Homework 7
Melissa Oberg
2022-12-03
1. Load the mpg excel file into R Markdown, and convert the dataset into a data frame.
library(readxl)
#mpg data file
mpgdata <- read_excel("G:/Other computers/My Laptop/Documents/Richard 621/Week 11/mpg.xlsx")
# made data set a data frame
mpgdata <- as.data.frame(mpgdata)2. Using all rows of the dataset as your training set, create a linear regression model to predict mpg based on acceleration. What is the adjusted R-squared for your model?
#linear regression model
mpgmodel1 <- lm(mpg~., data = mpgdata)
summary(mpgmodel1)##
## Call:
## lm(formula = mpg ~ ., data = mpgdata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -18.007 -5.636 -1.242 4.758 23.192
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.9698 2.0432 2.432 0.0154 *
## acceleration 1.1912 0.1292 9.217 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.101 on 396 degrees of freedom
## Multiple R-squared: 0.1766, Adjusted R-squared: 0.1746
## F-statistic: 84.96 on 1 and 396 DF, p-value: < 2.2e-16# Adjusted R-squared: 0.1746 for this model3. Determine if a Box-Cox transformation would be beneficial. If so, perform the transformation. What is your adjusted R-squared value for your model now? Did it improve after applying a Box-Cox transformation?
#Based on the plot below a box cox transformation would be beneficial since the data does not stay linear and
#instead travels away from the horizontal line
plot(mpgmodel1$fitted.values, mpgmodel1$residuals)
abline(h = 0)
# The center line below is closest to zero which means a log of the data should improve the model
library(MASS)
boxcox(mpgmodel1)
mpgmodel2 <- lm(I(log(mpg)) ~ ., data = mpgdata)
summary(mpgmodel2)##
## Call:
## lm(formula = I(log(mpg)) ~ ., data = mpgdata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.06515 -0.23641 -0.00943 0.23576 0.79343
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.24656 0.08759 25.648 <2e-16 ***
## acceleration 0.05491 0.00554 9.911 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3044 on 396 degrees of freedom
## Multiple R-squared: 0.1987, Adjusted R-squared: 0.1967
## F-statistic: 98.23 on 1 and 396 DF, p-value: < 2.2e-16# Initial adjusted Adjusted R-squared: 0.1746
# After transformation Adjusted R-squared: 0.1967, the model was improved after applying the box-cox
#transformation4. Create a scatter plot comparing mpg and acceleration. Does the relationship between these two variables appear to be perfectly linear? Or is there perhaps some slight nonlinearity in the relationship?
# The plots below do not show perfect linearity and that there is some non-linearity in the relationship
pairs(mpgdata)
5. Create a new regression model to predict the Box-Cox transformation of mpg using both acceleration and a transformation of acceleration. Look at the “Common Transformation on Covariates” document in the “Week 11” folder of our Google Drive to determine some appropriate transformations. Which of these 4 “common transformations” yields the highest adjusted R-squared?
mpgmodel 3 has the highest R-Squared at 0.2234
mpgmodel3 <- lm(I(log(mpg)) ~ acceleration + I(acceleration^2), data = mpgdata)
summary(mpgmodel3)##
## Call:
## lm(formula = I(log(mpg)) ~ acceleration + I(acceleration^2),
## data = mpgdata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.07126 -0.22527 -0.00066 0.21838 0.77803
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.023320 0.331575 3.086 0.002170 **
