STA 6543 Assignment 2
Abigail Davis mjv135
2025-02-26
Chapter 03: 2, 9, 10, 12
Chapter 3 Problem 2:
Carefully explain the differences between the KNN classifier and KNN regression methods
Answer:
The KNN classifiers are used for classification (categorical) tasks whereas the KNN regression is used for regression (continuous/numerical) tasks. While KNN classifier k favors majority classes when large leading to bias, KNN regression may have a smoother prediction but less variable fit. In KNN classification, accuracy, precision, recall, and F1-score are all used as performance metrics, while in KNN regression, MSE and R-squared are used.
Chapter 3 Problem 9:
This question involves the use of multiple linear regression on the Auto data set.
library(ISLR2)
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Boston(a) Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(Auto)
(b) Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, cor() which is qualitative.
Auto_quantitative <- Auto[, sapply(Auto, is.numeric)]
cor(Auto_quantitative)## 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 1.0000000(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. For instance:
mlr1 <- lm(mpg ~ ., data = Auto_quantitative)
summary(mlr1)##
## Call:
## lm(formula = mpg ~ ., data = Auto_quantitative)
##
## 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-16Answers:
i. Is there a relationship between the predictors and the response?
Yes, there are some statistically significant predictors in the model that explain ~82% of the variance in mpg based on the adjusted R-squared value.
ii. Which predictors appear to have a statistically significant relationship to the response?
Displacement, weight, year, and origin are all statistically significant because they have a p-value < 0.05.
iii. What does the coefficient for the year variable suggest?
The coefficient for the year variable suggests that as year increases by 1, the mpg increases by 0.75. This suggests that newer cars, have higher mpg.
(d) 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?
par(mfrow= c(2,2))
plot(mlr1)
Answer:
The residual vs. fitted plot suggests a slight patter as fitted values increase, may suggest possible heteroscedasticity.The Scale-location shows the same trend. The Normal Q-Q plot suggests that the points are not normally distributed because there is deviation from the line at the tails (right-skewness). The residual plot suggests that point 14 has a higher leverage than the rest of the points.
(e) Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
interaction1 <- lm(mpg~.+ displacement:horsepower, data = Auto_quantitative)
summary(interaction1)##
## Call:
## lm(formula = mpg ~ . + displacement:horsepower, data = Auto_quantitative)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.7010 -1.6009 -0.0967 1.4119 12.6734
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.894e+00 4.302e+00 -0.440 0.66007
## cylinders 6.466e-01 3.017e-01 2.143 0.03275 *
## displacement -7.487e-02 1.092e-02 -6.859 2.80e-11 ***
## horsepower -1.975e-01 2.052e-02 -9.624 < 2e-16 ***
## weight -3.147e-03 6.475e-04 -4.861 1.71e-06 ***
## acceleration -2.131e-01 9.062e-02 -2.351 0.01921 *
## year 7.379e-01 4.463e-02 16.534 < 2e-16 ***
## origin 6.891e-01 2.527e-01 2.727 0.00668 **
## displacement:horsepower 5.236e-04 4.813e-05 10.878 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.912 on 383 degrees of freedom
## Multiple R-squared: 0.8636, Adjusted R-squared: 0.8608
## F-statistic: 303.1 on 8 and 383 DF, p-value: < 2.2e-16In this model, adding the interaction term between displacement and horsepower is very significant. This suggests that displacement and mpg is influenced by the level of horsepower. The R-squared value is also higher in this model suggesting by adding the interaction term, the model improves.
interaction2 <- lm(mpg ~ . + cylinders*displacement + displacement*horsepower, data = Auto_quantitative)
summary(interaction2)##
## Call:
## lm(formula = mpg ~ . + cylinders * displacement + displacement *
## horsepower, data = Auto_quantitative)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.0692 -1.5453 -0.1215 1.4814 12.7232
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.006e+00 4.426e+00 -0.227 0.8203
## cylinders 2.309e-01 5.702e-01 0.405 0.6858
## displacement -8.149e-02 1.336e-02 -6.099 2.61e-09 ***
## horsepower -1.873e-01 2.367e-02 -7.915 2.71e-14 ***
## weight -3.235e-03 6.557e-04 -4.933 1.21e-06 ***
## acceleration -1.923e-01 9.382e-02 -2.050 0.0410 *
## year 7.402e-01 4.472e-02 16.550 < 2e-16 ***
## origin 6.435e-01 2.583e-01 2.492 0.0131 *
## cylinders:displacement 1.997e-03 2.325e-03 0.859 0.3908
## displacement:horsepower 4.812e-04 6.897e-05 6.976 1.34e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.913 on 382 degrees of freedom
## Multiple R-squared: 0.8639, Adjusted R-squared: 0.8607
## F-statistic: 269.4 on 9 and 382 DF, p-value: < 2.2e-16By adding cylinders:displacement, it did not change the model’s explanatory value much. This is because this interaction is not significant.
