16 Modeling

16.1 Objectives: Modeling

Understand the utility of a regression model
Demonstrate a basic understanding of the input components and model outputs of a regression model
Demonstrate the ability to write the code to create a regression model in R

16.2 Reading & resources: Modeling

Please refer to

R for Data Science, “Model Basics” for more information about this section. The chapters that follow this go deeper into modeling than we will in this course.
The chapter R Cookbook, 2nd ed., “Linear Regression and ANOVA” gives additional background on linear regression functions in R.

For information on the statistics of regression models, see the following:

Danielle Navarro, Learning statistics with R, Chapter 15 Linear regression (Navarro 2019)
Chester Ismay & Albert Y. Kim, Modern Dive: Statistical Inference via Data Science, Chapter 5 Basic Regression (Ismay and Kim 2019)
An interactive tool that visualizes different regression models can be found at Bivariate Correlation Simulation and Bivariate Scatter Plotting.
- The sheet allows users to explore different patterns. Note that in this tool, the statistics being reported are “Rho” (looks like a lower-case P), which is the correlation between X and Y in the population (i.e. all of the possible points from which our sample could have been drawn), and “r”, which is the correlation coefficient (to get the R-squared, we [you guessed it] can multiply the “r” value by itself).

16.3 Relationship between variables

When we talk about a “model” in the context of data analysis, we usually mean that we are summarizing the relationship between two or more variables. The question we are asking is “What’s the pattern in the data?”

In this example, we will be using linear regression to predict one value based on one or more values.

As it gets colder outside, we use more energy to heat our homes—can we predict energy consumption based on temperature? And in a more complex model, we could see if temperature plus relative humidity is a better predictor.

16.4 A simple example

This pair of plots shows how a simple linear model works: first we have the X-Y scatter plot.

In this example, the trend is fairly clear: as the value of “x” increases, so does “y”. We say that “x” is the independent variable, and “y” is the dependent variable. In a predictive model, we are estimating (or predicting) a value of “y” based on a known value of “x”. In the earlier home heating example, energy consumption depends on temperature; we are trying to predict energy consumption based on the temperature outside.

(Note that the data set sim1 comes from the package {modelr}, and is slightly altered¹¹.)

ggplot(sim1, aes(x = x, y = y)) +
  geom_point()

Then we can add the line of best fit that shows the trend, using the geom_smooth() function, where method = lm. This line shows the predicted values of “y”, based on the value of “x”.

ggplot(sim1, aes(x = x, y = y)) +
  geom_point() +
  geom_smooth(method = lm)

The line minimizes the distance between all of the points (more precisely, it’s the minimum squared distance). If we were to change the angle of the line in any way, it would increase the total distance. The plot below shows the distance between the measured points and the prediction line.

Being a statistics-based program, linear regression is a function in base R…it’s lm(), for “linear model”.

The syntax of the lm() function may look a bit different than what we’re used to.

lm(y ~ x, data)

This code creates a regression model, which we will assign to the object mod_sim1

mod_sim1 <- lm(y ~ x, data = sim1)

mod_sim1

## 
## Call:
## lm(formula = y ~ x, data = sim1)
## 
## Coefficients:
## (Intercept)            x  
##       4.124        2.052

The regression equation is the same as an equation for defining a line in geometry:

$y = b_{0} + b_{1}x_{1} + e$

How to read this:

the value of y (i.e. the point on the y axis) can be predicted by
the intercept value $b_{0}$, plus
the coefficient $b_{1}$ times the value of x

The distance between that predicted point and the actual point is the error term.

The equation calculated by the lm() function, to create the model model_mlb_pay_wl, is:

\[ \operatorname{\widehat{y}} = 4.1238 + 2.0522(\operatorname{x}) \]

The equation calculated by the lm() function above means that any predicted value of y (that is, a point on the line) is equal to 4.124 + 2.052 * the value of x.

The equation calculated by the lm() function above means that any predicted value of y (that is, the y-axis value of any point on the line) is equal to

4.12 + 2.0522 * the value of x.

