# Objectives

In this case study, we will fit a simple linear regression model. Later, we will also include one categorical variable into the model and check for interaction. The objective here will be transform data (if necessary), fit the model, analyse it, and write the fitted equations for the two studied groups. To do this, we will a the model, check the assumptions, re-fit it (if necessary), interpret it, visualize the estimated curves, and predict new values based on this fitted model for different classes for the categorical variable.

# 1 Introduction

For this problem, we will analyse data about the mileage per gallon performances of various cars. The data set was retrieved from this page (with changes). You can download the .csv file here.

This data set contains the following variables:

1. mpg (continuous)
2. cylinders (multi-valued discrete)
3. displacement (continuous)
4. horsepower (continuous)
5. weight (continuous)
6. accelaration (continuous)
7. model year (multi-valued discrete)
8. origin (multi-valued discrete)

# 2 Exploring the data set

In order to explore the data set and perform initial analyses, we have to read it (with R) first. Provided that the .csv file is saved within the datasets/ folder, one can read the file in the following way.

col.names <- c('mpg', 'cylinders', 'displacement', 'hp', 'weight', 'acceleration', 'year', 'origin')
car <- read.csv(file = 'datasets/car.csv', header = FALSE, sep = ',', col.names = col.names)
head(car, 5)
##   mpg cylinders displacement  hp weight acceleration year origin
## 1  18         8          307 130   3504         12.0   70      1
## 2  16         8          304 150   3433         12.0   70      1
## 3  17         8          302 140   3449         10.5   70      1
## 4  NA         8          350 165   4142         11.5   70      1
## 5  NA         8          351 153   4034         11.0   70      1

Now, let’s see a summary of our data.

summary(car)
##       mpg          cylinders      displacement         hp            weight
##  Min.   :12.00   Min.   :3.000   Min.   : 68.0   Min.   : 48.0   Min.   :1613
##  1st Qu.:18.00   1st Qu.:4.000   1st Qu.:107.0   1st Qu.: 77.5   1st Qu.:2237
##  Median :23.00   Median :4.000   Median :148.5   Median : 92.0   Median :2804
##  Mean   :23.74   Mean   :5.372   Mean   :187.9   Mean   :100.9   Mean   :2936
##  3rd Qu.:29.80   3rd Qu.:6.000   3rd Qu.:250.0   3rd Qu.:113.5   3rd Qu.:3456
##  Max.   :39.00   Max.   :8.000   Max.   :400.0   Max.   :190.0   Max.   :4746
##  NA's   :6                                       NA's   :2
##   acceleration        year           origin
##  Min.   : 8.00   Min.   :70.00   Min.   :1.000
##  1st Qu.:14.00   1st Qu.:73.00   1st Qu.:1.000
##  Median :15.50   Median :76.00   Median :1.000
##  Mean   :15.58   Mean   :76.36   Mean   :1.558
##  3rd Qu.:17.00   3rd Qu.:79.75   3rd Qu.:2.000
##  Max.   :23.50   Max.   :82.00   Max.   :3.000
## 

As one can see from the above table, some multi-valued discrete attributes are being interpreted as integer values; also, we have NA’s for the mpg and horsepower attributes. To verify (and change) the variable types, we can do the following

car$cylinders <- as.factor(car$cylinders)
car$year <- as.factor(car$year)
car$origin <- as.factor(car$origin)

Also, as there are too many classes for the year, and as a way to make our analyses simpler, let’s categorize the cars into old and new, such that cars from before 77 will be labeled as 1 and the remaining cars will be labeled as 2.

car$year <- as.factor(sapply(X = car$year, FUN = function (item) { ifelse(item %in% 70:76, 1, 2) }))

summary(car)
##       mpg        cylinders  displacement         hp            weight
##  Min.   :12.00   3:  3     Min.   : 68.0   Min.   : 48.0   Min.   :1613
##  1st Qu.:18.00   4:134     1st Qu.:107.0   1st Qu.: 77.5   1st Qu.:2237
##  Median :23.00   5:  1     Median :148.5   Median : 92.0   Median :2804
##  Mean   :23.74   6: 62     Mean   :187.9   Mean   :100.9   Mean   :2936
##  3rd Qu.:29.80   8: 58     3rd Qu.:250.0   3rd Qu.:113.5   3rd Qu.:3456
##  Max.   :39.00             Max.   :400.0   Max.   :190.0   Max.   :4746
##  NA's   :6                                 NA's   :2
##   acceleration   year    origin
##  Min.   : 8.00   1:132   1:167
##  1st Qu.:14.00   2:126   2: 38
##  Median :15.50           3: 53
##  Mean   :15.58
##  3rd Qu.:17.00
##  Max.   :23.50
## 