interaction3 <- lm(mpg ~ . + horsepower:cylinders + displacement*horsepower, data = Auto_quantitative)
summary(interaction3)##
## Call:
## lm(formula = mpg ~ . + horsepower:cylinders + displacement *
## horsepower, data = Auto_quantitative)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.891 -1.627 -0.089 1.353 12.506
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.7245533 5.4406102 0.868 0.38573
## cylinders -1.4390172 1.0981296 -1.310 0.19084
## displacement -0.0476961 0.0175411 -2.719 0.00684 **
## horsepower -0.2572895 0.0365449 -7.040 8.96e-12 ***
## weight -0.0033093 0.0006502 -5.089 5.65e-07 ***
## acceleration -0.2088022 0.0903086 -2.312 0.02130 *
## year 0.7378654 0.0444594 16.596 < 2e-16 ***
## origin 0.7459615 0.2533706 2.944 0.00344 **
## cylinders:horsepower 0.0173764 0.0087998 1.975 0.04903 *
## displacement:horsepower 0.0003226 0.0001125 2.867 0.00437 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.901 on 382 degrees of freedom
## Multiple R-squared: 0.865, Adjusted R-squared: 0.8618
## F-statistic: 271.9 on 9 and 382 DF, p-value: < 2.2e-16Adding the interaction of cylinders:horsepower, the adjusted r-squared value rose slightly. The interaction between cylinders:horsepower is significant suggesting that the effect of horsepower on mpg changes bassed on both cylinders and displacement.
(f) Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
Auto_quantitative$log_displacement <- log(Auto_quantitative$displacement)
Auto_quantitative$sqrt_displacement <- sqrt(Auto_quantitative$displacement)
Auto_quantitative$sq_displacement <- Auto_quantitative$displacement^2
Auto_quantitative$log_horsepower <- log(Auto_quantitative$horsepower)
Auto_quantitative$sqrt_horsepower <- sqrt(Auto_quantitative$horsepower)
Auto_quantitative$sq_horsepower <- Auto_quantitative$horsepower^2model_log <- lm(mpg ~ log_displacement + log_horsepower + weight + acceleration + year + origin + log_displacement:log_horsepower, data = Auto_quantitative)
summary(model_log)##
## Call:
## lm(formula = mpg ~ log_displacement + log_horsepower + weight +
## acceleration + year + origin + log_displacement:log_horsepower,
## data = Auto_quantitative)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.1112 -1.6669 -0.1139 1.4579 12.4210
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.159e+02 2.154e+01 10.026 < 2e-16 ***
## log_displacement -3.970e+01 4.212e+00 -9.426 < 2e-16 ***
## log_horsepower -4.983e+01 4.705e+00 -10.589 < 2e-16 ***
## weight -3.424e-03 6.294e-04 -5.440 9.49e-08 ***
## acceleration -2.279e-01 9.362e-02 -2.435 0.0154 *
## year 7.506e-01 4.421e-02 16.977 < 2e-16 ***
## origin 6.280e-01 2.599e-01 2.416 0.0162 *
## log_displacement:log_horsepower 8.249e+00 9.040e-01 9.125 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.896 on 384 degrees of freedom
## Multiple R-squared: 0.8648, Adjusted R-squared: 0.8623
## F-statistic: 350.9 on 7 and 384 DF, p-value: < 2.2e-16By taking the log of displacement and horsepower, every variable became significant in the model. The adjusted R-squared also increased a little bit.