## acceleration 0.213095 0.041764 5.102 5.22e-07 ***
## I(acceleration^2) -0.004959 0.001298 -3.820 0.000155 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2993 on 395 degrees of freedom
## Multiple R-squared: 0.2273, Adjusted R-squared: 0.2234
## F-statistic: 58.1 on 2 and 395 DF, p-value: < 2.2e-16#Adjusted R-squared: 0.2234mpgmodel4 <- lm(I(log(mpg)) ~ acceleration + I(1/acceleration), data = mpgdata)
summary(mpgmodel4)##
## Call:
## lm(formula = I(log(mpg)) ~ acceleration + I(1/acceleration),
## data = mpgdata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.05749 -0.22920 0.00108 0.22127 0.76895
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.26294 0.58605 7.274 1.89e-12 ***
## acceleration -0.01068 0.01963 -0.544 0.58682
## I(1/acceleration) -14.99800 4.31148 -3.479 0.00056 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3002 on 395 degrees of freedom
## Multiple R-squared: 0.2226, Adjusted R-squared: 0.2186
## F-statistic: 56.54 on 2 and 395 DF, p-value: < 2.2e-16#Adjusted R-squared: 0.2186mpgmodel5 <- lm(I(log(mpg)) ~ acceleration + (I(log(acceleration))), data = mpgdata)
summary(mpgmodel5)##
## Call:
## lm(formula = I(log(mpg)) ~ acceleration + (I(log(acceleration))),
## data = mpgdata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.06111 -0.22515 0.00151 0.21794 0.77069
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -1.59298 1.04365 -1.526 0.127724
## acceleration -0.09011 0.03966 -2.272 0.023624 *
## I(log(acceleration)) 2.23400 0.60516 3.692 0.000254 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2997 on 395 degrees of freedom
## Multiple R-squared: 0.2255, Adjusted R-squared: 0.2215
## F-statistic: 57.49 on 2 and 395 DF, p-value: < 2.2e-16#Adjusted R-squared: 0.2215 mpgmodel6 <- lm(I(log(mpg)) ~ acceleration + I(sqrt(acceleration)), data = mpgdata)
summary(mpgmodel6)##
## Call:
## lm(formula = I(log(mpg)) ~ acceleration + I(sqrt(acceleration)),
## data = mpgdata)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.06339 -0.22731 0.00077 0.21655 0.77214
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -2.38791 1.23506 -1.933 0.053897 .
## acceleration -0.24561 0.08008 -3.067 0.002310 **
## I(sqrt(acceleration)) 2.36968 0.62997 3.762 0.000194 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.2995 on 395 degrees of freedom
## Multiple R-squared: 0.2265, Adjusted R-squared: 0.2225
## F-statistic: 57.82 on 2 and 395 DF, p-value: < 2.2e-16#Adjusted R-squared: 0.22256. Using the “data.frame” function, create a new data frame with the following three variables:
• The Box-Cox transformation of mpg
• acceleration
• The transformation of acceleration that yielded the best adjusted R-squared in the preceding question
#data frame
mpg_data1 <- data.frame(log_mpg=log(mpgdata$mpg), acceleration = mpgdata$acceleration, acceleration_sq = (mpgdata$acceleration)^2)
mpg_data1## log_mpg acceleration acceleration_sq
## 1 2.890372 12.0 144.00
## 2 2.708050 11.5 132.25
## 3 2.890372 11.0 121.00
## 4 2.772589 12.0 144.00
## 5 2.833213 10.5 110.25
## 6 2.708050 10.0 100.00
## 7 2.639057 9.0 81.00
## 8 2.639057 8.5 72.25
## 9 2.639057 10.0 100.00
## 10 2.708050 8.5 72.25
## 11 2.708050 10.0 100.00
## 12 2.639057 8.0 64.00
## 13 2.708050 9.5 90.25
## 14 2.639057 10.0 100.00
## 15 3.178054 15.0 225.00
## 16 3.091042 15.5 240.25
## 17 2.890372 15.5 240.25
## 18 3.044522 16.0 256.00
## 19 3.295837 14.5 210.25
## 20 3.258097 20.5 420.25
## 21 3.218876 17.5 306.25
## 22 3.178054 14.5 210.25
## 23 3.218876 17.5 306.25
## 24 3.258097 12.5 156.25
## 25 3.044522 15.0 225.00
## 26 2.302585 14.0 196.00
## 27 2.302585 15.0 225.00
## 28 2.397895 13.5 182.25
## 29 2.197225 18.5 342.25
## 30 3.295837 14.5 210.25
## 31 3.332205 15.5 240.25
## 32 3.218876 14.0 196.00
## 33 3.218876 19.0 361.00
## 34 2.944439 13.0 169.00
## 35 2.772589 15.5 240.25
## 36 2.833213 15.5 