str(Auto)
## 'data.frame':    392 obs. of  9 variables:
##  $ mpg         : num  18 15 18 16 17 15 14 14 14 15 ...
##  $ cylinders   : num  8 8 8 8 8 8 8 8 8 8 ...
##  $ displacement: num  307 350 318 304 302 429 454 440 455 390 ...
##  $ horsepower  : num  130 165 150 150 140 198 220 215 225 190 ...
##  $ weight      : num  3504 3693 3436 3433 3449 ...
##  $ acceleration: num  12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
##  $ year        : num  70 70 70 70 70 70 70 70 70 70 ...
##  $ origin      : num  1 1 1 1 1 1 1 1 1 1 ...
##  $ name        : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...

9a

Produce a scatterplot matrix which include all the variables in the data set.

pairs(Auto)

#### 9b

Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the “name” variable, which is qualitative.

names(Auto)
## [1] "mpg"          "cylinders"    "displacement" "horsepower"   "weight"      
## [6] "acceleration" "year"         "origin"       "name"
cor(Auto[1:8])
##                     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

9c

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. Which predictors appear to have a statistically significant relationship to the response?
  2. What does the coefficient for the year variable suggest?
mlr <- lm(mpg ~ . - name, data = Auto)
summary(mlr)
## 
## 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. Is there a relationship between the predictors and the response?

we see the p-value corresponding to the F-statistic is very low and less than significance level 0.05, this means there is significant relationship between mpg which response variable and the other predictors.

  1. Which predictors appear to have a statistically significant relationship to the response ?

We can check the p-values associated with each predictor’s t-statistic. We see that all predictors are significant except for cylinders, horsepower and acceleration which have p value more than significant level 0.05.

  1. What does the coefficient for the “year” variable suggest ?

The coefficient for the year variable suggests that the average effect of an increase of 1 year is an increase of 0.7507727 in mpg with assumption of all other predictors remaining the same. In other words, cars become more fuel efficient every year by a factor of .75 mpg per year.

9d

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(mlr)

As before, the plot of residuals versus fitted values indicates presence of little non linearity in the data. The plot of standardized residuals versus leverage indicates the presence of a few outliers (as seen in plot higher than 2 or lower than -2 cooks distance) and one high leverage point 14.

9e

Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?

fit <- lm(mpg ~ cylinders * displacement+displacement * weight, data = Auto[, 1:8])
summary(fit)
## 
## Call:
## lm(formula = mpg ~ cylinders * displacement + displacement * 
##     weight, data = Auto[, 1:8])
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -13.2934  -2.5184  -0.3476   1.8399  17.7723 
## 
## Coefficients:
##                          Estimate Std. Error t value Pr(>|t|)    
## (Intercept)             5.262e+01  2.237e+00  23.519  < 2e-16 ***
## cylinders               7.606e-01  7.669e-01   0.992    0.322    
## displacement           -7.351e-02  1.669e-02  -4.403 1.38e-05 ***
## weight                 -9.888e-03  1.329e-03  -7.438 6.69e-13 ***
## cylinders:displacement -2.986e-03  3.426e-03  -0.872    0.384    
## displacement:weight     2.128e-05  5.002e-06   4.254 2.64e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.103 on 386 degrees of freedom
## Multiple R-squared:  0.7272, Adjusted R-squared:  0.7237 
## F-statistic: 205.8 on 5 and 386 DF,  p-value: < 2.2e-16

From p-values, we can see that the interaction between displacement and weight is signifcant and the interaction between cylinders and displacement is not significant.

9f

Try a few different transformations of the variables, such as log(X), √ X, X2. Comment on your findings.

par(mfrow = c(2, 2))
plot(log(Auto$horsepower), Auto$mpg)
plot(sqrt(Auto$horsepower), Auto$mpg)
plot((Auto$horsepower)^2, Auto$mpg)

We tried to use horsepower as a single predictor and based on the plots above we can conclude that the log transformation have maximum linearity.

10

This question should be answered using the Carseats data set.

(a) Fit a multiple regression model to predict Sales using Price, Urban, and US.

data(Carseats)
mlr1 <- lm(Sales ~ Price + Urban + US, data = Carseats)
summary(mlr1)
## 
## 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

##10b Provide an interpretation of each coefficient in the model. Be careful - some of the variables in the model are qualitative!