If x = 2.5, then y = 4.12 + (2.0522 * 2.5), which is 9.2505.

Look back at the line of best fit in our plot—if x = 2.5, the y point on the line is just below 10.

We can also see how well the data predicts the line, by examining the summary statistics of the model with the summary() function:

summary(mod_sim1)

## 
## Call:
## lm(formula = y ~ x, data = sim1)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -3.8750 -1.3729  0.1512  1.5461  4.5264 
## 
## Coefficients:
##             Estimate Std. Error t value           Pr(>|t|)    
## (Intercept)   4.1238     0.8861   4.654 0.0000714239635371 ***
## x             2.0522     0.1419  14.458 0.0000000000000163 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.229 on 28 degrees of freedom
## Multiple R-squared:  0.8819, Adjusted R-squared:  0.8777 
## F-statistic:   209 on 1 and 28 DF,  p-value: 0.00000000000001633

This isn’t a statistics class, so we won’t go into the details …

(For interpreting the regression model summary, see The R Cookbook, 2nd edition, “11.4 Understanding the Regression Summary”)

Can you find the coefficients for the equation?

model coefficients

The P value (the measure of statistical significance) is found in the column marked “Pr(>|t|)”.

The P values

The Multiple and Adjusted R-squared statistics tells us how well the line fits the points…this number ranges between 0 (no fit at all) and 1 (where all the X and Y values fall on the line). In this case, it’s 0.88, which is a very good fit.

R-squared value

The equation predicts a value of Y, based on what we know about X.

This has enormous practical applications.

16.5 A deeper dive

In the following example, we will use some data from another package, {datarium}. In this example, there are three types of advertising: youtube, facebook, and newspaper. What we are interested in predicting is sales—how much do they go up for each dollar spent on each of the three media?

Here’s what the first 6 rows of the 200 rows in the marketing table look like:

note that sales is measured in units of 1000 (a 5 means 5,000)

df_marketing <- datarium::marketing 

head(df_marketing)

##   youtube facebook newspaper sales
## 1  276.12    45.36     83.04 26.52
## 2   53.40    47.16     54.12 12.48
## 3   20.64    55.08     83.16 11.16
## 4  181.80    49.56     70.20 22.20
## 5  216.96    12.96     70.08 15.48
## 6   10.44    58.68     90.00  8.64

Let’s start with just the youtube and see how well it predicts sales. First, we will plot the relationship:

# solution
ggplot(df_marketing, aes(x = youtube, y = sales)) +
  geom_point() +
  geom_smooth(method = lm)

Then, run the regression model to create a new model object, and use summary() to get the model statistics:

# solution
mod_youtube <- lm(sales ~ youtube, df_marketing)

summary(mod_youtube)

## 
## Call:
## lm(formula = sales ~ youtube, data = df_marketing)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -10.0632  -2.3454  -0.2295   2.4805   8.6548 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 8.439112   0.549412   15.36   <2e-16 ***
## youtube     0.047537   0.002691   17.67   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.91 on 198 degrees of freedom
## Multiple R-squared:  0.6119, Adjusted R-squared:  0.6099 
## F-statistic: 312.1 on 1 and 198 DF,  p-value: < 2.2e-16

Y = 8.44 + (0.048 * X)

The way to read this is that we will get 8.44 units sold with zero youtube advertising, but with every dollar of youtube spending we would expect an increase of 0.048 in sales. So if we spent $1,000, we would see a total of 56.4 units sold (sales = 8.44 + 0.048 * 1000)

We can use our skills in writing functions. We can write one called ad_impact that takes a value called youtube that represents the amount spent on youtube advertising, and then calculates the predicted sales, using the values from the model.

# solution

ad_impact <- function(youtube) {
  8.439 + (0.048 * youtube)
}

Let’s test our function, seeing the sales impact of spending $1000 on youtube advertising.

ad_impact(1000)

## [1] 56.439

16.6 Multiple predictors

Regression equations also allow us to use multiple variables to predict our outcome. In this sales example, there’s not just youtube, but facebook and newspaper advertising happening at the same time.