Now, let’s deal with the missing values. Different approaches could have been taken here, and they highly depend on your problem (and your knowledge about the problem). For this particular example, suppose that we want to describe the mpg data as a function of the hp and year. Since we do not now much about this data, a simpler options would be to exclude the instances with missing values for the hp. Let’s do this.

car2 <- car[!is.na(car$hp), c('mpg', 'hp', 'year')] summary(car2) ## mpg hp year ## Min. :12.00 Min. : 48.0 1:131 ## 1st Qu.:18.00 1st Qu.: 77.5 2:125 ## Median :23.00 Median : 92.0 ## Mean :23.75 Mean :100.9 ## 3rd Qu.:29.80 3rd Qu.:113.5 ## Max. :39.00 Max. :190.0 ## NA's :6 Given this smaller data set, our goal might be to predict the missing values for mpg. However, to do this, we have to have a data set with no NA’s. Let’s name it car3. car3 <- car2[!is.na(car2$mpg), ]

As a last exploration step, let’s plot our data set.

plot(mpg ~ hp, pch = 19, col = (as.numeric(year) + 1), data = car3)
legend('topright', c('old', 'new'), col = unique(as.numeric(car3$year) + 1), pch = 19) # 3 Fitting a model From the previous plot, although we suspect that a linear model might not be appropriate for this data set as it is, let’s fit it and analyse the results. In particular, we will fit the following model $\texttt{mpg}_i = \beta_0 + \texttt{hp}_i \cdot \beta_1 + \epsilon_i; \text{ such that } \epsilon_i \overset{\text{i.i.d.}}{\sim} \text{Normal}(0, \sigma^2_{\epsilon})$ model <- lm(formula = mpg ~ hp, data = car3) summary(model) ## ## Call: ## lm(formula = mpg ~ hp, data = car3) ## ## Residuals: ## Min 1Q Median 3Q Max ## -9.3908 -3.3377 -0.0706 2.7713 11.7533 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 42.543005 0.941244 45.20 <2e-16 *** ## hp -0.188003 0.009001 -20.89 <2e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 4.362 on 248 degrees of freedom ## Multiple R-squared: 0.6376, Adjusted R-squared: 0.6361 ## F-statistic: 436.2 on 1 and 248 DF, p-value: < 2.2e-16 From the above summary, we have strong evidences that both $$\beta_0$$ and $$\beta_1$$ are different than 0. The residuals do not seem symmetric, though. Also, $$\text{R}^2 =$$ 0.6375558. Now, let’s plot the fitted model. plot(mpg ~ hp, pch = 19, col = (as.numeric(year) + 1), data = car3) abline(model) legend('topright', c('old', 'new'), col = unique(as.numeric(car3$year) + 1), pch = 19)

However, notice that the relation between hp and mpg does not seem to be linear, and the age might also provide information when describe the response variable. Thus, before taking any conclusions from the fitted model, let’s do an analysis of residuals. We will focus on the “Residuals vs Fitted” and “Normal Q-Q” plots.

plot(model, which = c(1, 2))

From the “Residuals vs Fitted” plots, we may see that a linear relationship does not correctly describe how mpg is written as a function of hp, since we can see a pattern for the residuals (as opposed to a “well spread and random” cloud of points around $$y = 0$$). Also, from the “Normal Q-Q” plot, the residuals seem to be normally distributed (we will test it).

To confirm these visual analyses, let’s conduct a proper test. To check for the assumption of equal variance, since we have a quantitative regressor, we can use the Score Test, available in the car package through the ncvTest() function. Also, to check for the normality of the residuals, we will use the Shapiro-Wilk test (shapiro.test()).

library('car')
ncvTest(model)
## Non-constant Variance Score Test
## Variance formula: ~ fitted.values
## Chisquare = 6.945645, Df = 1, p = 0.0084024
shapiro.test(resid(model))
##
##  Shapiro-Wilk normality test
##
## data:  resid(model)
## W = 0.98926, p-value = 0.06041

As the p-values are too small for the first test, we have strong evidences against equal variance. On the other hand, we fail to reject the hypothesis of normally distributed residuals (with a significance level of $$5\%$$). Thus, as at least one assumption for this model does not hold, the results might not be reliable.