Answer: The coefficient of the Price variable may be interpreted as that the average effect of a price increase of 1 dollar is a decrease of 54.4588492 units in Sales with assuming all other predictors remaining the same.

The coefficient of the Urban variable may be interpreted as that on average the unit sales in urban yes location are 21.9161508 units less than in urban no location with assuming all other predictors remaining the same.

The coefficient of the US variable may be interpreted as that on average the unit sales in a US store are 1200.5726978 units more than in a non US store with assuming all other predictors remaining the same.

10c. Write out the model in equation form, being careful to handle the qualitative variables properly.

Answer: The model can be written as below:

Sales=13.0434689+(−0.0544588)×Price+(−0.0219162)×Urban+(1.2005727)×US+ Error

Urban=1 if the store is in an urban location and 0 if not. US=1 if the store is in the US and 0 if not.

10d. For which of the predictors can you reject the null hypothesis H0 : βj = 0?

We can reject the null hypothesis for the Price and US variables as they have p values less than significant level of 0.05

10e. 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.

mlr2 <- lm(Sales ~ Price + US, data = Carseats)
summary(mlr2)
## 
## 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

10f. How well do the models in (a) and (e) fit the data?

The R Square of the smaller model is slightly better than for the bigger model with all variables. And about 23.93% of the variability can be explained by the model.

10g. Using the model from (e), obtain 95 % confidence intervals for the coefficient(s).

confint(mlr2)
##                   2.5 %      97.5 %
## (Intercept) 11.79032020 14.27126531
## Price       -0.06475984 -0.04419543
## USYes        0.69151957  1.70776632

10h. Is there evidence of outliers or high leverage observations in the model from (e)?

par(mfrow = c(2, 2))
plot(mlr2)

The standardized residuals versus leverage plot above indicates the presence of a few outliers (higher than 2 or lower than -2 cooks distance) and some leverage points as some points exceed (p+1)/n (0.01).

12. This problem involves simple linear regression without an intercept.

(a) Recall that the coefficient estimate βˆ 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?

(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)
x <- 1:100
sum(x^2)
## [1] 338350
y <- 2 * x + rnorm(100, sd = 0.1)
sum(y^2)
## [1] 1353606
fit.Y <- lm(y ~ x + 0)
fit.X <- lm(x ~ y + 0)
summary(fit.Y)
## 
## Call:
## lm(formula = y ~ x + 0)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.223590 -0.062560  0.004426  0.058507  0.230926 
## 
## Coefficients:
##    Estimate Std. Error t value Pr(>|t|)    
## x 2.0001514  0.0001548   12920   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.09005 on 99 degrees of freedom
## Multiple R-squared:      1,  Adjusted R-squared:      1 
## F-statistic: 1.669e+08 on 1 and 99 DF,  p-value: < 2.2e-16
summary(fit.X)
## 
## Call:
## lm(formula = x ~ y + 0)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.115418 -0.029231 -0.002186  0.031322  0.111795 
## 
## Coefficients:
##   Estimate Std. Error t value Pr(>|t|)    
## y 5.00e-01   3.87e-05   12920   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.04502 on 99 degrees of freedom
## Multiple R-squared:      1,  Adjusted R-squared:      1 
## F-statistic: 1.669e+08 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.

x <- 1:100
sum(x^2)
## [1] 338350
y <- 100:1
sum(y^2)
## [1] 338350
fit.Y <- lm(y ~ x + 0)
fit.X <- lm(x ~ y + 0)
summary(fit.Y)
## 
## Call:
## lm(formula = y ~ x + 0)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -49.75 -12.44  24.87  62.18  99.49 
## 
## Coefficients:
##   Estimate Std. Error t value Pr(>|t|)    
## x   0.5075     0.0866    5.86 6.09e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 50.37 on 99 degrees of freedom
## Multiple R-squared:  0.2575, Adjusted R-squared:   0.25 
## F-statistic: 34.34 on 1 and 99 DF,  p-value: 6.094e-08
summary(fit.X)
## 
## Call:
## lm(formula = x ~ y + 0)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -49.75 -12.44  24.87  62.18  99.49 
## 
## Coefficients:
##   Estimate Std. Error t value Pr(>|t|)    
## y   0.5075     0.0866    5.86 6.09e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 50.37 on 99 degrees of freedom
## Multiple R-squared:  0.2575, Adjusted R-squared:   0.25 
## F-statistic: 34.34 on 1 and 99 DF,  p-value: 6.094e-08