Simple Linear Regression - Categorical Predictor

Notes and in-class exercises

You can download the .qmd file for this activity here and open in R-studio. The rendered version is posted in the course website (Activities tab). I often experiment with the class activities (and see it in live!) and make updates, but I always post the final version before class starts. To be sure you have the most up-to-date copy, please download it once you’ve settled in before class begins.

Notes

Learning goals

By the end of this lesson, you should be able to:

Write a model formula for a simple linear regression model with a categorical predictor using indicator variables
Interpret the coefficients in a simple linear regression model with a categorical predictor

Readings and videos

Complete both the reading and the videos to go through before class.

Reading: Section 3.9 in the STAT 155 Notes only up through section 3.9.1 Indicator Variables
Videos:
- Simple linear regression: categorical predictor (slides)
- R Code for Categorical Predictors

File organization: Save this file in the “Activities” subfolder of your “STAT155” folder.

Exercises

Context: Today we’ll explore data on thousands of diamonds to understand how physical characteristics relate to price. Read in the data below.

# Load packages and import data
library(tidyverse)
data(diamonds)

# A little bit of data wrangling code - let's not focus on this for now
diamonds <- diamonds %>% 
    mutate(
        cut = factor(cut, ordered = FALSE),
        color = factor(color, ordered = FALSE),
        clarity = factor(clarity, ordered = FALSE)
    )

For the first several exercises, our focus will be on the relationship between diamond price and cut.

Exercise 1: Get to know the data

Write R code to answer the following:

1. How many cases and variables do we have? What does a case represent?
1. What do the first few rows of the data look like?
1. Construct and interpret two different visualizations of the price variable.
1. Construct and interpret a visualization of the cut variable.

dim(diamonds)
## [1] 53940    10

head(diamonds)
## # A tibble: 6 × 10
##   carat cut       color clarity depth table price     x     y     z
##   <dbl> <fct>     <fct> <fct>   <dbl> <dbl> <int> <dbl> <dbl> <dbl>
## 1  0.23 Ideal     E     SI2      61.5    55   326  3.95  3.98  2.43
## 2  0.21 Premium   E     SI1      59.8    61   326  3.89  3.84  2.31
## 3  0.23 Good      E     VS1      56.9    65   327  4.05  4.07  2.31
## 4  0.29 Premium   I     VS2      62.4    58   334  4.2   4.23  2.63
## 5  0.31 Good      J     SI2      63.3    58   335  4.34  4.35  2.75
## 6  0.24 Very Good J     VVS2     62.8    57   336  3.94  3.96  2.48

# Visualize price (outcome variable)
ggplot(diamonds, aes(x = price)) +
    geom_histogram()

ggplot(diamonds, aes(y = price)) +
    geom_boxplot()

diamonds %>%
    summarize(mean(price), median(price), sd(price))
## # A tibble: 1 × 3
##   `mean(price)` `median(price)` `sd(price)`
##           <dbl>           <dbl>       <dbl>
## 1         3933.            2401       3989.

# Visualize cut (predictor variable)
ggplot(diamonds, aes(x = cut)) +
    geom_bar()

diamonds %>% 
    count(cut)
## # A tibble: 5 × 2
##   cut           n
##   <fct>     <int>
## 1 Fair       1610
## 2 Good       4906
## 3 Very Good 12082
## 4 Premium   13791
## 5 Ideal     21551

Exercise 2: Visualizations

Start by visualizing this relationship of interest, that between price and cut.

The appropriate plot depends upon the type of variables we’re plotting. When exploring the relationship between a quantitative response and a quantitative predictor, a scatterplot was an effective choice. After running the code below, explain why a scatterplot is not effective for exploring the relationship between the outcome price and categorical cut predictor.

# Try a scatterplot
ggplot(diamonds, aes(y = price, x = cut)) + 
    geom_point()

Response: Put your response here.

Separately run each chunk below, with two plots. Comment (#) on what changes in the code / output.

b.1.

# Univariate boxplot
ggplot(diamonds, aes(y = price)) + 
    geom_boxplot()

Response: Put your response here.

b.2.

# ???
ggplot(diamonds, aes(y = price, x = cut)) + 
    geom_boxplot()

Response: Put your response here.

b.3.

# Univariate density plot
ggplot(diamonds, aes(x = price)) + 
    geom_density()

Response: Put your response here.

b.4.

# Comparisons in density plot
ggplot(diamonds, aes(x = price, color = cut)) + 
    geom_density()

Response: Put your response here.

b.5.

