Data Mining Assignment 2
Tim Harrison
2023-02-09
Problem 2
Carefully Explain KNN Classifier and KNN Regressor methods.
KNN calssifier is going to be used on group variable where KNN Regression methods will be used on continuous variables. With KNN Classifier will choose a category based off your neighbors where KNN Regressor will take the average of your neighbors.
Problem 9
This question involves the use of multiple linear regression on the Auto data set.
Produce a scatterplot matrix which includes all of the variables in the data set.
Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, cor() which is qualitative.
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. For instance:
- Is there a relationship between the predictors and the response?
- Which predictors appear to have a statistically significant relationship to the response?
- What does the coefficient for the year variable suggest?
There is a relationship between the predictors and the model is significant because the F-statsitic generats a low p-vlaue of <2.2e-16.The predictors that have a statistically significant relationship are displacment, weight, year and origin. The coefficient of the year vairable suggests that for every year newer your car is your gas mileage will increase by 3/4 of a gallon.
- Use the plot() function to produce diagnostic plots of the linear regression fit. Comment on any problems you see with the fit. Do the residual plots suggest any unusually large outliers? Does the leverage plot identify any observations with unusually high leverage?
The values that are large outliers are observations 327 and 394. Observation 14 has an unusally high leverage.
- Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
I just used the ’*’ styntax because it answered everything at once. Interactions that are significan’t are indicated by **. They are Displanement:origin, Acceleration:year and Acceleration:origin.
- Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
plot(Auto)
Auto1 <- Auto
Auto1$name=NULL
cor(Auto1)## 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 1.0000000AutoRegress<- lm(mpg ~., data=Auto1)
summary(AutoRegress)##
## Call:
## lm(formula = mpg ~ ., data = Auto1)
##
## 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-16par(mfrow=c(2,2))
plot(AutoRegress)
AutoInteract<- lm(mpg~ .*., Auto1)
summary(AutoInteract)##
## Call:
## lm(formula = mpg ~ . * ., data = Auto1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -7.6303 -1.4481 0.0596 1.2739 11.1386
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.548e+01 5.314e+01 0.668 0.50475
