 8 8 8 ...
## $ displacement: num 307 350 318 304 302 429 454 440 455 390 ...
## $ horsepower : int 130 165 150 150 140 198 220 215 225 190 ...
## $ weight : int 3504 3693 3436 3433 3449 4341 4354 4312 4425 3850 ...
## $ acceleration: num 12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
## $ year : int 70 70 70 70 70 70 70 70 70 70 ...
## $ origin : int 1 1 1 1 1 1 1 1 1 1 ...
## $ name : chr "chevrolet chevelle malibu" "buick skylark 320" "plymouth satellite" "amc rebel sst" ...
## - attr(*, "na.action")= 'omit' Named int [1:5] 33 127 331 337 355
## ..- attr(*, "names")= chr [1:5] "33" "127" "331" "337" ...9. (a) Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(auto[,1:7]) # can't produce scatterplot matrix with non numeric variables, name is excluded
9. (b) Compute the matrix of correlations between the variables using the function cor().
cor(auto[,1:7])## 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
## acceleration year
## mpg 0.4233285 0.5805410
## cylinders -0.5046834 -0.3456474
## displacement -0.5438005 -0.3698552
## horsepower -0.6891955 -0.4163615
## weight -0.4168392 -0.3091199
## acceleration 1.0000000 0.2903161
## year 0.2903161 1.00000009. (c) Use the lm() function to perform a multiple linear regression with mpg as the response and all other variables except name as the predictors. Use the summary() function to print the results. Comment on the output.
auto_lm = lm(mpg ~ . -name, data = auto)
summary(auto_lm)##
## 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-16i. Is there a relationship between the predictors and the response?
- The F-statistic (p-value = 2.2e-16) indicates that the model is significant and look at the summary of the model shows that there is relationship between the predictors and the response. Cylinders, hoursepower and acceleration are not statistically significant.
ii. Which predictors appear to have a statistically significant relationship to the response?
- Displacement, weight, year, origin
iii. What does the coefficient for the “year” variable suggest?
- The coefficient for the variable year is 0.750773 and it suggests that fuel efficiency (mpg) increases by 0.750773 mpg every year(one year increase), given all variables are constant.
9.(d) Use the plot() function to produce diagnostic plots of the linear regression fit.
par(mfrow = c(2,2))
plot(auto_lm)
Comment on any problems you see with the fit
- the Residual vs Fitted plot suggest that there is visible curve pattern in the residuals plots indicating non-linearity in the data and possible presence of heteroscedasticity nature.
- Normal Q-Q plot shows that the data deviates from the straight line for larger standard residuals values and is right skewed.
Do the residual plots suggest any unusually large outliers?
- The plot of standardized residuals versus leverage and sqrt(standardized residuals) vs Fitted values, indicate the presence of outliers.
- Possible outlier observations
- standardized residuals higher than 2 or lower than -2 on Residual vs Leverage plot
- sqrt(standardized residuals) > 1.5) on Scale-Location
Does the leverage plot identify any observations with unusually high leverage?
- yes, observation 14 has high leverage.
9. (e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
- Fit model with interaction term :cylinder:displacement
fit.int = (lm(mpg ~ .-name + cylinders:displacement, data = auto))
summary(fit.int)##
## Call:
## lm(formula = mpg ~ . - name + cylinders:displacement, data = auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -11.6081 -1.7833 -0.0465 1.6821 12.2617
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.7096590 4.6858582 -0.578 0.563426
## cylinders -2.6962123 0.4094916 -6.584 1.51e-10 ***
## displacement -0.0774797 0.0141535 -5.474 7.96e-08 ***
## horsepower -0.0476026 0.0133736 -3.559 0.000418 ***
## weight -0.0052339 0.0006253 -8.370 1.10e-15 ***
## acceleration 0.0597997 0.0918038 0.651 0.515188
## year 0.7594500 0.0473354 16.044 < 2e-16 ***
## origin 0.7087399 0.2736917 2.590 0.009976 **
## cylinders:displacement 0.0136081 0.0017209 7.907 2.84e-14 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.089 on 383 degrees of freedom
## Multiple R-squared: 0.8465, Adjusted R-squared: 0.8433
## F-statistic: 264.1 on 8 and 383 DF, p-value: < 2.2e-16- In the linear model with cylinder and displacement interaction added, all but acceleration are statistically significant. The interaction term seems to improve the model. R2 improved from 82.15% to 84.65%.