# Univariate histogram
ggplot(diamonds, aes(x = price)) + 
    geom_histogram()

Response: Put your response here.

b.6. What’s the difference between this and b4 density plot comparison? Can we now interpret the relationship between price and cut as we could in density plot? Why not?

# ???
ggplot(diamonds, aes(x = price)) + 
    geom_histogram() + 
    facet_wrap(~ cut)

Response: Put your response here.

Do you notice anything interesting about the relationship between price and cut? What do you think might be happening here?

Response: Put your response here.

Exercise 3: Numerical summaries

Let’s follow up our plots with some numerical summaries.

To warm up, first calculate the mean price across all diamonds.

diamonds %>% 
     summarize(mean(price))
## # A tibble: 1 × 1
##   `mean(price)`
##           <dbl>
## 1         3933.

To summarize the trends we observed in the grouped plots above, we can calculate the mean price for each type of cut. This requires the inclusion of the group_by() function:

# Calculate mean price by cut
diamonds %>% 
    group_by(cut) %>% 
    summarize(mean(price))
## # A tibble: 5 × 2
##   cut       `mean(price)`
##   <fct>             <dbl>
## 1 Fair              4359.
## 2 Good              3929.
## 3 Very Good         3982.
## 4 Premium           4584.
## 5 Ideal             3458.

Examine the group mean measurements. can you match these numbers up with what you see in the plots?
Based on the results above, we can see that, on average, diamonds with a “Fair” cut tend to cost more than higher-quality cuts. Let’s construct a new variable named cutFair, using on the following criteria:

cutFair = 1 if the diamond is of Fair cut
cutFair = 0 otherwise (any other value of cut (Good, Very Good, Premium, Ideal))

The ifelse function allows to create a new variable from an existing one, based on whether or not the values in that variable meet a certain “condition” (remember, you can always look up function documentation in R by typing ?ifelse in the Console, and hitting enter!).

Fill in the following code to create cutFair. The condition was given to you already. Try to use this to complete the code.

# In the first blank, put what value cutFair should have if the condition is "met", or TRUE
# In the second blank, put what value cutFair should have if the condition is "not met", or FALSE
diamonds <- diamonds %>%
  mutate(cutFair=ifelse(cut == "Fair", 1, 0))

Variables like cutFair that are coded as 0/1 to numerically indicate if a categorical variable is at a particular state are known as an indicator variable. You will sometimes see these referred to as a “binary variable” or “dichotomous variable”; you may also encounter the term “dummy variable” in older statistical literature.

Now, let’s calculate the group means based on the new cutFair indicator variable:

diamonds %>% 
    group_by(cutFair) %>% 
    summarize(mean(price))
## # A tibble: 2 × 2
##   cutFair `mean(price)`
##     <dbl>         <dbl>
## 1       0         3920.
## 2       1         4359.

Exercise 4: Modeling trend using a categorical predictor with exactly 2 categories

Next, let’s model the trend in the relationship between the cutFair and price variables using a simple linear regression model:

# Construct the model
diamond_mod0 <- lm(price ~ cutFair, data = diamonds)

# Summarize the model
summary(diamond_mod0)
## 
## Call:
## lm(formula = price ~ cutFair, data = diamonds)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
##  -4022  -2977  -1529   1391  14903 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  3919.69      17.44  224.80  < 2e-16 ***
## cutFair       439.06     100.93    4.35 1.36e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3989 on 53938 degrees of freedom
## Multiple R-squared:  0.0003507,  Adjusted R-squared:  0.0003322 
## F-statistic: 18.93 on 1 and 53938 DF,  p-value: 1.362e-05

Compare these results to the output of exercise 3e. What do you notice? How do you interpret the intercept and cutFair coefficient terms from this model?

Exercise 5: Modeling trend using a categorical predictor with >2 categories

Using a single binary predictor like the cutFair indicator variable is useful when there are two clearly delineated categories. However, the cut variable actually contains 5 categories! Because we’ve collapsed all non-Fair classifications into a single category (i.e. cutFair = 0), the model above can’t tell us anything about the difference in expected price between, say, Premium and Ideal cuts. The good news is that it is very straightforward to model categorical predictors with >2 categories. We can do this by using the cut variable as our predictor:

# Construct the model
diamond_mod <- lm(price ~ cut, data = diamonds)

# Summarize the model
summary(diamond_mod)
## 
## Call:
## lm(formula = price ~ cut, data = diamonds)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
##  -4258  -2741  -1494   1360  15348 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   4358.76      98.79  44.122  < 2e-16 ***
## cutGood       -429.89     113.85  -3.776 0.000160 ***
## cutVery Good  -377.00     105.16  -3.585 0.000338 ***
## cutPremium     225.50     104.40   2.160 0.030772 *  
## cutIdeal      -901.22     102.41  -8.800  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3964 on 53935 degrees of freedom
## Multiple R-squared:  0.01286,    Adjusted R-squared:  0.01279 
## F-statistic: 175.7 on 4 and 53935 DF,  p-value: < 2.2e-16

Even though we specified a single predictor variable in the model, we are seeing 4 coefficient estimates–why do you think this is the case?

