Problem 2 Carefully explain the differences between the KNN classifier and KNN#regression methods. KNN classifier uses k nearest neighbors to output a class (think….type/thing/noun). KNN regression uses k nearest neighbors to output a numerical value (think…stock rices/miles per gallon/number of widgets/East|West of Charles# River/0|1/Yes|No/Pass|Fail). Both are machine learning methods that use the k nearest neighbor to build a model for prediction purposes.

options(repos = list(CRAN="http://cran.rstudio.com/"))

install.packages("ISLR2")
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install.packages("ISLR")
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install.packages("GGally")
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install.packages("ggplot2")
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install.packages("tidyverse")
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install.packages("rsconnect")
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install.packages("knitr")
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library(ISLR)
library(ISLR2)
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## Attaching package: 'ISLR2'
## The following objects are masked from 'package:ISLR':
## 
##     Auto, Credit
library(GGally)
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library(ggplot2)
library(tidyverse)
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## ✔ forcats   1.0.0     ✔ stringr   1.5.1
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library(rsconnect)
library(knitr)
data("Auto")


head(Auto)
##   mpg cylinders displacement horsepower weight acceleration year origin
## 1  18         8          307        130   3504         12.0   70      1
## 2  15         8          350        165   3693         11.5   70      1
## 3  18         8          318        150   3436         11.0   70      1
## 4  16         8          304        150   3433         12.0   70      1
## 5  17         8          302        140   3449         10.5   70      1
## 6  15         8          429        198   4341         10.0   70      1
##                        name
## 1 chevrolet chevelle malibu
## 2         buick skylark 320
## 3        plymouth satellite
## 4             amc rebel sst
## 5               ford torino
## 6          ford galaxie 500
  1. This question involves the use of multiple linear regression on the Auto data set.
  1. Produce a scatterplot matrix which includes all of the variables in the data set.
pairs(Auto, main = "Scatterplot Auto Matrix", col="orange", pch = 19)

  1. Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, which is qualitative. cor()
auto <- as_tibble(Auto)

autoNoName <- select(Auto, -name)

cor(autoNoName)
##                     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
  1. 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:
  1. Is there a relationship between the predictors and the response? Yes there are 5 significatn relationships. The base mpg, the displacement, weight, year, and origin.
  2. Which predictors appear to have a statistically significant relationship to the response? see above
  3. What does the coefficient for the year variable suggest? That the age of the vehicle weighs a lot in the MPG of the vehicle. (This could be due to maintenance or efficiency or technology in newer models. That would be a better question for the MEs at the auto maker.) But every year, the mpg goes up by 0.75 miles.
NineC <- lm(mpg ~ . -name, data = Auto)

summary(NineC)
## 
## 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-16
  1. 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? Yes, around the 320s datapoints are outliers.
par(mfrow = c(2,2))

plot(NineC)

  1. Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
interactionPlot <- lm(mpg ~weight*displacement + . -name, data = Auto)

summary(interactionPlot)
## 
## Call:
## lm(formula = mpg ~ weight * displacement + . - name, 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    
## weight              -1.064e-02  7.136e-04 -14.915  < 2e-16 ***
## displacement        -6.837e-02  1.104e-02  -6.193 1.52e-09 ***
## cylinders            1.175e-01  2.943e-01   0.399   0.6899    
## horsepower          -3.280e-02  1.238e-02  -2.649   0.0084 ** 
## 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 *  
## weight:displacement  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
interactionPlot2 <- lm(mpg ~ year*origin + . -name, data = Auto)

summary(interactionPlot2)
## 
## Call:
## lm(formula = mpg ~ year * origin + . - name, data = Auto)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -8.6072 -2.0439 -0.0596  1.7121 12.3368 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   8.492e+00  9.044e+00   0.939 0.348353    
## year          4.189e-01  1.125e-01   3.723 0.000226 ***
## origin       -1.405e+01  4.699e+00  -2.989 0.002978 ** 
## cylinders    -5.042e-01  3.192e-01  -1.579 0.115082    
## displacement  1.567e-02  7.530e-03   2.081 0.038060 *  
## horsepower   -1.399e-02  1.364e-02  -1.025 0.305786    
## weight       -6.352e-03  6.449e-04  -9.851  < 2e-16 ***
## acceleration  9.185e-02  9.766e-02   0.941 0.347546    
## year:origin   1.989e-01  6.030e-02   3.298 0.001064 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.286 on 383 degrees of freedom
## Multiple R-squared:  0.8264, Adjusted R-squared:  0.8228 
## F-statistic: 227.9 on 8 and 383 DF,  p-value: < 2.2e-16
  1. Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings.
interactionPlot3 <- lm(mpg ~ log(year*origin) + . -name, data = Auto)