- Fit model with interaction term: weight*displacement
fit.int1 = (lm(mpg ~ .-name + weight * displacement, data = auto))
summary(fit.int1)##
## Call:
## lm(formula = mpg ~ . - name + weight * displacement, data = auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.9027 -1.8092 -0.0946 1.5549 12.1687
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.389e+00 4.301e+00 -1.253 0.2109
## cylinders 1.175e-01 2.943e-01 0.399 0.6899
## displacement -6.837e-02 1.104e-02 -6.193 1.52e-09 ***
## horsepower -3.280e-02 1.238e-02 -2.649 0.0084 **
## weight -1.064e-02 7.136e-04 -14.915 < 2e-16 ***
## acceleration 6.724e-02 8.805e-02 0.764 0.4455
## year 7.852e-01 4.553e-02 17.246 < 2e-16 ***
## origin 5.610e-01 2.622e-01 2.139 0.0331 *
## displacement:weight 2.269e-05 2.257e-06 10.054 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.964 on 383 degrees of freedom
## Multiple R-squared: 0.8588, Adjusted R-squared: 0.8558
## F-statistic: 291.1 on 8 and 383 DF, p-value: < 2.2e-16- In the linear model with weight and displacement interaction term added, all but the variables cylinders and acceleration are statistically significant. Adding the weight:displacement interaction term seems to improve the model. The goodness of fit(R2) improved from 82.15% to 85.88%. This interaction term have improved the original model more than the cylinder:displacement interaction term.
9. (f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
- poly(cylinders, 3) transformation of cylinder.
fit_cyl = lm(mpg ~ cylinders, auto)
fit_ptrans = lm(mpg ~ poly(cylinders,3), auto)
summary(fit_cyl)##
## Call:
## lm(formula = mpg ~ cylinders, data = auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -14.2413 -3.1832 -0.6332 2.5491 17.9168
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 42.9155 0.8349 51.40 <2e-16 ***
## cylinders -3.5581 0.1457 -24.43 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.914 on 390 degrees of freedom
## Multiple R-squared: 0.6047, Adjusted R-squared: 0.6037
## F-statistic: 596.6 on 1 and 390 DF, p-value: < 2.2e-16summary(fit_ptrans)##
## Call:
## lm(formula = mpg ~ poly(cylinders, 3), data = auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -11.2869 -2.9058 -0.9627 2.3403 18.0218
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 23.446 0.237 98.919 < 2e-16 ***
## poly(cylinders, 3)1 -120.013 4.693 -25.574 < 2e-16 ***
## poly(cylinders, 3)2 8.113 4.693 1.729 0.0846 .
## poly(cylinders, 3)3 28.379 4.693 6.047 3.46e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.693 on 388 degrees of freedom
## Multiple R-squared: 0.6413, Adjusted R-squared: 0.6385
## F-statistic: 231.2 on 3 and 388 DF, p-value: < 2.2e-16anova(fit_cyl,fit_ptrans)- The linear regression model with the polynomial transformation (3) on cylinders included, shows that x2 transformation is not significant, but the x3 transformation is significant. The anova test suggest that this model is superior to the original model with cylinder only. The R2 also increased showing that transformation of the variable improved the fit of the model.
- log trasformation of horsepower
lm.mha = lm(mpg ~ horsepower, auto)
lm.trans = lm(mpg ~ horsepower + log(horsepower), auto)
summary(lm.mha)##
## Call:
## lm(formula = mpg ~ horsepower, data = auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -13.5710 -3.2592 -0.3435 2.7630 16.9240
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 39.935861 0.717499 55.66 <2e-16 ***
## horsepower -0.157845 0.006446 -24.49 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.906 on 390 degrees of freedom
## Multiple R-squared: 0.6059, Adjusted R-squared: 0.6049
## F-statistic: 599.7 on 1 and 390 DF, p-value: < 2.2e-16summary(lm.trans)##
## Call:
## lm(formula = mpg ~ horsepower + log(horsepower), data = auto)
##
## Residuals:
## Min 1Q Median 3Q Max
## -14.5118 -2.5018 -0.2533 2.4446 15.3102
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 156.04057 12.08267 12.914 < 2e-16 ***
## horsepower 0.11846 0.02929 4.044 6.34e-05 ***
## log(horsepower) -31.59815 3.28363 -9.623 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.415 on 389 degrees of freedom
## Multiple R-squared: 0.6817, Adjusted R-squared: 0.6801
## F-statistic: 416.6 on 2 and 389 DF, p-value: < 2.2e-16test which model is superior
anova(lm.mha, lm.trans)- In the full model with horsepower and log transformation of horsepower, both variables are significant. The anova test suggest that this model is superior to the model with only horsepower. The R2 also improved from 61 to 68 %.
10. This question should be answered using the Carseats data set.
10. (a) Fit a multiple regression model to predict Sales using Price, Urban, and US.
cseats_lm = lm(Sales ~ Price + Urban + US, Carseats)
summary(cseats_lm)##
## 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-1610. (b) Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
Price is significant and is inversely (negatively) related to Sales and this indicates that for every dollar price increase, sales decreases by $54.
The dummy variable USYes is significant and sales in stores located in US are $1,200 higher than sales in stores located outside US.
Urban is not significant and doesn’t have effect on sales.
10. (c) Write out the model in equation form, being careful to handle the qualitative variables properly.
- Sales = 13.04 - 0.055* Price - 0.022* UrbanYes + 1.20* USYes
10. (d) For which of the predictors can you reject the null hypothesis H0 : βj = 0?
- Based on the p-value, we can not reject the null hypothesis for the Price and USYes.