We are seeing 4 coefficient estimates because each category is being assigned to a separate indicator variable–cutGood = 1 when cut == "Good" and 0 otherwise, cutVery Good = 1 when `cut == “Very Good” and 0 otherwise, and so on.

NOTE: We see 4 indicator variables (for Good, Very Good, Premium, and Ideal), but we do not see cutFair in the model output. This is because Fair is the reference level of the cut variable (it’s first alphabetically).

After examining the summary table output from the code chunk above, complete the model formula:

E[price | cut] = 4358.7578 - 429.8933 cutGood - 376.9979 cutVery Good + 225.4999 cutPremium - 901.2158 cutIdeal

Exercise 6: Making sense of the model

Recall our model: E[price | cut] = 4358.7578 - 429.8933 cutGood - 376.9979 cutVery Good + 225.4999 cutPremium - 901.2158 cutIdeal

Use the model formula to calculate the expected/typical price for diamonds of Good cut.

Expected/typical price for diamonds of Good cut:

E[price | cut] = 4358.7578 - 429.8933 * 1 - 376.9979 * 0 + 225.4999 * 0 - 901.2158 * 0 = 4358.7578 - 429.8933 = $3928.865

predict(diamond_mod, newdata = data.frame(cut = "Good"))
##        1 
## 3928.864

Similarly, calculate the expected/typical price for diamonds of Fair cut.

Expected/typical price for diamonds of Fair cut:

E[price | cut] = 4358.7578 - 429.8933 * 0 - 376.9979 * 0 + 225.4999 * 0 - 901.2158 * 0 = $4358.7578

predict(diamond_mod, newdata = data.frame(cut = "Fair"))
##        1 
## 4358.758

Re-examine these 2 calculations. Where have you seen these numbers before?!

These come from our group mean calculations in Exercise 3b! The predicted value for diamonds of Fair cut is also the same as what we obtained using the SLR model in exercise 4 with only a single cutFair indicator variable.

Exercise 7: Interpreting coefficients

Recall that our model formula is not a formula for a line. Thus we can’t interpret the coefficients as “slopes” as we have before. Taking this into account and reflecting upon your calculations above…

Interpret the intercept coefficient (4358.7578) in terms of the data context. Make sure to use non-causal language, include units, and talk about averages rather than individual cases.

The average price of a Fair cut diamonds is $4358.7578.

Interpret the cutGood and cutVery Good coefficients (-429.8933 and -376.9979) in terms of the data context. Hint: where did you use these value in the prediction calculations above?
- Interpretation of cutGood coefficient: On average, Good cut diamonds are worth $429.89 less than Fair cut diamonds.
- Interpretation of cutVery Good coefficient: On average, Very Good cut diamonds are worth $377.00 less than Fair cut diamonds.

Exercise 8: Modeling choices (CHALLENGE)

Why do we fit this model in this way (using 4 indicator variables cutGood, cutVery Good, cutPremium, cutIdeal)? Instead, suppose that we created a single variable cutCat that gave each category a numerical value: 0 for Fair, 1 for Good, 2 for Very Good, 3 for Premium, and 4 for Ideal.

How would this change things? What are the pros and cons of each approach?

Render your work

Click the “Render” button in the menu bar for this pane (blue arrow pointing right). This will create an HTML file containing all of the directions, code, and responses from this activity. A preview of the HTML will appear in the browser.
Scroll through and inspect the document to check that your work translated to the HTML format correctly.
Close the browser tab.
Go to the “Background Jobs” pane in RStudio and click the Stop button to end the rendering process.
Navigate to your “Activities” subfolder within your “STAT155” folder and locate the HTML file. You can open it again in your browser to double check.

Additional Practice

Exercise 9: The least squares criterion

The coefficient estimates diamond_mod were selected in the same way as when our predictor is quantitative: by minimizing the sum of the squared residuals. Use this model to calculate the residual for the first diamond in the dataset:

# Observed data on the first diamond
diamonds %>% 
  select(price, cut) %>% 
  head(1)
## # A tibble: 1 × 2
##   price cut  
##   <int> <fct>
## 1   326 Ideal

Exercise 10: Diamond color

Consider modeling price by color.