A full regression equation looks like this:

$y = b_{0} + b_{1}x_{1} + b_{2}x_{2} + \cdots + b_{n}x_{n} + e$

In R syntax, it is quite simple:

# solution
mod_all <- lm(sales ~ youtube + facebook + newspaper, df_marketing)

summary(mod_all)

## 
## Call:
## lm(formula = sales ~ youtube + facebook + newspaper, data = df_marketing)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -10.5932  -1.0690   0.2902   1.4272   3.3951 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  3.526667   0.374290   9.422   <2e-16 ***
## youtube      0.045765   0.001395  32.809   <2e-16 ***
## facebook     0.188530   0.008611  21.893   <2e-16 ***
## newspaper   -0.001037   0.005871  -0.177     0.86    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.023 on 196 degrees of freedom
## Multiple R-squared:  0.8972, Adjusted R-squared:  0.8956 
## F-statistic: 570.3 on 3 and 196 DF,  p-value: < 2.2e-16

Note that the value for newspaper is very small (and less important, negative). With such a small number, it’s time to introduce the statistical significance measure, shown on the right with the handy asterix column. The “P value” is the probability that the estimate is not significant, so the smaller the better. There’s no asterix associated with the newspaper line, so it’s not signficant, but the other two variables get the full 3-star rating.

This means that we can re-run the model, but without newspaper as a predictive variable:

# solution
mod_all <- lm(sales ~ youtube + facebook, df_marketing)

summary(mod_all)

## 
## Call:
## lm(formula = sales ~ youtube + facebook, data = df_marketing)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -10.5572  -1.0502   0.2906   1.4049   3.3994 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  3.50532    0.35339   9.919   <2e-16 ***
## youtube      0.04575    0.00139  32.909   <2e-16 ***
## facebook     0.18799    0.00804  23.382   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.018 on 197 degrees of freedom
## Multiple R-squared:  0.8972, Adjusted R-squared:  0.8962 
## F-statistic: 859.6 on 2 and 197 DF,  p-value: < 2.2e-16

Interpretation:

Y = 3.52 + 0.046 (youtube) + 0.188 (facebook)

For a given predictor (youtube or facebook), the coefficient is the effect on Y of a one unit increase in that predictor, holding the other predictors constant.

So to see the impact of a $1,000 increase in facebook advertising, we’d evaluate the equation as

3.52 + (0.046 * 1) + (0.188 * 1000) = 191.57

We would expect to see an increase of sales of 191.57 units, for each additional expenditure of $1000 of facebook adverstising.

Rather than calculating this long-hand, we can write a short function to calculate this for us:

# solution

ad_impact2 <- function(youtube, facebook) {
  3.52 + (0.046 * youtube) + (0.188 * facebook)
}

Let’s test: calculate the impact of a $1000 spend on facebook

# solution

ad_impact2(1, 1000)

## [1] 191.566

What if we split our spending? What’s the impact of a $500 spend on youtube and a $500 spend on facebook?

# solution

ad_impact2(500, 500)

## [1] 120.52

16.7 Supplement: Working with model data

As you get deeper into this, you’ll quickly realize that you may want to extract the various model values programmatically, rather than reading the numbers on the screen and typing them in to our code.

For these examples, we will use the data set sim1 (from the package {modelr}) that we saw earlier in this chapter.

mod_sim1 <- lm(y ~ x, data = sim1)

16.7.1 Extracting the coefficients

The first thing we might want to do is extract the coefficients that are calculated by the lm() function.