par(mfrow = c(2,2))
plot(model_log)
model_sqrt <- lm(mpg ~ sqrt_displacement + sqrt_horsepower + weight + acceleration + year + origin + sqrt_displacement:sqrt_horsepower, data = Auto_quantitative)
summary(model_sqrt)##
## Call:
## lm(formula = mpg ~ sqrt_displacement + sqrt_horsepower + weight +
## acceleration + year + origin + sqrt_displacement:sqrt_horsepower,
## data = Auto_quantitative)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.047 -1.572 -0.115 1.433 12.452
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 36.5625343 6.3010566 5.803 1.37e-08 ***
## sqrt_displacement -3.0920551 0.3129674 -9.880 < 2e-16 ***
## sqrt_horsepower -5.5087076 0.4903298 -11.235 < 2e-16 ***
## weight -0.0032206 0.0006332 -5.086 5.72e-07 ***
## acceleration -0.2244416 0.0914854 -2.453 0.01460 *
## year 0.7438615 0.0441035 16.866 < 2e-16 ***
## origin 0.6717378 0.2533701 2.651 0.00835 **
## sqrt_displacement:sqrt_horsepower 0.2792940 0.0250597 11.145 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.884 on 384 degrees of freedom
## Multiple R-squared: 0.8659, Adjusted R-squared: 0.8634
## F-statistic: 354.2 on 7 and 384 DF, p-value: < 2.2e-16By taking the square root of displacement and horsepower, every variable became significant in the model. The adjusted R-squared also increased a little bit more from the log model.
par(mfrow = c(2,2))
plot(model_sqrt)
model_sq <- lm(mpg ~ sq_displacement + sq_horsepower + weight + acceleration + year + origin + sq_displacement:sq_horsepower, data = Auto_quantitative)
summary(model_sq)##
## Call:
## lm(formula = mpg ~ sq_displacement + sq_horsepower + weight +
## acceleration + year + origin + sq_displacement:sq_horsepower,
## data = Auto_quantitative)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.4488 -1.8420 -0.0478 1.7963 12.4703
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.799e+01 4.047e+00 -4.445 1.15e-05 ***
## sq_displacement -1.639e-05 1.560e-05 -1.051 0.294
## sq_horsepower -3.518e-04 8.012e-05 -4.392 1.46e-05 ***
## weight -5.445e-03 6.511e-04 -8.363 1.14e-15 ***
## acceleration 4.939e-02 9.036e-02 0.547 0.585
## year 7.611e-01 4.821e-02 15.789 < 2e-16 ***
## origin 1.122e+00 2.523e-01 4.447 1.14e-05 ***
## sq_displacement:sq_horsepower 2.342e-09 4.455e-10 5.258 2.43e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 3.14 on 384 degrees of freedom
## Multiple R-squared: 0.8411, Adjusted R-squared: 0.8382
## F-statistic: 290.3 on 7 and 384 DF, p-value: < 2.2e-16By taking the square of displacement and horsepower, the square of displacement and acceleration become insignificant. The adjusted R-squared also decreased lower than any model except for the original mlr1 model.
par(mfrow = c(2,2))
plot(model_sq)
None of the plots had very significant changes in any of the transformations.
Chapter 3 Problem 10:
This question should be answered using the Carseats data set.
library(ISLR2)
attach(Carseats)
head(Carseats)## Sales CompPrice Income Advertising Population Price ShelveLoc Age Education
## 1 9.50 138 73 11 276 120 Bad 42 17
## 2 11.22 111 48 16 260 83 Good 65 10
## 3 10.06 113 35 10 269 80 Medium 59 12
## 4 7.40 117 100 4 466 97 Medium 55 14
## 5 4.15 141 64 3 340 128 Bad 38 13
## 6 10.81 124 113 13 501 72 Bad 78 16
## Urban US
## 1 Yes Yes
## 2 Yes Yes
## 3 Yes Yes
## 4 Yes Yes
## 5 Yes No
## 6 No Yessummary(Carseats)## Sales CompPrice Income Advertising
## Min. : 0.000 Min. : 77 Min. : 21.00 Min. : 0.000
## 1st Qu.: 5.390 1st Qu.:115 1st Qu.: 42.75 1st Qu.: 0.000
## Median : 7.490 Median :125 Median : 69.00 Median : 5.000
## Mean : 7.496 Mean :125 Mean : 68.66 Mean : 6.635
## 3rd Qu.: 9.320 3rd Qu.:135 3rd Qu.: 91.00 3rd Qu.:12.000
## Max. :16.270 Max. :175 Max. :120.00 Max. :29.000
## Population Price ShelveLoc Age Education
## Min. : 10.0 Min. : 24.0 Bad : 96 Min. :25.00 Min. :10.0
## 1st Qu.:139.0 1st Qu.:100.0 Good : 85 1st Qu.:39.75 1st Qu.:12.0
## Median :272.0 Median :117.0 Medium:219 Median :54.50 Median :14.0
## Mean :264.8 Mean :115.8 Mean :53.32 Mean :13.9
## 3rd Qu.:398.5 3rd Qu.:131.0 3rd Qu.:66.00 3rd Qu.:16.0
## Max. :509.0 Max. :191.0 Max. :80.00 Max. :18.0
## Urban US
## No :118 No :142
## Yes:282 Yes:258
##
##
##
## (a) Fit a multiple regression model to predict Sales using Price, Urban, and US.
model1 <- lm(Sales ~ Price + Urban + US, data = Carseats)
summary(model1)##
## 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-16(b) Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
After looking at the table above, we can tell that Price and USYes are significant predictors of Sales.