Before creating a visualization that shows the relationship between price and color, write down what you expect the plot to look like. Then construct and interpret an appropriate plot.
Compute the average price for each color.
Fit an appropriate linear model with lm() and display a short summary of the model.
Write out the model formula from the above summary.
Which color is the reference level? How can you tell from the model summary?
Interpret the intercept and two other coefficients from the model in terms of the data context.

Exercise 11: Diamond clarity

Repeat the steps from the previous exercise for the clarity variable.

Exercise 12: Evaluating model strength

Let’s study some penguin data!

data(penguins)
penguins <- penguins %>% 
  filter(!is.na(sex), !is.na(species))

We’ll focus on 3 variables with the goal of predicting flipper_len:

flipper_len = the length of the penguin’s flippers (arms) in mm
species = Adelie, Chinstrap, or Gentoo
sex = female, male

And build 2 models of flipper_len:

# Model of flipper_len by sex
flipper_model_1 <- lm(flipper_len ~ sex, data = penguins)
flipper_model_2 <- lm(flipper_len ~ species, data = penguins)

Just as with models that use quantitative predictors, it’s important to evaluate our (eventual) models of flipper_len by species and sex: are they correct? strong? fair?

Let’s start with strength. How strong are these models? What’s the best predictor? Let’s explore.

Based on the boxplots below, what is the stronger predictor of flipper_len: sex or species?

penguins %>% 
  ggplot(aes(y = flipper_len, x = sex)) + 
  geom_boxplot()

penguins %>% 
  ggplot(aes(y = flipper_len, x = species)) + 
  geom_boxplot()

We used 2 approaches to measuring the strength of the simple linear regression models in previous activities: correlation and R-squared. Unfortunately, we get errors when we try to calculate the correlation between flipper_len and sex, and between flipper_len and species. Why?

penguins %>% 
  summarize(cor(sex, flipper_len))
## Error in `summarize()`:
## ℹ In argument: `cor(sex, flipper_len)`.
## Caused by error in `cor()`:
## ! 'x' must be numeric
penguins %>% 
  summarize(cor(species, flipper_len))
## Error in `summarize()`:
## ℹ In argument: `cor(species, flipper_len)`.
## Caused by error in `cor()`:
## ! 'x' must be numeric

Luckily, R-squared works no matter whether a predictor is quantitative or categorical! Interpret and compare the R-squared values for our separate models of flipper_len by sex and species. Which is the stronger predictor? Does this match your answer in part a?

# Model of flipper_len by sex
summary(flipper_model_1)
## 
## Call:
## lm(formula = flipper_len ~ sex, data = penguins)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -26.506 -10.364  -4.364  12.636  26.494 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  197.364      1.057 186.792  < 2e-16 ***
## sexmale        7.142      1.488   4.801 2.39e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 13.57 on 331 degrees of freedom
## Multiple R-squared:  0.06511,    Adjusted R-squared:  0.06229 
## F-statistic: 23.05 on 1 and 331 DF,  p-value: 2.391e-06

# Model of flipper_len by species
summary(flipper_model_2)
## 
## Call:
## lm(formula = flipper_len ~ species, data = penguins)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -18.1027  -4.8235  -0.1027   4.7647  19.8973 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)      190.1027     0.5522  344.25  < 2e-16 ***
## speciesChinstrap   5.7208     0.9796    5.84 1.25e-08 ***
## speciesGentoo     27.1326     0.8241   32.92  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.673 on 330 degrees of freedom
## Multiple R-squared:  0.7747, Adjusted R-squared:  0.7734 
## F-statistic: 567.4 on 2 and 330 DF,  p-value: < 2.2e-16

Exercise 13: Evaluating model correctness

Let’s just focus on the stronger of our 2 models, that of flipper_len by species (flipper_model_2), and ask: is it correct (not wrong)? Recall that when our predictor was quantitative, residual plots provided some insight. Check out the residual plots below. They look a little goofy! Explain the goofiness (what’s happening here) and describe what you learn about the model’s “correctness”. Is the model “correct”?

# Residual plot
flipper_model_2 %>% 
  ggplot(aes(x = .fitted, y = .resid)) + 
  geom_point() + 
  geom_hline(yintercept = 0)

# Residual plot using boxes!
flipper_model_2 %>% 
  ggplot(aes(x = .fitted, y = .resid, group = .fitted)) + 
  geom_boxplot() + 
  geom_hline(yintercept = 0)

Exercise 14: Really understand how the coefficients work

Next, consider the model of flipper_len by island. The average flipper_len of penguins on each island is calculated below:

penguins %>% 
  group_by(island) %>% 
  summarize(mean(flipper_len))
## # A tibble: 3 × 2
##   island    `mean(flipper_len)`
##   <fct>                   <dbl>
## 1 Biscoe                   210.
## 2 Dream                    193.
## 3 Torgersen                192.