Earlier, we used the summary() function to show the coefficients and other aspects of the model:

summary(mod_sim1)

## 
## Call:
## lm(formula = y ~ x, data = sim1)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -3.8750 -1.3729  0.1512  1.5461  4.5264 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   4.1238     0.8861   4.654 7.14e-05 ***
## x             2.0522     0.1419  14.458 1.63e-14 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.229 on 28 degrees of freedom
## Multiple R-squared:  0.8819, Adjusted R-squared:  0.8777 
## F-statistic:   209 on 1 and 28 DF,  p-value: 1.633e-14

We can also see the values by accessing the coefficients that are within the mod_sim1 object.

mod_sim1$coefficients

## (Intercept)           x 
##    4.123849    2.052166

To access them, we need our square brackets:

# the Y intercept
mod_sim1$coefficients[1]

## (Intercept) 
##    4.123849

# the X coefficient
mod_sim1$coefficients[2]

##        x 
## 2.052166

These can be included in an equation, if we want to estimate a specific predicted Y value. In this code, our value of X is set to 7.5, and the regression equation evaluated to return the estimated value of Y.

our_value_of_x <- 7.5

estimated_y <- mod_sim1$coefficients[1] + # the Y intercept
  mod_sim1$coefficients[2] * our_value_of_x

estimated_y

## (Intercept) 
##    19.51509

We can also use inline R code to write a sentence with these values. For this purpose, I’ve rounded the values; here is the example of what the code looks like for the Y intercept, rounded to 2 decimal places.

round(mod_sim1$coefficients[1], 2)

The regression equation from the sim1 data is Y = 4.12 + 2.0522 * the value of x.

16.7.2 Using {broom} to tidy (and then plot) the predicted values

To “tidy” the data, you can use the package {broom}.

broom hex

In this example, we will

run the lm() model
extract the predicted values
plot those predicted values as a line on our scatter plot

(These are the steps that the geom_smooth(method = lm) function in {ggplot2} is doing behind the scenes.)

Using the {broom} package, we use the augment() function to create a new dataframe with the fitted values (also known as the predicted values) and residuals for each of the original points in the data:

mod_sim1_table <- broom::augment(mod_sim1)

mod_sim1_table

## # A tibble: 30 × 8
##        y     x .fitted  .resid   .hat .sigma   .cooksd .std.resid
##    <dbl> <dbl>   <dbl>   <dbl>  <dbl>  <dbl>     <dbl>      <dbl>
##  1  4.20 1.17     6.52 -2.32   0.111    2.22 0.0760       -1.10  
##  2  7.51 1.17     6.53  0.976  0.111    2.26 0.0134        0.464 
##  3  2.13 0.914    6.00 -3.87   0.120    2.13 0.235        -1.85  
##  4  8.99 2.13     8.50  0.489  0.0806   2.27 0.00230       0.229 
##  5 10.2  2.06     8.34  1.90   0.0827   2.24 0.0357        0.889 
##  6 11.3  2.01     8.24  3.05   0.0841   2.19 0.0940        1.43  
##  7  7.36 3.09    10.5  -3.12   0.0577   2.18 0.0636       -1.44  
##  8 10.5  2.85     9.98  0.525  0.0627   2.27 0.00198       0.243 
##  9 10.5  3.06    10.4   0.102  0.0583   2.27 0.0000694     0.0473
## 10 12.4  4.08    12.5  -0.0663 0.0420   2.27 0.0000202    -0.0304
## # ℹ 20 more rows

The variable we want to plot as the line is “.fitted”.

The code to create our original scatterplot is as follows:

pl_sim1 <- ggplot(sim1, aes(x = x, y = y)) +
  geom_point()

pl_sim1

To our object “pl_sim1” (for “plot_sim1”) we can add the line using the geom_line() function. This is pulling data from a second dataframe—{ggplot2} gives you the power to plot data points from multiple data frames.

Note that in the first solution, the data is specified using the df$var syntax.

pl_sim1 +
  geom_line(aes(x = mod_sim1_table$x, y = mod_sim1_table$.fitted))

# syntax that specifies the data
pl_sim1 +
  geom_line(data = mod_sim1_table, aes(x = x, y = .fitted))

Here’s another version of the syntax that skips the creation of an object.

# embedding the data and aes in the `geom_` functions
# note that `data = ` has to be specified`
ggplot() +
  geom_point(data = sim1, aes(x = x, y = y)) +
  geom_line(data = mod_sim1_table, aes(x = x, y = .fitted))

-30-

15 Write your own functions

Assignment 3 - week 4 - Moneyball