βœ”οΈ Formative Problem Set 01 | Solutions

DS202 - Data Science for Social Scientists

Author

DS202 2022MT Teaching Group

Published

25 October 2022

βš™οΈ Setup

Import required libraries.No need to change if you are not planning to use other packages.

library(car)
library(ISLR2)
library(tidyverse)
# If you use other packages, add them here

library(ggcorrplot)
library(tidymodels) ## to use the tidy() function to extract data from lm models

🎯 Questions

Use the Carseats data set in the ISLR2 package to answer the following questions:

Q1. Variables

List the names of the variables in this dataset and whether they are quantitative or qualitative. (Worth 1/100 mock marks)

Quantitative:

  • Sales

  • CompPrice

  • Income

  • Advertising

  • Population

  • Price

  • Age

  • Education

Qualitative:

  • ShelveLoc

  • Urban

  • US

Q2. Dimension

Use R to list the number of rows in the dataset. (Worth 1/100 mock marks)

nrow(Carseats)
[1] 400

Q3. Visual

Selecting only the quantitative variables, plot the correlation between variables (Worth 5/100 mock marks)

Tip: If you want to use other R packages, add them to the list of libraries at the top.

Carseats_quant <- Carseats %>% select(-c(ShelveLoc, Urban, US))
corr_Carseats <- cor(Carseats_quant)

ggcorrplot(corr_Carseats, lab=TRUE, digits=2, lab_size=3)

Q4. Initial Variable Identification

Based on just an initial inspection of the data, which variables would you select to train an algorithm to predict Sales? Why? (Worth 7/100 mock marks)

Price is clearly the best first candidate. It is the variable most highly correlated with the dependent variable (Sales) in absolute terms. The correlation between Price and Sales is -0.44. As price increases, sales decreases in an approximately linear fashion.

g <- (
  ggplot(Carseats, aes(x=Price, y=Sales))
  
  + geom_point(size=3, alpha=0.5)
  
  + theme_bw()
  
  )

g

If we want to add more variables, the next candidates would be Age and Advertising, as these variables also exhibit a correlation with Sales that is not too close from zero, in absolute terms. Another interesting point about these two variables is the fact that they are not correlated to each other; that is, correlation Age vs Advertising and Age vs Price and Price vs Advertising is close to zero.

g <- (
  ggplot(Carseats, aes(x=Age, y=Sales))
  
  + geom_point(size=3, alpha=0.5)
  
  + theme_bw()
  
  )

g

g <- (
  ggplot(Carseats, aes(x=Advertising, y=Sales))
  
  + geom_point(size=3, alpha=0.5)
  
  + theme_bw()
  
  )

g

Q5. Simple Linear Regression

Chose ONE SINGLE variable – any variable – and fit a linear regression to predict Sales. Show the summary of this linear model. (Worth 3/100 mock marks)

simple_model <- lm(Sales ~ Price, data=Carseats)

summary(simple_model)

Call:
lm(formula = Sales ~ Price, data = Carseats)

Residuals:
    Min      1Q  Median      3Q     Max 
-6.5224 -1.8442 -0.1459  1.6503  7.5108 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 13.641915   0.632812  21.558   <2e-16 ***
Price       -0.053073   0.005354  -9.912   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.532 on 398 degrees of freedom
Multiple R-squared:  0.198, Adjusted R-squared:  0.196 
F-statistic: 98.25 on 1 and 398 DF,  p-value: < 2.2e-16

Q6. Simple Linear Regression - Interpretation

Provide an interpretation of each coefficient in the model, stating their values, whether they are significant and what they represent. Be carefulβ€”some of the variables in the model are qualitative! (Worth 10/100 mock marks)

When Price = 0, the model predicts Sales = 13.64.

For every $100 increase in Price, Sales decrease by approximately 5.3 units.

Both the intercept and the regression coefficient for Price are statistically significant, their p-values are close to zero.