Using just the above averages and your understanding of categorical predictors, fill in the model coefficients and indicator variables below. Do NOT use lm() yet!!!

E[flipper_len | island] = ___ +/- ___ island??? +/- ___ island???

Check your work to Part a.

flipper_model_3 <- lm(flipper_len ~ island, penguins)
summary(flipper_model_3)
## 
## Call:
## lm(formula = flipper_len ~ island, data = penguins)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -37.558  -5.532   1.468   7.442  21.442 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)      209.558      0.879 238.410   <2e-16 ***
## islandDream      -16.371      1.340 -12.214   <2e-16 ***
## islandTorgersen  -18.026      1.858  -9.702   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 11.22 on 330 degrees of freedom
## Multiple R-squared:  0.3628, Adjusted R-squared:  0.3589 
## F-statistic: 93.94 on 2 and 330 DF,  p-value: < 2.2e-16

Solutions

Exercise 1: Get to know the data

A case represents a single diamond.
The distribution of price is right skewed with considerable high outliers. The right skew is evidenced by the mean price ($3932) being much higher than the median price ($2401).
Most diamonds in this data are of Good cut or better. Ideal cut diamonds are the most common with each succesive grade being the next most common.

dim(diamonds)
## [1] 53940    11

head(diamonds)
## # A tibble: 6 × 11
##   carat cut       color clarity depth table price     x     y     z cutFair
##   <dbl> <fct>     <fct> <fct>   <dbl> <dbl> <int> <dbl> <dbl> <dbl>   <dbl>
## 1  0.23 Ideal     E     SI2      61.5    55   326  3.95  3.98  2.43       0
## 2  0.21 Premium   E     SI1      59.8    61   326  3.89  3.84  2.31       0
## 3  0.23 Good      E     VS1      56.9    65   327  4.05  4.07  2.31       0
## 4  0.29 Premium   I     VS2      62.4    58   334  4.2   4.23  2.63       0
## 5  0.31 Good      J     SI2      63.3    58   335  4.34  4.35  2.75       0
## 6  0.24 Very Good J     VVS2     62.8    57   336  3.94  3.96  2.48       0

# Visualize price (outcome variable)
ggplot(diamonds, aes(x = price)) +
    geom_histogram()

ggplot(diamonds, aes(y = price)) +
    geom_boxplot()

diamonds %>%
    summarize(mean(price), median(price), sd(price))
## # A tibble: 1 × 3
##   `mean(price)` `median(price)` `sd(price)`
##           <dbl>           <dbl>       <dbl>
## 1         3933.            2401       3989.

# Visualize cut (predictor variable)
ggplot(diamonds, aes(x = cut)) +
    geom_bar()

diamonds %>% 
    count(cut)
## # A tibble: 5 × 2
##   cut           n
##   <fct>     <int>
## 1 Fair       1610
## 2 Good       4906
## 3 Very Good 12082
## 4 Premium   13791
## 5 Ideal     21551

Exercise 2: Visualizations

Start by visualizing this relationship of interest, that between price and cut.

We just don’t see anything clearly on a scatterplot. With the small number of unique values of the predictor variable, all of the points are bunched up on each other.

# Try a scatterplot
ggplot(diamonds, aes(y = price, x = cut)) + 
    geom_point()

Separately run each chunk below, with two plots. Comment (#) on what changes in the code / output.

# Univariate boxplot
ggplot(diamonds, aes(y = price)) + 
    geom_boxplot()

# Separate boxes by category
ggplot(diamonds, aes(y = price, x = cut)) + 
    geom_boxplot()

# Univariate density plot
ggplot(diamonds, aes(x = price)) + 
    geom_density()

# Separate density plots by category
ggplot(diamonds, aes(x = price, color = cut)) + 
    geom_density()

# Univariate histogram
ggplot(diamonds, aes(x = price)) + 
    geom_histogram()

# Separate histograms by category
ggplot(diamonds, aes(x = price)) + 
    geom_histogram() + 
    facet_wrap(~ cut)

The relationship between price and cut seems to be opposite what we would expect. The diamonds with the best cut (Ideal) have the lowest average price, and the ones with the worst cut (Fair) are woth the most. Maybe something else is different between the diamonds with the best and worst cuts…size maybe?

