Homework 7
A. Franco
2025-04-08
library(tidyverse)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## âś” dplyr 1.1.4 âś” readr 2.1.5
## âś” forcats 1.0.0 âś” stringr 1.5.1
## âś” ggplot2 3.5.1 âś” tibble 3.2.1
## âś” lubridate 1.9.4 âś” tidyr 1.3.1
## âś” purrr 1.0.2
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## âś– dplyr::filter() masks stats::filter()
## âś– dplyr::lag() masks stats::lag()
## â„ą Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorslibrary(readxl)
library(MASS)##
## Attaching package: 'MASS'
##
## The following object is masked from 'package:dplyr':
##
## selectlibrary(lmtest)## Loading required package: zoo
##
## Attaching package: 'zoo'
##
## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numericcars_data<-read.csv("cars_data(1).csv")cars1<-lm(wt~hp+am,data=cars_data)
plot(cars1,which=1)
Here we see that the red line is mostly close to the straight line which
shows there is a linear relationship where horsepower and transmission
affect the weight.
raintest(cars1)##
## Rainbow test
##
## data: cars1
## Rain = 0.9008, df1 = 16, df2 = 13, p-value = 0.5844In the Rainbow Test, a low p-value means that the data is NOT linear. As we can see, this model is just barely linear.
library(car)## Loading required package: carData##
## Attaching package: 'car'## The following object is masked from 'package:dplyr':
##
## recode## The following object is masked from 'package:purrr':
##
## somedurbinWatsonTest(cars1)## lag Autocorrelation D-W Statistic p-value
## 1 0.4261432 1.101482 0.01
## Alternative hypothesis: rho != 0A p-value greater than 0.05 means that the errors ARE independent. In this case, the errors are not independent since the p-value=0.008. Also, the D-W statistic is not close to “2”, which is not good.
plot(cars1,which=3)
The red line should be almost totally straight, and the points should
cluster evenly around it. This red line is starts straight but then gets
wavy, so this likely violates the assumption of Homoscedasticity.
bptest(cars1)##
## studentized Breusch-Pagan test
##
## data: cars1
## BP = 5.0175, df = 2, p-value = 0.08137In this case the, p-Value > 0.05 so the model is not homoscedatic, therefore heterscedasticity exists. I think.
plot(cars1,which=2)
There is some deviation from the dotted line but its not much as seen, lots of the plots land on the line when reviewing for the Normality of Residuals.
shapiro.test(cars1$residuals)##
## Shapiro-Wilk normality test
##
## data: cars1$residuals
## W = 0.92104, p-value = 0.02219Since the p-value is less than 0.05, this means that the residuals are significantly different from a normal distribution, which means the are not normal.
vif(cars1)## hp am
## 1.062867 1.062867Since neither of these are over 10 means that the variables is not strongly correlated with some other variable. There is no multicolinarity.
My model is not fair well. It failed/violated several test including the Shapiro Test, and was not normally distributive. Also it was heterostatic. This would require some mitigation.
cars1<-lm(wt~hp+am,data=cars_data)
cars1log<-lm(log(wt)~log(hp)+am,data=cars_data)
raintest(cars1log)##
## Rainbow test
##
## data: cars1log
## Rain = 1.7267, df1 = 16, df2 = 13, p-value = 0.1628raintest(cars1log)##
## Rainbow test
##
## data: cars1log
## Rain = 1.7267, df1 = 16, df2 = 13, p-value = 0.1628Logging the information and re-running a rainbow test increased the p-value which now makes it linear.
cars1_robust<-rlm(wt~hp,data=cars_data)
summary(cars1)##
## Call:
## lm(formula = wt ~ hp + am, data = cars_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.83338 -0.24390 -0.05175 0.15592 1.24801
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.576956 0.255062 10.103 5.22e-11 ***
## hp 0.007437 0.001406 5.289 1.14e-05 ***
## am -1.109360 0.193215 -5.742 3.24e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5207 on 29 degrees of freedom
## Multiple R-squared: 0.7351, Adjusted R-squared: 0.7168
## F-statistic: 40.24 on 2 and 29 DF, p-value: 4.315e-09summary(cars1_robust)##
## Call: rlm(formula = wt ~ hp, data = cars_data)
## Residuals:
## Min 1Q Median 3Q Max
## -1.3739796 -0.5225638 -0.0004306 0.4365530 1.6001726
##
## Coefficients:
## Value Std. Error t value
## (Intercept) 1.8491 0.3076 6.0110
## hp 0.0092 0.0019 4.8210
##
## Residual standard error: 0.6837 on 30 degrees of freedombptest(cars1)##
## studentized Breusch-Pagan test
##
## data: cars1
## BP = 5.0175, df = 2, p-value = 0.08137bptest(cars1log)##
## studentized Breusch-Pagan test
##
## data: cars1log
## BP = 0.27048, df = 2, p-value = 0.8735plot(cars1,which=2)
plot(cars1log,which=2)
Although my model stayed heteroscedasic, it slightly improved more than it was before.
vif(cars1)## hp am
## 1.062867 1.062867This didn’t change, it remained below 2.
cars1<-lm(wt~hp+am,data=cars_data)
summary(cars1)##
## Call:
## lm(formula = wt ~ hp + am, data = cars_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.83338 -0.24390 -0.05175 0.15592 1.24801
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.576956 0.255062 10.103 5.22e-11 ***
## hp 0.007437 0.001406 5.289 1.14e-05 ***
## am -1.109360 0.193215 -5.742 3.24e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5207 on 29 degrees of freedom
## Multiple R-squared: 0.7351, Adjusted R-squared: 0.7168
## F-statistic: 40.24 on 2 and 29 DF, p-value: 4.315e-09The mitigation helped some. So yes, weight is affected by both horsepower and transmission choices, but is not in a direct way. A heavier vehicle will most likely require more horsepower to make it perform, transmission can be adjusted to the cars weight to optimize performance like a lower weight car. Its accurate to say that horse power and transmission are significant variables.