Q7. Simple Linear Regression - Formula

Write the model in equation form, carefully handling the qualitative variables properly. (Worth 5/100 mock marks)

\[Sales = - 0.053073 \times Price + 13.641915 + \epsilon\]

Q8. Multiple Linear Regression

Chose ONLY THREE variables and fit a linear regression to predict Sales. Show the summary of this linear model. (Worth 3/100 mock marks)

multiple_model <- lm(Sales ~ Price + Age + Advertising, data=Carseats)

summary(multiple_model)

Call:
lm(formula = Sales ~ Price + Age + Advertising, data = Carseats)

Residuals:
    Min      1Q  Median      3Q     Max 
-6.6247 -1.5288  0.0148  1.5220  6.2925 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 16.003472   0.718754  22.266  < 2e-16 ***
Price       -0.058028   0.004827 -12.022  < 2e-16 ***
Age         -0.048846   0.007047  -6.931 1.70e-11 ***
Advertising  0.123106   0.017095   7.201 3.02e-12 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.269 on 396 degrees of freedom
Multiple R-squared:  0.3595,    Adjusted R-squared:  0.3547 
F-statistic:  74.1 on 3 and 396 DF,  p-value: < 2.2e-16

Q9. Multiple Linear Regression - Interpretation

Provide an interpretation of each coefficient in the model, stating their values, whether they are significant and what they represent. Be carefulβ€”some of the variables in the model are qualitative! (Worth 10/100 mock marks)

This multiple regression model is statistically significant (p-value associated with the F-statistic is very small). In addition, all regression coefficients were also found to be statistically significant (p-value close to zero).

In the hypothetical scenario in which \(Price = 0\) and \(Age = 0\) and \(Advertising = 0\) , the multiple regression model predicts \(Sales \approx 16\) units.

  • For every fixed combination of Age and Advertising, an increase in \(100\) dollars in Price is associated with a Sales decrease by approximately 5.8 units.

  • If Price and Advertising are fixed, the Age of a car impacts negatively the number of sales. The regression coefficient is \(-0.048846\) , which means we would expect Sales to decrease by \(1\) unit for every \(\approx 20\) years of age of the Car (since \(\frac{1}{0.048846} \approx 20\) ) .

  • Advertising, on the other hand, is associated with an increase in the number of Sales. If Price and Age are fixed, the model predicts that 12 more items of Sales for each increase in \(100\) Advertising units (dollars?)

Q10. Multiple Linear Regression - Formula

Write the model in equation form, carefully handling the qualitative variables properly. (Worth 5/100 mock marks)

\[Sales = 16.003472 - 0.058028 \times Price - 0.048846 \times Age + 0.123106 \times Advertising + \epsilon\]

Q11. Model comparison

Which of the two models you created, in questions Q5 and Q8 provide a better fit? (Worth 10/100 mock marks)

The multiple model in Q8 fitted the Sales data better.

The adjusted R-squared is higher in the Q8 versus the Q5 model, that is, adding Age and Advertising explains more of the variance in Sales than the model with just Price.

We also see a reduction in the Residual Standard Error.

This is also somewhat apparent from the diagnostic plot of residual vs fitted (shown below). The standardized residuals look more concentrated around zero, on the multiple model.

plot_df <- data.frame(fitted_vals=predict(simple_model),
                      residuals=rstudent(simple_model))

g <- (
  ggplot(plot_df, aes(x=fitted_vals, y=residuals))

  # Add dots
  + geom_point(alpha=0.4, size=3.5) 
  + xlab("Fitted values") 
  + ylab("Residuals") 
  + ylim(c(-4,4))
  + ggtitle("Residuals vs Fitted (Simple Model)")

  # Add lines
  + geom_hline(yintercept=0, size=1.5, color='yellow') 
  + geom_hline(yintercept=3, size=1.5, color='red')

  # Customising the plot +
  + theme_bw()
)
Warning: Using `size` aesthetic for lines was deprecated in ggplot2 3.4.0.
β„Ή Please use `linewidth` instead.
g

plot_df <- data.frame(fitted_vals=predict(multiple_model),
                      residuals=rstudent(multiple_model))

g <- (
  ggplot(plot_df, aes(x=fitted_vals, y=residuals))

  # Add dots
  + geom_point(alpha=0.4, size=3.5) 
  + xlab("Fitted values") 
  + ylab("Residuals") 
  + ylim(c(-4, 4))
  + ggtitle("Residuals vs Fitted (Multiple Model)")

  # Add lines
  + geom_hline(yintercept=0, size=1.5, color='yellow')
  + geom_hline(yintercept=3, size=1.5, color='red')

  # Customising the plot +
  + theme_bw()
)

g

Q12. Collinearity

What is the Variance Inflation Factor (VIF) of each variable? (Worth 3/100 mock marks)