Exercise 3: Numerical summaries

Let’s follow up our plots with some numerical summaries.

Mean price across all diamonds:

diamonds %>% 
    summarize(mean(price))
## # A tibble: 1 × 1
##   `mean(price)`
##           <dbl>
## 1         3933.

Mean price for each type of cut:

diamonds %>% 
    group_by(cut) %>% 
    summarize(mean(price))
## # A tibble: 5 × 2
##   cut       `mean(price)`
##   <fct>             <dbl>
## 1 Fair              4359.
## 2 Good              3929.
## 3 Very Good         3982.
## 4 Premium           4584.
## 5 Ideal             3458.

Group means should reflect what you see in the plots (easiest to see in the boxplots)
Create our new cutFair variable:

diamonds <- diamonds %>%
  mutate(cutFair=ifelse(cut == "Fair", 1, 0))

Calculate the group means based on this new variable

diamonds %>% 
    group_by(cutFair) %>% 
    summarize(mean(price))
## # A tibble: 2 × 2
##   cutFair `mean(price)`
##     <dbl>         <dbl>
## 1       0         3920.
## 2       1         4359.

Exercise 4: Modeling trend using a categorical predictor with exactly 2 categories

# Construct the model
diamond_mod0 <- lm(price ~ cutFair, data = diamonds)

# Summarize the model
coef(summary(diamond_mod0))
##              Estimate Std. Error    t value     Pr(>|t|)
## (Intercept) 3919.6946    17.4367 224.795616 0.000000e+00
## cutFair      439.0632   100.9269   4.350309 1.361951e-05

The intercept is the expected value (mean) of the price for all diamonds with a cut quality that isn’t Fair (Good, Very Good, Premium, or Ideal, i.e. when cutFair = 0)–the same as we saw in exercise 3e.

When we add the intercept and coefficient for cutFair, we get 3919.69 + 439.06 = 4358.75–this is the mean price for all diamonds with a Fair cut quality that we saw in exercise 3e! Therefore, the coefficient of cutFair (439.06) is interpreted as the difference between the mean value of diamonds with a Fair cut quality and the mean value of diamonds with a higher cut quality.

Exercise 5: Modeling trend using a categorical predictor with >2 categories

# Construct the model
diamond_mod <- lm(price ~ cut, data = diamonds)

# Summarize the model
coef(summary(diamond_mod))
##               Estimate Std. Error   t value     Pr(>|t|)
## (Intercept)  4358.7578   98.78795 44.122361 0.000000e+00
## cutGood      -429.8933  113.84940 -3.775982 1.595493e-04
## cutVery Good -376.9979  105.16422 -3.584849 3.375707e-04
## cutPremium    225.4999  104.39521  2.160060 3.077240e-02
## cutIdeal     -901.2158  102.41155 -8.799943 1.408406e-18

We are seeing 4 coefficient estimates because each category is being assigned to a separate indicator variable–cutGood = 1 when cut == "Good" and 0 otherwise, cutVery Good = 1 when `cut == “Very Good” and 0 otherwise, and so on.
E[price | cut] = 4358.7578 - 429.8933 cutGood - 376.9979 cutVery Good + 225.4999 cutPremium - 901.2158 cutIdeal

Exercise 6: Making sense of the model

Expected/typical price for diamonds of Good cut:

E[price | cut] = 4358.7578 - 429.8933 * 1 - 376.9979 * 0 + 225.4999 * 0 - 901.2158 * 0 = 4358.7578 - 429.8933 = $3928.865

predict(diamond_mod, newdata = data.frame(cut = "Good"))
##        1 
## 3928.864

Expected/typical price for diamonds of Fair cut:

E[price | cut] = 4358.7578 - 429.8933 * 0 - 376.9979 * 0 + 225.4999 * 0 - 901.2158 * 0 = $4358.7578

predict(diamond_mod, newdata = data.frame(cut = "Fair"))
##        1 
## 4358.758

These come from our group mean calculations in Exercise 3b! The predicted value for diamonds of Fair cut is also the same as what we obtained using the SLR model in exercise 4 with only a single cutFair indicator variable.

Exercise 7: Interpreting coefficients

The average price of a Fair cut diamonds is $4358.7578.
- Interpretation of cutGood coefficient: On average, Good cut diamonds are worth $429.89 less than Fair cut diamonds.
- Interpretation of cutVery Good coefficient: On average, Very Good cut diamonds are worth $377.00 less than Fair cut diamonds.