240.25
## 37 2.944439 15.5 240.25
## 38 2.890372 15.5 240.25
## 39 2.639057 12.0 144.00
## 40 2.639057 11.5 132.25
## 41 2.639057 13.5 182.25
## 42 2.639057 13.0 169.00
## 43 2.484907 11.5 132.25
## 44 2.564949 12.0 144.00
## 45 2.564949 12.0 144.00
## 46 2.890372 13.5 182.25
## 47 3.091042 19.0 361.00
## 48 2.944439 15.0 225.00
## 49 2.890372 14.5 210.25
## 50 3.135494 14.0 196.00
## 51 3.332205 14.0 196.00
## 52 3.401197 19.5 380.25
## 53 3.401197 14.5 210.25
## 54 3.433987 19.0 361.00
## 55 3.555348 18.0 324.00
## 56 3.295837 19.0 361.00
## 57 3.258097 20.5 420.25
## 58 3.178054 15.5 240.25
## 59 3.218876 17.0 289.00
## 60 3.135494 23.5 552.25
## 61 2.995732 19.5 380.25
## 62 3.044522 16.5 272.25
## 63 2.564949 12.0 144.00
## 64 2.639057 12.0 144.00
## 65 2.708050 13.5 182.25
## 66 2.639057 13.0 169.00
## 67 2.833213 11.5 132.25
## 68 2.397895 11.0 121.00
## 69 2.564949 13.5 182.25
## 70 2.484907 13.5 182.25
## 71 2.564949 12.5 156.25
## 72 2.944439 13.5 182.25
## 73 2.708050 12.5 156.25
## 74 2.564949 14.0 196.00
## 75 2.564949 16.0 256.00
## 76 2.639057 14.0 196.00
## 77 2.890372 14.5 210.25
## 78 3.091042 18.0 324.00
## 79 3.044522 19.5 380.25
## 80 3.258097 18.0 324.00
## 81 3.091042 16.0 256.00
## 82 3.332205 17.0 289.00
## 83 3.135494 14.5 210.25
## 84 3.332205 15.0 225.00
## 85 3.295837 16.5 272.25
## 86 2.564949 13.0 169.00
## 87 2.639057 11.5 132.25
## 88 2.564949 13.0 169.00
## 89 2.639057 14.5 210.25
## 90 2.708050 12.5 156.25
## 91 2.484907 11.5 132.25
## 92 2.564949 12.0 144.00
## 93 2.564949 13.0 169.00
## 94 2.639057 14.5 210.25
## 95 2.564949 11.0 121.00
## 96 2.484907 11.0 121.00
## 97 2.564949 11.0 121.00
## 98 2.890372 16.5 272.25
## 99 2.772589 18.0 324.00
## 100 2.890372 16.0 256.00
## 101 2.890372 16.5 272.25
## 102 3.135494 16.0 256.00
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## 116 2.708050 13.0 169.00
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## 378 3.433987 17.6 309.76
## 379 3.637586 14.7 216.09
## 380 3.583519 17.3 299.29
## 381 3.583519 14.5 210.25
## 382 3.583519 14.5 210.25
## 383 3.526361 16.9 285.61
## 384 3.637586 15.0 225.00
## 385 3.465736 15.7 246.49
## 386 3.637586 16.2 262.44
## 387 3.218876 16.4 268.96
## 388 3.637586 17.0 289.00
## 389 3.258097 14.5 210.25
## 390 3.091042 14.7 216.09
## 391 3.465736 13.9 193.21
## 392 3.583519 13.0 169.00
## 393 3.295837 17.3 299.29
## 394 3.295837 15.6 243.36
## 395 3.784190 24.6 605.16
## 396 3.465736 11.6 134.56
## 397 3.332205 18.6 345.96
## 398 3.433987 19.4 376.367. Apply unit normal scaling to the new data frame that you created in Question 6. Is acceleration or its transformation more influential in predicting the Box-Cox transformation of mpg?
#After performing unit normal scaling acceleration without it's transformation is more influential as
#it's estimate (1.730) is larger than the absolute value of the transformation of accelration (1.295)
mpgdata1_unit_normal = as.data.frame(apply(mpg_data1, 2, function(x){(x - mean(x))/sd(x)}))
modelmpg_unit_normal <- lm(log_mpg~., data = mpgdata1_unit_normal)
summary(modelmpg_unit_normal)##
## Call:
## lm(formula = log_mpg ~ ., data = mpgdata1_unit_normal)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.15395 -0.66322 -0.00194 0.64295 2.29063
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.683e-16 4.417e-02 0.000 1.000000
## acceleration 1.730e+00 3.391e-01 5.102 5.22e-07 ***
## acceleration_sq -1.295e+00 3.391e-01 -3.820 0.000155 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.8813 on 395 degrees of freedom
## Multiple R-squared: 0.2273, Adjusted R-squared: 0.2234
## F-statistic: 58.1 on 2 and 395 DF, p-value: < 2.2e-16