Considering only the model built on Q8:

vif(multiple_model)
      Price         Age Advertising 
   1.012538    1.010550    1.001987

Alternatively, if you interpreted the question to refer to all variables, the following would also be accepted:

vif(lm(Sales ~ ., data=Carseats))
                GVIF Df GVIF^(1/(2*Df))
CompPrice   1.554618  1        1.246843
Income      1.024731  1        1.012290
Advertising 2.103136  1        1.450219
Population  1.145534  1        1.070296
Price       1.537068  1        1.239785
ShelveLoc   1.033891  2        1.008367
Age         1.021051  1        1.010471
Education   1.026342  1        1.013086
Urban       1.022705  1        1.011289
US          1.980720  1        1.407380

Q13. Collinearity (cont.)

Based on your responses to Q3 and the output of Q12, would you consider ignoring any variables when building a linear model? Why/Why not? (Worth 7/100 mock marks)

In terms of vif – which measures the linear dependency of each predictor to ALL the other predictors – there are no β€œproblematic variables”. If there were, we would find variables with vif above 5 or 10.

In terms of pairwise collinearity (plot in Q3), we also don’t find any problematic colinearities.

Q14. Modelling

Considering ALL possible combinations of TWO variables in this dataset, find the one linear model that has the smallest Residual Standard Error. Explain how you reached that conclusion, show us the summary of that model and write the model in equation form. (Worth 15/100 mock marks)

πŸ’‘Tip: Here I show an β€œelegant” way of solving this question by selecting variables by their names and by using dataframes, the pipe and other R functions. There are simpler ways to solve this question, the simplest perhaps would be to solve it by iterating over the column indices instead of column names. If you did it like that, your answer will be accepted, if done correctly.

# Select all columns in the dataset, except Sales and get a list of their names
all_predictors <- Carseats %>% select(-Sales) %>% names()
all_predictors
 [1] "CompPrice"   "Income"      "Advertising" "Population"  "Price"      
 [6] "ShelveLoc"   "Age"         "Education"   "Urban"       "US"

Use the function combn to produce a list of all possible combination of pairs of predictors. In addition to that, let’s reshape the data and transform it to a nicely formatted dataframe with two columns. Since these columns do not have names, the function as.data.frame() automatically name them V1 and V2.

πŸ’‘ Tip: type ?combn and hit ENTER in the R console to read more about the combn function.

all_combn_predictors <- combn(all_predictors, 2) %>% t() %>% as.data.frame() 

all_combn_predictors
            V1          V2
1    CompPrice      Income
2    CompPrice Advertising
3    CompPrice  Population
4    CompPrice       Price
5    CompPrice   ShelveLoc
6    CompPrice         Age
7    CompPrice   Education
8    CompPrice       Urban
9    CompPrice          US
10      Income Advertising
11      Income  Population
12      Income       Price
13      Income   ShelveLoc
14      Income         Age
15      Income   Education
16      Income       Urban
17      Income          US
18 Advertising  Population
19 Advertising       Price
20 Advertising   ShelveLoc
21 Advertising         Age
22 Advertising   Education
23 Advertising       Urban
24 Advertising          US
25  Population       Price
26  Population   ShelveLoc
27  Population         Age
28  Population   Education
29  Population       Urban
30  Population          US
31       Price   ShelveLoc
32       Price         Age
33       Price   Education
34       Price       Urban
35       Price          US
36   ShelveLoc         Age
37   ShelveLoc   Education
38   ShelveLoc       Urban
39   ShelveLoc          US
40         Age   Education
41         Age       Urban
42         Age          US
43   Education       Urban
44   Education          US
45       Urban          US

So, there are a total of 45 combinations of pairs of features. Now, we need to fit linear regression models for all combinations and compute the Residual Standard Error of each.

How do we get the Residual Standard Error of a model? The R documentation says I can do it with the sigma() function. Read more about it here (or just type ?sigma in the R console and hit ENTER):

sigma(simple_model)
[1] 2.532326

How do we create these 45 models ? We can iterate over each line of the all_combn_predictors using a for loop, grab the names of the variables and create a linear model with them.