As a way to overcome this issue, we will transform the data according to the following rule

$w(\lambda) = \begin{cases} (y^{\lambda} - 1)/\lambda &, \text{ if } \lambda \neq 0 \\ \log(y) &, \text{ if } \lambda = 0. \end{cases}$

This can be achieved by using the boxCox() function from the car package. Based on it, we will retrieve the value of lambda and will apply the above transformation.

bc <- boxCox(model)

(lambda <- bc$x[which.max(bc$y)])
## [1] -0.5454545

Now, let’s create a function to transform our data based on the value of $$\lambda$$ and based on the above rule. We will name it tranfData(). Also, we will create a function to transform our data back to the original scale. We will name it tranfData_back(). This second function, if $$\lambda \neq 0$$, will be given by $$y(\lambda) = (w\lambda + 1)^{1/\lambda}$$.

transfData <- function (data, lambda) { ((data ^ lambda - 1) / lambda) }
transfData_back <- function (data, lambda) { ((data * lambda + 1) ^ (1 / lambda)) }

Therefore, we can easily transform our data, given $$\lambda =$$ -0.55, in the following way

car3$mpg_transf <- sapply(X = car3$mpg, FUN = transfData, lambda = lambda)
head(car3, 5)
##   mpg  hp year mpg_transf
## 1  18 130    1   1.454413
## 2  16 150    1   1.429271
## 3  17 140    1   1.442414
## 8  14 160    1   1.398742
## 9  24  95    1   1.509442
plot(mpg_transf ~ hp, pch = 19, col = (as.numeric(year) + 1), data = car3)
legend('topright', c('old', 'new'), col = unique(as.numeric(car3$year) + 1), pch = 19) Finally, we can fit again a linear model for the transformed data. model2 <- lm(formula = mpg_transf ~ hp, data = car3) summary(model2) ## ## Call: ## lm(formula = mpg_transf ~ hp, data = car3) ## ## Residuals: ## Min 1Q Median 3Q Max ## -0.079235 -0.021397 0.001967 0.020464 0.074192 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 1.656e+00 6.607e-03 250.69 <2e-16 *** ## hp -1.622e-03 6.318e-05 -25.68 <2e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.03062 on 248 degrees of freedom ## Multiple R-squared: 0.7267, Adjusted R-squared: 0.7256 ## F-statistic: 659.3 on 1 and 248 DF, p-value: < 2.2e-16 plot(mpg_transf ~ hp, pch = 19, col = (as.numeric(year) + 1), data = car3) abline(model2) legend('topright', c('old', 'new'), col = unique(as.numeric(car3$year) + 1), pch = 19)

Also, we can analyse the diagnostic plots, as before. As well as conduct the appropriate tests.

plot(model2, which = c(1, 2))

ncvTest(model2)
## Non-constant Variance Score Test
## Variance formula: ~ fitted.values
## Chisquare = 1.087962, Df = 1, p = 0.29692
shapiro.test(resid(model2))
##
##  Shapiro-Wilk normality test
##
## data:  resid(model2)
## W = 0.98934, p-value = 0.06252

As we would expect, the results look much better now. However, we can still use information about year, which seems to play a role in explaining the response variable. That being said, let’s fit this new model.

Notice that we will consider a model with interaction (an interaction occurs when an independent variable has a different effect on the outcome depending on the values of another independent variable). For an extensive discussion on this topic, one can refer to this link.

model3 <- lm(formula = mpg_transf ~ hp * year, data = car3)
summary(model3)
##
## Call:
## lm(formula = mpg_transf ~ hp * year, data = car3)
##
## Residuals:
##       Min        1Q    Median        3Q       Max
## -0.061671 -0.017084  0.001801  0.018128  0.065467
##
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)
## (Intercept)  1.620e+00  7.767e-03 208.563  < 2e-16 ***
## hp          -1.435e-03  6.886e-05 -20.836  < 2e-16 ***
## year2        4.570e-02  1.159e-02   3.945 0.000104 ***
## hp:year2    -1.118e-04  1.133e-04  -0.986 0.325015
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02561 on 246 degrees of freedom
## Multiple R-squared:  0.8103, Adjusted R-squared:  0.808
## F-statistic: 350.3 on 3 and 246 DF,  p-value: < 2.2e-16