Exercise 8: Modeling choices (CHALLENGE)

If we used 0-4 instead of creating indicator variables, we would be constraining the change from 0 to 1, from 1 to 2, etc. to always be of the same magnitude. That is, a 1 unit change in the cut variable would always have the same change in price in our model.
Using separate indicator variables allows the difference between subsequent categories to be different, which allows our model to be a bit more nuanced. It is possible to take nuance too far though. For example, in our previous investigations of bikeshare data, we modeled ridership versus temperature. We treated temperature as a quantitative predictor. Imagine if we had created an indicator variable for each unique temperature in the data—that would be so many variables! Having so many variables creates a very complex model which can be hard to make sense of. (These ideas are addressed further in STAT 253: Statistical Machine Learning!)

Exercise 9: The least squares criterion

# Observed price = 326
diamonds %>% 
  select(price, cut) %>% 
  head(1)
## # A tibble: 1 × 2
##   price cut  
##   <int> <fct>
## 1   326 Ideal

# Predicted price = 3457.542
predict(diamond_mod, newdata = data.frame(cut = "Ideal"))
##        1 
## 3457.542

# Residual
326 - 3457.542
## [1] -3131.542

Exercise 10: Diamond color

Consider modeling price by color.

The best color diamonds are J, and worst are D. We would expect D diamonds to have the lowest price and increase steadily as we get to J. This is in fact what we see in the boxplots.

ggplot(diamonds, aes(x = color, y = price)) +
    geom_boxplot()


diamonds %>% 
    group_by(color) %>% 
    summarize(mean(price))
## # A tibble: 7 × 2
##   color `mean(price)`
##   <fct>         <dbl>
## 1 D             3170.
## 2 E             3077.
## 3 F             3725.
## 4 G             3999.
## 5 H             4487.
## 6 I             5092.
## 7 J             5324.

We fit a linear model and obtain the model formula: E[price | color] = 3169.95 - 93.20 colorE + 554.93 colorF + 829.18 colorG + 1316.72 colorH + 1921.92 colorI + 2153.86 colorJ

diamond_mod2 <- lm(price ~ color, data = diamonds)

coef(summary(diamond_mod2))
##               Estimate Std. Error   t value      Pr(>|t|)
## (Intercept) 3169.95410   47.70694 66.446391  0.000000e+00
## colorE       -93.20162   62.04724 -1.502107  1.330752e-01
## colorF       554.93230   62.38527  8.895246  6.004834e-19
## colorG       829.18158   60.34470 13.740751  6.836340e-43
## colorH      1316.71510   64.28715 20.481777  7.074714e-93
## colorI      1921.92086   71.55308 26.860072 7.078041e-158
## colorJ      2153.86392   88.13203 24.439060 3.414906e-131

Color D is the reference level because we don’t see its indicator variable in the model output.
Interpretation of the intercept: Diamonds with D color cost $3169.95 on average.
Interpretation of the colorE coefficient: Diamonds with E color cost $93.20 less than D color diamonds on average.
Interpretation of the colorF coefficient: Diamonds with F color cost $554.93 more than D color diamonds on average.

Exercise 11: Diamond clarity

We see the unexpected result that diamonds of better clarity (VS1 and higher) have lower average prices. In fact the best clarity diamonds (VVS1 and IF) have the lowest average prices. What might be going on? What if the most clear diamonds were also quite small…

ggplot(diamonds, aes(x = clarity, y = price)) +
    geom_boxplot()


diamonds %>% 
    group_by(clarity) %>% 
    summarize(mean(price))
## # A tibble: 8 × 2
##   clarity `mean(price)`
##   <fct>           <dbl>
## 1 I1              3924.
## 2 SI2             5063.
## 3 SI1             3996.
## 4 VS2             3925.
## 5 VS1             3839.
## 6 VVS2            3284.
## 7 VVS1            2523.
## 8 IF              2865.

diamond_mod3 <- lm(price ~ clarity, data = diamonds)

coef(summary(diamond_mod3))
##                  Estimate Std. Error      t value      Pr(>|t|)
## (Intercept)  3924.1686910   144.5619 27.145247517 3.513547e-161
## claritySI2   1138.8599147   150.2746  7.578526239  3.550711e-14
## claritySI1     71.8324571   148.6049  0.483378837  6.288287e-01
## clarityVS2      0.8207037   148.8672  0.005512992  9.956013e-01
## clarityVS1    -84.7132999   150.9746 -0.561109670  5.747251e-01
## clarityVVS2  -640.4316203   154.7737 -4.137858008  3.510944e-05
## clarityVVS1 -1401.0540535   158.5401 -8.837224284  1.010097e-18
## clarityIF   -1059.3295848   171.8990 -6.162510636  7.210567e-10