Linear models in R are typically defined as a formula, but how do I build a formula without typing the name of the variables manually? We can do it by using the as.formula as illustrated in this Stackoverflow response.

First, let me show you how the as.formula combined with paste works. Look at all the formulas produced in this loop:

for(i in 1:nrow(all_combn_predictors)){
  
   # get the i-th row of the dataframe
  selected_vars <- all_combn_predictors[i,]
  
  # build the formula
  model_formula <- as.formula(paste("Sales ~", paste(selected_vars, collapse="+")))
  print(model_formula)
}
Sales ~ CompPrice + Income
Sales ~ CompPrice + Advertising
Sales ~ CompPrice + Population
Sales ~ CompPrice + Price
Sales ~ CompPrice + ShelveLoc
Sales ~ CompPrice + Age
Sales ~ CompPrice + Education
Sales ~ CompPrice + Urban
Sales ~ CompPrice + US
Sales ~ Income + Advertising
Sales ~ Income + Population
Sales ~ Income + Price
Sales ~ Income + ShelveLoc
Sales ~ Income + Age
Sales ~ Income + Education
Sales ~ Income + Urban
Sales ~ Income + US
Sales ~ Advertising + Population
Sales ~ Advertising + Price
Sales ~ Advertising + ShelveLoc
Sales ~ Advertising + Age
Sales ~ Advertising + Education
Sales ~ Advertising + Urban
Sales ~ Advertising + US
Sales ~ Population + Price
Sales ~ Population + ShelveLoc
Sales ~ Population + Age
Sales ~ Population + Education
Sales ~ Population + Urban
Sales ~ Population + US
Sales ~ Price + ShelveLoc
Sales ~ Price + Age
Sales ~ Price + Education
Sales ~ Price + Urban
Sales ~ Price + US
Sales ~ ShelveLoc + Age
Sales ~ ShelveLoc + Education
Sales ~ ShelveLoc + Urban
Sales ~ ShelveLoc + US
Sales ~ Age + Education
Sales ~ Age + Urban
Sales ~ Age + US
Sales ~ Education + Urban
Sales ~ Education + US
Sales ~ Urban + US

Now, let’s actually create the linear models and identify the one with the smallest RSE:

best_RSE        <- Inf   # we don't know yet 
best_predictors <- c()   # start empty, fill later
best_model      <- NULL  # we don't know yet

for(i in 1:nrow(all_combn_predictors)){
  
   # get the i-th row of the dataframe
  selected_vars <- all_combn_predictors[i,]
  
  # build the formula
  model_formula <- as.formula(paste("Sales ~", paste(selected_vars, collapse="+")))
  
  fitted_model      <- lm(model_formula, data=Carseats)
  current_model_RSE <- sigma(fitted_model)
  
  if(current_model_RSE < best_RSE){
    best_RSE <- current_model_RSE
    best_predictors <- selected_vars
    best_model <- fitted_model
    cat("Found a better model!\n")
    cat(paste0("  Vars: [", paste(selected_vars, collapse=","), "]\n"))
    cat(paste0("  RSE:", best_RSE, "\n\n"))
  }
  
}
Found a better model!
  Vars: [CompPrice,Income]
  RSE:2.7899309103294

Found a better model!
  Vars: [CompPrice,Advertising]
  RSE:2.71911905676241

Found a better model!
  Vars: [CompPrice,Price]
  RSE:2.26880676843526

Found a better model!
  Vars: [Advertising,ShelveLoc]
  RSE:2.24418017219941

Found a better model!
  Vars: [Price,ShelveLoc]
  RSE:1.91725333683979

As we can see from above, the best model – if we consider only the minimum RSE – is:

summary(best_model)

Call:
lm(formula = model_formula, data = Carseats)

Residuals:
    Min      1Q  Median      3Q     Max 
-5.8229 -1.3930 -0.0179  1.3868  5.0780 

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
(Intercept)     12.001802   0.503447  23.839  < 2e-16 ***
Price           -0.056698   0.004059 -13.967  < 2e-16 ***
ShelveLocGood    4.895848   0.285921  17.123  < 2e-16 ***
ShelveLocMedium  1.862022   0.234748   7.932 2.23e-14 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.917 on 396 degrees of freedom
Multiple R-squared:  0.5426,    Adjusted R-squared:  0.5391 
F-statistic: 156.6 on 3 and 396 DF,  p-value: < 2.2e-16

\[Sales = 12.001802 - 0.056698\times Price + 4.895848 \times ShelveLoc[Good] + 1.862022 \times ShelveLoc[Medium] + \epsilon\]

Teeechnically, the model above has more than just two variables, it uses three variables (two combinations of ShelveLoc. This is because R converts categorical variables to independent binary variables automatically. Since we didn’t distinguish this case in the question, this is an acceptable response.