10. (e) On the basis of your response to the previous question, fit a smaller model that only uses the predictors for which there is evidence of association with the outcome.
- Sales = 13.04 - 0.055* Price + 1.20* USYes
cseats_lm2 = lm(Sales ~ Price + US, Carseats)
summary(cseats_lm2)##
## 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-1610. (f) How well do the models in (a) and (e) fit the data?
Based on the F-statistics and its smaller p-value both models are significant. Both models fit the data similarly, with the model from (e) fitting the data slightly better (adjusted R2 - 23.54 vs 23.35%).
However, the R-square and Adjust R-squared values for both models are relatively small and only 23.54 (23.35) % of the variation is explained by the models.
10. (g) Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(cseats_lm2, level = 0.95 )## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.7077663210. (h) Is there evidence of outliers or high leverage observations in the model from (e)?
par(mfrow=c(2,2))
plot(cseats_lm2)
The influence.measures() function provides potential influential observation of the model.
summary(influence.measures(cseats_lm2))## Potentially influential observations of
## lm(formula = Sales ~ Price + US, data = Carseats) :
##
## dfb.1_ dfb.Pric dfb.USYs dffit cov.r cook.d hat
## 26 0.24 -0.18 -0.17 0.28_* 0.97_* 0.03 0.01
## 29 -0.10 0.10 -0.10 -0.18 0.97_* 0.01 0.01
## 43 -0.11 0.10 0.03 -0.11 1.05_* 0.00 0.04_*
## 50 -0.10 0.17 -0.17 0.26_* 0.98 0.02 0.01
## 51 -0.05 0.05 -0.11 -0.18 0.95_* 0.01 0.00
## 58 -0.05 -0.02 0.16 -0.20 0.97_* 0.01 0.01
## 69 -0.09 0.10 0.09 0.19 0.96_* 0.01 0.01
## 126 -0.07 0.06 0.03 -0.07 1.03_* 0.00 0.03_*
## 160 0.00 0.00 0.00 0.01 1.02_* 0.00 0.02
## 166 0.21 -0.23 -0.04 -0.24 1.02 0.02 0.03_*
## 172 0.06 -0.07 0.02 0.08 1.03_* 0.00 0.02
## 175 0.14 -0.19 0.09 -0.21 1.03_* 0.02 0.03_*
## 210 -0.14 0.15 -0.10 -0.22 0.97_* 0.02 0.01
## 270 -0.03 0.05 -0.03 0.06 1.03_* 0.00 0.02
## 298 -0.06 0.06 -0.09 -0.15 0.97_* 0.01 0.00
## 314 -0.05 0.04 0.02 -0.05 1.03_* 0.00 0.02_*
## 353 -0.02 0.03 0.09 0.15 0.97_* 0.01 0.00
## 357 0.02 -0.02 0.02 -0.03 1.03_* 0.00 0.02
## 368 0.26 -0.23 -0.11 0.27_* 1.01 0.02 0.02_*
## 377 0.14 -0.15 0.12 0.24 0.95_* 0.02 0.01
## 384 0.00 0.00 0.00 0.00 1.02_* 0.00 0.02
## 387 -0.03 0.04 -0.03 0.05 1.02_* 0.00 0.02
## 396 -0.05 0.05 0.08 0.14 0.98_* 0.01 0.00- It shows that there are a few high leverage or outlier points in the model. These points can be removed and refitted.
12. This problem involves simple linear regression without an intercept.
12. (a) Recall that the coefficient estimate β for the linear regression of Y onto X without an intercept is given by (3.38). Under what circumstance is the coefficient estimate for the regression of X onto Y the same as the coefficient estimate for the regression of Y onto X?
- The coefficient estimate for the regression of X onto Y are the same as the coefficient estimate for the regression of Y onto , if the sum(x^2) and sum(y^2) are equal.
12. (b) Generate an example in R with n = 100 observations in which the coefficient estimate for the regression of X onto Y is different from the coefficient estimate for the regression of Y onto X.
x = rnorm(100)
y = 2^x + rnorm(100)
sum(x^2)## [1] 104.7884sum(y^2)## [1] 339.4378lm(x ~ y - 1) # -1 to ensure, there will be no intercept##
## Call:
## lm(formula = x ~ y - 1)
##
## Coefficients:
## y
## 0.3314lm(y ~ x - 1)##
## Call:
## lm(formula = y ~ x - 1)
##
## Coefficients:
## x
## 1.07412. (c) Generate an example in R with n = 100 observations in which the coefficient estimate for the regression of X onto Y is the same as the coefficient estimate for the regression of Y onto X.
set.seed(1)
x <- rnorm(100)
y <- -sample(x)
sum(x^2)## [1] 81.05509sum(y^2)## [1] 81.05509lm(x ~ y - 1)##
## Call:
## lm(formula = x ~ y - 1)
##
## Coefficients:
## y
## 0.07768lm(y ~ x - 1)##
## Call:
## lm(formula = y ~ x - 1)
##
## Coefficients:
## x
## 0.07768