Suppose that you want to build a regression model that predicts the price of cars using a data set named cars.

Q1. Per the scatter plot and the computed correlation coefficient, describe relationships between the two variables - price and weight.

Make sure to interpret the direction and the magnitude of the relationship. In addition, keep in mind that correlation (or regression) coefficients do not show causation but only association.

ANS: There appears to be a strong positive relationship between weight and price. (the coefficient’s absolute value is about 0.8, which is > 0.6) (its sign is positive)

Create scatterplots

## 'data.frame':    54 obs. of  6 variables:
##  $ type      : Factor w/ 3 levels "large","midsize",..: 3 2 2 2 2 1 1 2 1 2 ...
##  $ price     : num  15.9 33.9 37.7 30 15.7 20.8 23.7 26.3 34.7 40.1 ...
##  $ mpgCity   : int  25 18 19 22 22 19 16 19 16 16 ...
##  $ driveTrain: Factor w/ 3 levels "4WD","front",..: 2 2 2 3 2 2 3 2 2 2 ...
##  $ passengers: int  5 5 6 4 6 6 6 5 6 5 ...
##  $ weight    : int  2705 3560 3405 3640 2880 3470 4105 3495 3620 3935 ...

## [1] 0.758112

Interpretation

Run a regression model for price with one explanatory variable, weight, and answer Q2 through Q5.

Q2. Is the coefficient of weight statistically significant at 5%? Interpret the coefficient.

ANS: Looking at the significant codes, it indicates three dots (***) which means that the weight is statisically significant at 5%. Changes in the weight explains changes in price. (Increasing)

Q3. What price does the model predict for a car that weighs 4000 pounds?

Hint: Check the units of the variables in the openintro manual.

ANS: Based off the math that I did, the price predicts to be $43,000 when a car weighs 4000 lbs. W = 2171 + (43 * P) 4000 lbs = 2171 + (43 * P) P = 43

Q4. What is the reported residual standard error? What does it mean?

ANS: The reported residual standard error is listed as 433 lbs. This means that the difference between the actual weight and the predicted weight is 433 lbs. The model misses the actual weight by 433 lbs.

Q5. What is the reported adjusted R squared? What does it mean?

Run a second regression model for price with two explanatory variables: weight and passengers, and answer Q6.

ANS: The adjusted R squared is 0.7209 which means that 72% of the variability in weight can be explained by price.

Q6. Which of the two models better fits the data? Discuss your answer by comparing the residual standard error and the adjusted R squared between the two models.

ANS: The second model better fits the data because the second model has a smaller residual error The first model has a residual error or 433 and the second model has a reidual error of 347.

Build regression model

## 
## Call:
## lm(formula = weight ~ price, data = cars)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1328.29  -228.09    10.92   258.19   924.27 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 2171.113    118.956  18.251  < 2e-16 ***
## price         43.331      5.169   8.383 3.17e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 433 on 52 degrees of freedom
## Multiple R-squared:  0.5747, Adjusted R-squared:  0.5666 
## F-statistic: 70.28 on 1 and 52 DF,  p-value: 3.173e-11
## 
## Call:
## lm(formula = weight ~ price + passengers, data = cars)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -976.81 -201.56    6.13  151.33  799.88 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   294.25     356.98   0.824    0.414    
## price          35.99       4.36   8.256 5.80e-11 ***
## passengers    395.91      72.56   5.456 1.44e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 347.4 on 51 degrees of freedom
## Multiple R-squared:  0.7315, Adjusted R-squared:  0.7209 
## F-statistic: 69.46 on 2 and 51 DF,  p-value: 2.748e-15

Interpretation