## cylinders 6.989e+00 8.248e+00 0.847 0.39738
## displacement -4.785e-01 1.894e-01 -2.527 0.01192 *
## horsepower 5.034e-01 3.470e-01 1.451 0.14769
## weight 4.133e-03 1.759e-02 0.235 0.81442
## acceleration -5.859e+00 2.174e+00 -2.696 0.00735 **
## year 6.974e-01 6.097e-01 1.144 0.25340
## origin -2.090e+01 7.097e+00 -2.944 0.00345 **
## cylinders:displacement -3.383e-03 6.455e-03 -0.524 0.60051
## cylinders:horsepower 1.161e-02 2.420e-02 0.480 0.63157
## cylinders:weight 3.575e-04 8.955e-04 0.399 0.69000
## cylinders:acceleration 2.779e-01 1.664e-01 1.670 0.09584 .
## cylinders:year -1.741e-01 9.714e-02 -1.793 0.07389 .
## cylinders:origin 4.022e-01 4.926e-01 0.816 0.41482
## displacement:horsepower -8.491e-05 2.885e-04 -0.294 0.76867
## displacement:weight 2.472e-05 1.470e-05 1.682 0.09342 .
## displacement:acceleration -3.479e-03 3.342e-03 -1.041 0.29853
## displacement:year 5.934e-03 2.391e-03 2.482 0.01352 *
## displacement:origin 2.398e-02 1.947e-02 1.232 0.21875
## horsepower:weight -1.968e-05 2.924e-05 -0.673 0.50124
## horsepower:acceleration -7.213e-03 3.719e-03 -1.939 0.05325 .
## horsepower:year -5.838e-03 3.938e-03 -1.482 0.13916
## horsepower:origin 2.233e-03 2.930e-02 0.076 0.93931
## weight:acceleration 2.346e-04 2.289e-04 1.025 0.30596
## weight:year -2.245e-04 2.127e-04 -1.056 0.29182
## weight:origin -5.789e-04 1.591e-03 -0.364 0.71623
## acceleration:year 5.562e-02 2.558e-02 2.174 0.03033 *
## acceleration:origin 4.583e-01 1.567e-01 2.926 0.00365 **
## year:origin 1.393e-01 7.399e-02 1.882 0.06062 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.695 on 363 degrees of freedom
## Multiple R-squared: 0.8893, Adjusted R-squared: 0.8808
## F-statistic: 104.2 on 28 and 363 DF, p-value: < 2.2e-16AutoTransform1<-lm(mpg~weight+I((weight)^2),Auto1)
summary(AutoTransform1)##
## Call:
## lm(formula = mpg ~ weight + I((weight)^2), data = Auto1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.6246 -2.7134 -0.3485 1.8267 16.0866
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.226e+01 2.993e+00 20.800 < 2e-16 ***
## weight -1.850e-02 1.972e-03 -9.379 < 2e-16 ***
## I((weight)^2) 1.697e-06 3.059e-07 5.545 5.43e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.176 on 389 degrees of freedom
## Multiple R-squared: 0.7151, Adjusted R-squared: 0.7137
## F-statistic: 488.3 on 2 and 389 DF, p-value: < 2.2e-16plot(AutoTransform1)
AutoTransform2<-lm(mpg~weight+I((weight)^(1/2)),Auto1)
summary(AutoTransform2)##
## Call:
## lm(formula = mpg ~ weight + I((weight)^(1/2)), data = Auto1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.5660 -2.6552 -0.4161 1.7373 16.1001
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 109.218284 11.573797 9.437 < 2e-16 ***
## weight 0.013191 0.003828 3.446 0.000631 ***
## I((weight)^(1/2)) -2.314535 0.424250 -5.456 8.7e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.181 on 389 degrees of freedom
## Multiple R-squared: 0.7145, Adjusted R-squared: 0.713
## F-statistic: 486.7 on 2 and 389 DF, p-value: < 2.2e-16plot(AutoTransform2)
AutoTransform3<-lm(mpg~weight+log(weight),Auto1)
summary(AutoTransform3)##
## Call:
## lm(formula = mpg ~ weight + log(weight), data = Auto1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -12.5329 -2.7031 -0.4016 1.7038 16.0835
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 263.812407 40.366256 6.535 1.99e-10 ***
## weight 0.002582 0.001914 1.349 0.178
## log(weight) -31.166013 5.780558 -5.392 1.21e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.185 on 389 degrees of freedom
## Multiple R-squared: 0.714, Adjusted R-squared: 0.7125
## F-statistic: 485.6 on 2 and 389 DF, p-value: < 2.2e-16plot(AutoTransform3)
Problem 10
This question should be answered using the Carseats data set.
Fit a multiple regression model to predict Sales using Price, Urban, and US.
Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
The Urban and US variables are coded as dummy or indicatior variables. By coding one as a 0 and the other as 1 we can contrast the two. So we set carseats sold in non-urban environments as our baseline and compare carseats sold in urban environments to it. UrbanYes is a negative coeeficient because sales in urban environments are lower. The same is applied to USYes, but now sales sold outside of the US are the baseline. More carseats are sold in the US so it has a positive coeeficinet.
- Write out the model in equation form, being careful to handle the qualitative variables properly.
Sales=13.043469 - 0.054459(Price) - (0.021916)(1 ,if Urban is Yes 0 if not yes) + 1.200573 (1,if US is Yes 0 if not yes)
- For which of the predictors can you reject the null hypothesis H0 : βj = 0?
Price and USYes
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.
How well do the models in (a) and (e) fit the data?
The smaller model has a slightly higher adjusted R-square, but eveything else is pretty much the same.
Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
Is there evidence of outliers or high leverage observations in the model from (e)?
CarSeatRegress<-lm(Sales~Price+Urban+US,data=Carseats)
summary(CarSeatRegress)##
## 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-16sales_var<-lm(Sales~Price+US,data=Carseats)
summary## function (object, ...)
## UseMethod("summary")
## <bytecode: 0x000001ae5a75b6a0>
## <environment: namespace:base>confint(sales_var)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632par(mfrow=c(2,2))
plot(sales_var)
### Problem 12 This problem involves simple linear regression without an
intercept.
- Recall that the coefficient estimate βˆ 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?
They will have the same coefficient if they have the same distribution.
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.
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.
x=rnorm(100)
y=rbinom(100,2,0.3)
example1<-lm(y~x+0)
summary(example1)##
## Call:
## lm(formula = y ~ x + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.09583 0.00647 0.08036 1.00574 2.07139
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## x 0.04605 0.08137 0.566 0.573
##
## Residual standard error: 0.8274 on 99 degrees of freedom
## Multiple R-squared: 0.003224, Adjusted R-squared: -0.006844
## F-statistic: 0.3202 on 1 and 99 DF, p-value: 0.5728example2<-lm(x~y+0)
summary(example2)##
## Call:
## lm(formula = x ~ y + 0)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.2519 -0.8132 -0.1691 0.6185 2.0812
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## y 0.07002 0.12373 0.566 0.573
##
## Residual standard error: 1.02 on 99 degrees of freedom
## Multiple R-squared: 0.003224, Adjusted R-squared: -0.006844
## F-statistic: 0.3202 on 1 and 99 DF, p-value: 0.5728x=1:100
y=100:1
example3<-lm(y~x+0)
summary(example3)##
## 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-08example4<-lm(x~y+0)
summary(example4)##
## 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