The coefficient for Price is -0.054459 which means for
every unit of increase in the price of the carseat, the store’s sales
decrease by ~54 units on average.
The coefficient for US = Yes is 1.200573 which means US
stores sell, on average, $1,200 more worth of carseats compared to
stores outside the US.
(c) Write out the model in equation form, being careful to handle the qualitative variables properly.
\(Sales = 13.04 - 0.05Price - 0.02Urban + 1.2USYes\)
(d) For which of the predictors can you reject the null hypothesis \(H_0 : \beta_j = 0\)?
Price and US = Yes are significant thus we
can reject the null hypothesis \(H_0 : \beta_j
= 0\).
(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.
model_significant <- lm(Sales ~ Price + US, data = Carseats)
summary(model_significant)##
## 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-16(f) How well do the models in (a) and (e) fit the data?
They are not great fit for the data because the adjusted R-square values are not good. The Adj R-squared for part (a) is 0.2335, whereas the Adj R-squared for part(e) is 0.2354.This means each model only explains about 23% variance in Sales. Preferrably the Adj R-squared value would be greater than 0.7.
(g) Using the model from (e), obtain 95% confidence intervals for the coefficient(s).
confint(model_significant)## 2.5 % 97.5 %
## (Intercept) 11.79032020 14.27126531
## Price -0.06475984 -0.04419543
## USYes 0.69151957 1.70776632(h) Is there evidence of outliers or high leverage observations in the model from (e)?
par(mfrow = c(2,2))
plot(model_significant)
summary(influence.measures(model_significant))## 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 0.00There are quiet a few observations that are outliers in the data and influence the measures.
Chapter 3 Problem 12:
This problem involves simple linear regression without an intercept.
(a) Recall that the coefficient estimate ˆ 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
Answer:
The coefficient estimate for the regression of X onto Y is the same as the coefficient estimate for the regression of Y onto X when the sum of squared values of xi is equal to the sum of the squaared values of yi (the total variance is the same for X and Y).
(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.
set.seed(1)
n <- 100
X <- rnorm(n)
Y <- 3 * X + rnorm(n)
XonY <- lm(X~Y-1)
summary(XonY)##
## Call:
## lm(formula = X ~ Y - 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.63420 -0.16066 0.07099 0.18507 0.59841
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## Y 0.29684 0.01056 28.12 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3019 on 99 degrees of freedom
## Multiple R-squared: 0.8887, Adjusted R-squared: 0.8876
## F-statistic: 790.6 on 1 and 99 DF, p-value: < 2.2e-16YonX <- lm(Y~X-1)
summary(YonX)##
## Call:
## lm(formula = Y ~ X - 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.9154 -0.6472 -0.1771 0.5056 2.3109
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## X 2.9939 0.1065 28.12 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.9586 on 99 degrees of freedom
## Multiple R-squared: 0.8887, Adjusted R-squared: 0.8876
## F-statistic: 790.6 on 1 and 99 DF, p-value: < 2.2e-16(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<-X
XonY2 <-lm(X~Y-1)
summary(XonY2)## Warning in summary.lm(XonY2): essentially perfect fit: summary may be
## unreliable##
## Call:
## lm(formula = X ~ Y - 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.888e-16 -1.689e-17 1.339e-18 3.057e-17 2.552e-16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## Y 1.000e+00 6.479e-18 1.543e+17 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.833e-17 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 2.382e+34 on 1 and 99 DF, p-value: < 2.2e-16YonX2<- lm(Y~X-1)
summary(YonX2)## Warning in summary.lm(YonX2): essentially perfect fit: summary may be
## unreliable##
## Call:
## lm(formula = Y ~ X - 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.888e-16 -1.689e-17 1.339e-18 3.057e-17 2.552e-16
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## X 1.000e+00 6.479e-18 1.543e+17 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.833e-17 on 99 degrees of freedom
## Multiple R-squared: 1, Adjusted R-squared: 1
## F-statistic: 2.382e+34 on 1 and 99 DF, p-value: < 2.2e-16