summary(interactionPlot3)
## 
## Call:
## lm(formula = mpg ~ log(year * origin) + . - name, data = Auto)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -9.0421 -2.0758 -0.0992  1.9924 13.3481 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)        -4.142e+01  1.141e+01  -3.631 0.000321 ***
## log(year * origin)  7.966e+00  3.434e+00   2.320 0.020872 *  
## cylinders          -4.916e-01  3.215e-01  -1.529 0.127026    
## displacement        2.384e-02  7.663e-03   3.110 0.002008 ** 
## horsepower         -1.800e-02  1.372e-02  -1.312 0.190233    
## weight             -6.707e-03  6.561e-04 -10.223  < 2e-16 ***
## acceleration        7.950e-02  9.829e-02   0.809 0.419116    
## year                6.710e-01  6.125e-02  10.955  < 2e-16 ***
## origin             -2.949e+00  1.906e+00  -1.547 0.122641    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.309 on 383 degrees of freedom
## Multiple R-squared:  0.824,  Adjusted R-squared:  0.8203 
## F-statistic: 224.1 on 8 and 383 DF,  p-value: < 2.2e-16
  1. This question should be answered using the Carseats data set.
  1. Fit a multiple regression model to predict Sales using Price, Urban, and US.
data("Carseats")


str(Carseats)
## 'data.frame':    400 obs. of  11 variables:
##  $ Sales      : num  9.5 11.22 10.06 7.4 4.15 ...
##  $ CompPrice  : num  138 111 113 117 141 124 115 136 132 132 ...
##  $ Income     : num  73 48 35 100 64 113 105 81 110 113 ...
##  $ Advertising: num  11 16 10 4 3 13 0 15 0 0 ...
##  $ Population : num  276 260 269 466 340 501 45 425 108 131 ...
##  $ Price      : num  120 83 80 97 128 72 108 120 124 124 ...
##  $ ShelveLoc  : Factor w/ 3 levels "Bad","Good","Medium": 1 2 3 3 1 1 3 2 3 3 ...
##  $ Age        : num  42 65 59 55 38 78 71 67 76 76 ...
##  $ Education  : num  17 10 12 14 13 16 15 10 10 17 ...
##  $ Urban      : Factor w/ 2 levels "No","Yes": 2 2 2 2 2 1 2 2 1 1 ...
##  $ US         : Factor w/ 2 levels "No","Yes": 2 2 2 2 1 2 1 2 1 2 ...
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 Yes
carseat_Sales <- lm(Sales ~Price + Urban + US, data = Carseats)

carseat_Sales
## 
## Call:
## lm(formula = Sales ~ Price + Urban + US, data = Carseats)
## 
## Coefficients:
## (Intercept)        Price     UrbanYes        USYes  
##    13.04347     -0.05446     -0.02192      1.20057
set.seed(997)


carseat_AOV_Sales <- anova(carseat_Sales)



print(carseat_AOV_Sales)
## Analysis of Variance Table
## 
## Response: Sales
##            Df  Sum Sq Mean Sq  F value    Pr(>F)    
## Price       1  630.03  630.03 103.0603 < 2.2e-16 ***
## Urban       1    0.10    0.10   0.0158    0.9001    
## US          1  131.31  131.31  21.4802  4.86e-06 ***
## Residuals 396 2420.83    6.11                       
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
  1. Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!

For each dollar raised on the price of the carseat, the number of carseats sold goes down by 5%. If the car seat is sold in an Urban area, the number of carseats sold goes down by 2% (possible causes being greater use of mass transportation/fewer personal vehicles, or fewer young children). For every car seat placed in a US store, the number of car seats sold go up by 120%.

  1. Write out the model in equation form, being careful to handle the qualitative variables properly.
#Sales = 13.04347 - 0.05446(Price) - 0.02192(Urban-0|1) + 1.20057(US-0|1)
  1. For which of the predictors can you reject the null hypothesis H0 : βj = 0?

You can reject null hypothesis for the ANOVA statistically significant Price and US factors.

  1. 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.
smaller_Carseats <- lm(Sales ~ Price + US, data = Carseats)

smaller_Carseats
## 
## Call:
## lm(formula = Sales ~ Price + US, data = Carseats)
## 
## Coefficients:
## (Intercept)        Price        USYes  
##    13.03079     -0.05448      1.19964
smaller_CarseatsAOV <- anova(smaller_Carseats)

smaller_CarseatsAOV
## Analysis of Variance Table
## 
## Response: Sales
##            Df  Sum Sq Mean Sq F value    Pr(>F)    
## Price       1  630.03  630.03 103.319 < 2.2e-16 ***
## US          1  131.37  131.37  21.543 4.707e-06 ***
## Residuals 397 2420.87    6.10                      
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
reduced_Carseats <- anova(smaller_Carseats, carseat_Sales)

reduced_Carseats
## Analysis of Variance Table
## 
## Model 1: Sales ~ Price + US
## Model 2: Sales ~ Price + Urban + US
##   Res.Df    RSS Df Sum of Sq      F Pr(>F)
## 1    397 2420.9                           
## 2    396 2420.8  1   0.03979 0.0065 0.9357
summary(smaller_Carseats)
## 
## 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
  1. How well do the models in (a) and (e) fit the data?
AIC(smaller_Carseats)
## [1] 1863.319
AIC(carseat_Sales)
## [1] 1865.312
BIC(smaller_Carseats)
## [1] 1879.285
BIC(carseat_Sales)
## [1] 1885.269

Neither fit the data well. The R-squared and adjusted R-squared only predicts ~23% of the model. I noticed that none of the models included number of children < 4yo & <1yo as a factor (i.e. possible demand). Might want to look into that instead of Urban.