Q15. Interaction Effects

Use the * and : symbols to fit linear regression models with interaction effects. Could you find any model with interactions that fit better than the models you built in Questions Q5 & Q8 & Q14? Feel free to use as many interactions and variables as you prefer. Justify your answer. Explain how you reached that conclusion, show us the summary of that model and write the model in equation form. (Worth 15/100 mock marks)

πŸ’‘ Here, you were not expected to test ALL combinations. You could solve this by trial and error, combining multiple combinations of variables until you reached a model with better RSE, R-squared or other metric.

For didactic purposes, I will re-use the solution from Q14 and test all possible linear models with only two variables, this time considering * and : operator instead of the + operator.

I will look for the model that optimizes for RSE.

# Using the tidyverse function crossing
new_combn_predictors <- crossing(all_combn_predictors, tibble(interaction_type=c("*", ":")))
new_combn_predictors
# A tibble: 90 Γ— 3
   V1          V2         interaction_type
   <chr>       <chr>      <chr>           
 1 Advertising Age        :               
 2 Advertising Age        *               
 3 Advertising Education  :               
 4 Advertising Education  *               
 5 Advertising Population :               
 6 Advertising Population *               
 7 Advertising Price      :               
 8 Advertising Price      *               
 9 Advertising ShelveLoc  :               
10 Advertising ShelveLoc  *               
# … with 80 more rows

This time there are 90 possible combinations:

for(i in 1:nrow(new_combn_predictors)){
  
   # get the i-th row of the dataframe
  selected_vars <- new_combn_predictors[i,c("V1", "V2")]
  
  # build the formula
  # Notice that this time I will use whatever is in the column `interaction_type`
  model_formula <- as.formula(paste("Sales ~", paste(selected_vars, collapse=new_combn_predictors[i, ]$interaction_type)))
  
  print(model_formula)
  
}
Sales ~ Advertising:Age
Sales ~ Advertising * Age
Sales ~ Advertising:Education
Sales ~ Advertising * Education
Sales ~ Advertising:Population
Sales ~ Advertising * Population
Sales ~ Advertising:Price
Sales ~ Advertising * Price
Sales ~ Advertising:ShelveLoc
Sales ~ Advertising * ShelveLoc
Sales ~ Advertising:Urban
Sales ~ Advertising * Urban
Sales ~ Advertising:US
Sales ~ Advertising * US
Sales ~ Age:Education
Sales ~ Age * Education
Sales ~ Age:Urban
Sales ~ Age * Urban
Sales ~ Age:US
Sales ~ Age * US
Sales ~ CompPrice:Advertising
Sales ~ CompPrice * Advertising
Sales ~ CompPrice:Age
Sales ~ CompPrice * Age
Sales ~ CompPrice:Education
Sales ~ CompPrice * Education
Sales ~ CompPrice:Income
Sales ~ CompPrice * Income
Sales ~ CompPrice:Population
Sales ~ CompPrice * Population
Sales ~ CompPrice:Price
Sales ~ CompPrice * Price
Sales ~ CompPrice:ShelveLoc
Sales ~ CompPrice * ShelveLoc
Sales ~ CompPrice:Urban
Sales ~ CompPrice * Urban
Sales ~ CompPrice:US
Sales ~ CompPrice * US
Sales ~ Education:Urban
Sales ~ Education * Urban
Sales ~ Education:US
Sales ~ Education * US
Sales ~ Income:Advertising
Sales ~ Income * Advertising
Sales ~ Income:Age
Sales ~ Income * Age
Sales ~ Income:Education
Sales ~ Income * Education
Sales ~ Income:Population
Sales ~ Income * Population
Sales ~ Income:Price
Sales ~ Income * Price
Sales ~ Income:ShelveLoc
Sales ~ Income * ShelveLoc
Sales ~ Income:Urban
Sales ~ Income * Urban
Sales ~ Income:US
Sales ~ Income * US
Sales ~ Population:Age
Sales ~ Population * Age
Sales ~ Population:Education
Sales ~ Population * Education
Sales ~ Population:Price
Sales ~ Population * Price
Sales ~ Population:ShelveLoc
Sales ~ Population * ShelveLoc
Sales ~ Population:Urban
Sales ~ Population * Urban
Sales ~ Population:US
Sales ~ Population * US
Sales ~ Price:Age
Sales ~ Price * Age
Sales ~ Price:Education
Sales ~ Price * Education
Sales ~ Price:ShelveLoc
Sales ~ Price * ShelveLoc
Sales ~ Price:Urban
Sales ~ Price * Urban
Sales ~ Price:US
Sales ~ Price * US
Sales ~ ShelveLoc:Age
Sales ~ ShelveLoc * Age
Sales ~ ShelveLoc:Education
Sales ~ ShelveLoc * Education
Sales ~ ShelveLoc:Urban
Sales ~ ShelveLoc * Urban
Sales ~ ShelveLoc:US
Sales ~ ShelveLoc * US
Sales ~ Urban:US
Sales ~ Urban * US
best_RSE        <- Inf   # we don't know yet 
best_predictors <- c()   # start empty, fill later
best_model      <- NULL  # we don't know yet