Exercise 12: Evaluating model strength

flipper_model_1 <- lm(flipper_len ~ sex, data = penguins)
flipper_model_2 <- lm(flipper_len ~ species, data = penguins)

Part a

species appears to be the stronger predictor - the flipper_len values are more distinct between species (there’s less overlap in the boxes)

data(penguins)
penguins <- penguins %>% 
  filter(!is.na(sex), !is.na(species))

penguins %>% 
  ggplot(aes(y = flipper_len, x = sex)) + 
  geom_boxplot()


penguins %>% 
  ggplot(aes(y = flipper_len, x = species)) + 
  geom_boxplot()

Part b

Correlation only works when both variables in the pair are quantitative, but sex and species are categorical.

Part c

Using sex: R-squared = 0.06511

Using species: R-squared = 0.7782. Thus species is the stronger predictor of flipper_len. It explains roughly 78% of the variability in flipper length from penguin to penguin.

summary(flipper_model_1)
## 
## Call:
## lm(formula = flipper_len ~ sex, data = penguins)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -26.506 -10.364  -4.364  12.636  26.494 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  197.364      1.057 186.792  < 2e-16 ***
## sexmale        7.142      1.488   4.801 2.39e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 13.57 on 331 degrees of freedom
## Multiple R-squared:  0.06511,    Adjusted R-squared:  0.06229 
## F-statistic: 23.05 on 1 and 331 DF,  p-value: 2.391e-06
summary(flipper_model_2)
## 
## Call:
## lm(formula = flipper_len ~ species, data = penguins)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -18.1027  -4.8235  -0.1027   4.7647  19.8973 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)      190.1027     0.5522  344.25  < 2e-16 ***
## speciesChinstrap   5.7208     0.9796    5.84 1.25e-08 ***
## speciesGentoo     27.1326     0.8241   32.92  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.673 on 330 degrees of freedom
## Multiple R-squared:  0.7747, Adjusted R-squared:  0.7734 
## F-statistic: 567.4 on 2 and 330 DF,  p-value: < 2.2e-16

Exercise 13: Evaluating model correctness

Since there are only 3 different species, there are only 3 possible predictions of flipper_len in this model (one per species). Thus each of the 3 groups on the x-axis corresponds to a species. Within each species, the observed flipper_len deviates from the species-based prediction, thus so too do the residuals. That’s why the points in each group have different y coordinates.

Overall, our model doesn’t seem too wrong! No matter the x value (species!), the residuals are normally balanced above and below 0 with roughly equal variance in each species.

# Residual plot
flipper_model_2 %>% 
  ggplot(aes(x = .fitted, y = .resid)) + 
  geom_point() + 
  geom_hline(yintercept = 0)

# Residual plot using boxes!
flipper_model_2 %>% 
  ggplot(aes(x = .fitted, y = .resid, group = .fitted)) + 
  geom_boxplot() + 
  geom_hline(yintercept = 0)

Exercise 14: Really understand how the coefficients work

penguins %>% 
  group_by(island) %>% 
  summarize(mean(flipper_len))
## # A tibble: 3 × 2
##   island    `mean(flipper_len)`
##   <fct>                   <dbl>
## 1 Biscoe                   210.
## 2 Dream                    193.
## 3 Torgersen                192.

E[flipper_len | island] = 209.56 - 16.37 islandDream - 18.03 islandTorgersen

Why?

The island terms are islandDream and islandTorgersen since Biscoe is the reference.
The intercept is the average flipper length for the reference, Biscoe.
The islandDream coefficient reflects the fact that the average flipper length on Dream is 16.37mm lower than that on Biscoe (193.1870 - 209.5583)
The islandTorgersen coefficient reflects the fact that the average flipper length on Dream is 18.03mm lower than that on Biscoe (191.5319 - 209.5583)

flipper_model_3 <- lm(flipper_len ~ island, penguins)
summary(flipper_model_3)
## 
## Call:
## lm(formula = flipper_len ~ island, data = penguins)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -37.558  -5.532   1.468   7.442  21.442 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)      209.558      0.879 238.410   <2e-16 ***
## islandDream      -16.371      1.340 -12.214   <2e-16 ***
## islandTorgersen  -18.026      1.858  -9.702   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 11.22 on 330 degrees of freedom
## Multiple R-squared:  0.3628, Adjusted R-squared:  0.3589 
## F-statistic: 93.94 on 2 and 330 DF,  p-value: < 2.2e-16