  1. Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).
confint(smaller_Carseats, level = 0.95)
##                   2.5 %      97.5 %
## (Intercept) 11.79032020 14.27126531
## Price       -0.06475984 -0.04419543
## USYes        0.69151957  1.70776632
  1. Is there evidence of outliers or high leverage observations in the model from (e)?
par(mfrow = c(2,2))

plot(smaller_Carseats)

residuals <- resid(smaller_Carseats)

outliers <- which(abs(residuals) > 2*sd(residuals))

print(outliers)
##  18  26  29  31  50  51  58  63  69  83  99 107 144 210 259 298 299 305 317 329 
##  18  26  29  31  50  51  58  63  69  83  99 107 144 210 259 298 299 305 317 329 
## 353 377 396 
## 353 377 396
#Below are the outlier observation numbers.
# Calculate Cook's Distance
cooks_dist <- cooks.distance(smaller_Carseats)

# Identify influential points
outliers2 <- which(cooks_dist > 0.75)
print(outliers2)
## named integer(0)

This means that none of the outlier observations have significant leverage on the model.

  1. This problem involves simple linear regression without an intercept.
  1. 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?

If there is no error term and/or a 1:1 relationship between X and Y.

  1. 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(997)

x = rnorm(100)

y = 1.5*x + rnorm(100)
modelx_y <- lm(x~y+0)
coefx_y <- coef(modelx_y)


modely_x <- lm(y~x+0)
coefy_x <- coef(modely_x)

summary(modelx_y)
## 
## Call:
## lm(formula = x ~ y + 0)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.27251 -0.28998  0.01147  0.36325  1.04423 
## 
## Coefficients:
##   Estimate Std. Error t value Pr(>|t|)    
## y   0.5001     0.0276   18.12   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5044 on 99 degrees of freedom
## Multiple R-squared:  0.7683, Adjusted R-squared:  0.766 
## F-statistic: 328.3 on 1 and 99 DF,  p-value: < 2.2e-16
summary(modely_x)
## 
## Call:
## lm(formula = y ~ x + 0)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.8367 -0.5946  0.0378  0.5903  2.0377 
## 
## Coefficients:
##   Estimate Std. Error t value Pr(>|t|)    
## x  1.53631    0.08479   18.12   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.884 on 99 degrees of freedom
## Multiple R-squared:  0.7683, Adjusted R-squared:  0.766 
## F-statistic: 328.3 on 1 and 99 DF,  p-value: < 2.2e-16
print(coefx_y)
##         y 
## 0.5001115
print(coefy_x)
##        x 
## 1.536306
  1. 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(997)

x = rnorm(100)

y = x
model2y_x <- lm(y ~ x + 0)
model2x_y <- lm(x ~ y + 0)


coef2y_x <- coef(model2y_x)
coef2x_y <- coef(model2x_y)

summary(model2y_x)
## Warning in summary.lm(model2y_x): essentially perfect fit: summary may be
## unreliable
## 
## Call:
## lm(formula = y ~ x + 0)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -2.091e-16 -2.670e-17  1.060e-17  5.000e-17  3.650e-15 
## 
## Coefficients:
##    Estimate Std. Error   t value Pr(>|t|)    
## x 1.000e+00  3.608e-17 2.771e+16   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.762e-16 on 99 degrees of freedom
## Multiple R-squared:      1,  Adjusted R-squared:      1 
## F-statistic: 7.68e+32 on 1 and 99 DF,  p-value: < 2.2e-16
summary(model2x_y)
## Warning in summary.lm(model2x_y): essentially perfect fit: summary may be
## unreliable
## 
## Call:
## lm(formula = x ~ y + 0)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -2.091e-16 -2.670e-17  1.060e-17  5.000e-17  3.650e-15 
## 
## Coefficients:
##    Estimate Std. Error   t value Pr(>|t|)    
## y 1.000e+00  3.608e-17 2.771e+16   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.762e-16 on 99 degrees of freedom
## Multiple R-squared:      1,  Adjusted R-squared:      1 
## F-statistic: 7.68e+32 on 1 and 99 DF,  p-value: < 2.2e-16
print(coef2y_x)
## x 
## 1
print(coef2x_y)
## y 
## 1