for(i in 1:nrow(new_combn_predictors)){
  
   # get the i-th row of the dataframe
  selected_vars <- new_combn_predictors[i,c("V1", "V2")]
  
  # build the formula
  # Notice that this time I will use whatever is in the column `interaction_type`
  model_formula <- as.formula(paste("Sales ~", paste(selected_vars, collapse=new_combn_predictors[i, ]$interaction_type)))
  
  fitted_model      <- lm(model_formula, data=Carseats)
  current_model_RSE <- sigma(fitted_model)
  
  if(current_model_RSE < best_RSE){
    best_RSE <- current_model_RSE
    best_predictors <- selected_vars
    best_model <- fitted_model
    cat("Found a better model!\n")
    cat(paste0(model_formula, "\n"))
    cat(paste0("  RSE:", best_RSE, "\n\n"))
  }
  
}
Found a better model!
~
 Sales
 Advertising:Age
  RSE:2.77861087594811

Found a better model!
~
 Sales
 Advertising * Age
  RSE:2.65041673414083

Found a better model!
~
 Sales
 Advertising * Price
  RSE:2.40120656790555

Found a better model!
~
 Sales
 Advertising * ShelveLoc
  RSE:2.24885583915041

Found a better model!
~
 Sales
 Price:ShelveLoc
  RSE:1.96438627687265

Found a better model!
~
 Sales
 Price * ShelveLoc
  RSE:1.91821983511205

We didn’t find a better model, but we found a model that leads to almost the same RSE as the one in Q14:

summary(best_model)

Call:
lm(formula = model_formula, data = Carseats)

Residuals:
    Min      1Q  Median      3Q     Max 
-5.9037 -1.3461 -0.0595  1.3679  4.9037 

Coefficients:
                       Estimate Std. Error t value Pr(>|t|)    
(Intercept)           11.832984   0.965788  12.252  < 2e-16 ***
Price                 -0.055220   0.008276  -6.672 8.57e-11 ***
ShelveLocGood          6.135880   1.392844   4.405 1.36e-05 ***
ShelveLocMedium        1.630481   1.171616   1.392    0.165    
Price:ShelveLocGood   -0.010564   0.011742  -0.900    0.369    
Price:ShelveLocMedium  0.001984   0.010007   0.198    0.843    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.918 on 394 degrees of freedom
Multiple R-squared:  0.5444,    Adjusted R-squared:  0.5386 
F-statistic: 94.17 on 5 and 394 DF,  p-value: < 2.2e-16

\[ \begin{eqnarray} Sales &=& 11.832984 - 0.055220 \times Price \\ &&+ 6.135880 \times ShelveLoc[Good] \\ &&+1.630481 \times ShelveLoc[Medium] \\ &&-0.010564 \times (Price \times ShelveLoc[Good] ) \\ &&+0.001984 \times (Price \times ShelveLoc[Medium] ) \end{eqnarray} \]