From the above table, we can see that the interaction (hp:year2) is not significant; therefore, we can fit a simpler model.

model4 <- lm(formula = mpg_transf ~ hp + year, data = car3)
coeffi <- model4$coefficients summary(model4) ## ## Call: ## lm(formula = mpg_transf ~ hp + year, data = car3) ## ## Residuals: ## Min 1Q Median 3Q Max ## -0.061994 -0.017143 0.001694 0.017929 0.065681 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 1.624e+00 6.323e-03 256.91 <2e-16 *** ## hp -1.476e-03 5.469e-05 -26.99 <2e-16 *** ## year2 3.477e-02 3.353e-03 10.37 <2e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.02561 on 247 degrees of freedom ## Multiple R-squared: 0.8096, Adjusted R-squared: 0.808 ## F-statistic: 525 on 2 and 247 DF, p-value: < 2.2e-16 plot(mpg_transf ~ hp, pch = 19, col = (as.numeric(year) + 1), data = car3) abline(coeffi[[1]], coeffi[[2]], col = 2) abline(coeffi[[1]] + coeffi[[3]], coeffi[[2]], col = 3) legend('topright', c('old', 'new'), col = unique(as.numeric(car3$year) + 1), pch = 19)

Again, we can analyse the diagnostic plots and conduct the appropriate tests.

plot(model4, which = c(1, 2))

ncvTest(model4)
## Non-constant Variance Score Test
## Variance formula: ~ fitted.values
## Chisquare = 2.838162, Df = 1, p = 0.092049
shapiro.test(resid(model4))
##
##  Shapiro-Wilk normality test
##
## data:  resid(model4)
## W = 0.99006, p-value = 0.08516

Thus, for a significance level of $$5\%$$ we fail to reject the hypotheses of equal variance and normality. Meaning that this might be an appropriate model for our data. However, recall that we are modelling a transformed data set. We can get a model for our original data by doing the following. For a transformation $$f$$, we have the

\begin{align*} \texttt{mpg}_i &= f^{-1}(1.624 -0.001\texttt{hp}_i)&, \text{ if } \texttt{year} = 1 \\ \texttt{mpg}_i &= f^{-1}((1.624 + 0.035) -0.001\texttt{hp}_i)&, \text{ if } \texttt{year} = 2 \end{align*}

And we can plot it in the following way

plot(mpg ~ hp, pch = 19, col = (as.numeric(year) + 1), data = car3)
curve(transfData_back(coeffi[[1]] + coeffi[[2]] * x, lambda = lambda), from = 0, to = 250, add = TRUE, col = 2)
curve(transfData_back((coeffi[[1]] + coeffi[[3]]) + coeffi[[2]] * x, lambda = lambda), from = 0, to = 250, add = TRUE, col = 3)
legend('topright', c('old', 'new'), col = unique(as.numeric(car3$year) + 1), pch = 19) # 4 Predicting unknown values Now that we have a “good” fitted model, we can predict, as suggested before, the values of mpg for which we had NA’s before. We can do this in the following way pos_unk <- which(is.na(car2$mpg))
unknown <- car2[is.na(car2$mpg), ] (predicted_values <- sapply(X = predict(object = model4, newdata = data.frame(hp = unknown$hp, year = unknown$year)), FUN = transfData_back, lambda = lambda)) ## 1 2 3 4 5 6 ## 13.00153 13.98905 12.25847 12.25847 31.38660 22.37227 car2[is.na(car2$mpg), 'mpg'] <- predicted_values

pch <- rep(19, nrow(car2)); pch[pos_unk] <- 9
plot(mpg ~ hp, pch = pch, col = (as.numeric(year) + 1), data = car2)
curve(transfData_back(coeffi[[1]] + coeffi[[2]] * x, lambda = lambda), from = 0, to = 250, add = TRUE, col = 2)
curve(transfData_back((coeffi[[1]] + coeffi[[3]]) + coeffi[[2]] * x, lambda = lambda), from = 0, to = 250, add = TRUE, col = 3)
legend('topright', c('old', 'new'), col = unique(as.numeric(car3\$year) + 1), pch = 19)