Introduction to linear regression

The Human Freedom Index is a report that attempts to summarize the idea of “freedom” through a bunch of different variables for many countries around the globe. It serves as a rough objective measure for the relationships between the different types of freedom - whether it’s political, religious, economical or personal freedom - and other social and economic circumstances. The Human Freedom Index is an annually co-published report by the Cato Institute, the Fraser Institute, and the Liberales Institut at the Friedrich Naumann Foundation for Freedom.

In this lab, you’ll be analyzing data from Human Freedom Index reports from 2008-2016. Your aim will be to summarize a few of the relationships within the data both graphically and numerically in order to find which variables can help tell a story about freedom.

Getting Started

Load packages

In this lab, you will explore and visualize the data using the tidyverse suite of packages. The data can be found in the companion package for OpenIntro resources, openintro.

Let’s load the packages.

The data

The data we’re working with is in the openintro package and it’s called hfi, short for Human Freedom Index.

Exercise 1 - What are the dimensions of the dataset?

print(paste0('There are : ', dim(hfi)[1],' observations and ', dim(hfi)[2],' variables in the dataset'))
## [1] "There are : 1458 observations and 123 variables in the dataset"

Exercise 2: What type of plot would you use to display the relationship between the personal freedom score, pf_score, and one of the other numerical variables?

Plot this relationship using the variable pf_expression_control as the predictor. Does the relationship look linear? If you knew a country’s pf_expression_control, or its score out of 10, with 0 being the most, of political pressures and controls on media content, would you be comfortable using a linear model to predict the personal freedom score?

# Scatterplot of pf_score vs pf_expression_control using the HFI dataset
ggplot(hfi, aes(x = pf_expression_control, y = pf_score)) +
  geom_point(alpha = 0.6) +
  geom_smooth(method = "lm", se = FALSE, color = "blue") +
  labs(title = "Personal Freedom Score vs Expression Control",
       x = "Expression Control (0 = most control, 10 = least)",
       y = "Personal Freedom Score")
## `geom_smooth()` using formula = 'y ~ x'
## Warning: Removed 80 rows containing non-finite outside the scale range
## (`stat_smooth()`).
## Warning: Removed 80 rows containing missing values or values outside the scale range
## (`geom_point()`).

positive linear relationship between pf_expression_control and pf_score. As expression control becomes less strict, personal freedom scores tend to rise. The relationship appears fairly linear, with a consistent upward trend. While there is some spread, especially at mid-levels of control, the linear model seems appropriate for predicting personal freedom from expression control.

If the relationship looks linear, we can quantify the strength of the relationship with the correlation coefficient.

hfi %>%
  summarise(cor(pf_expression_control, pf_score, use = "complete.obs"))

Here, we set the use argument to “complete.obs” since there are some observations of NA.

Sum of squared residuals

In this section, you will use an interactive function to investigate what we mean by “sum of squared residuals”. You will need to run this function in your console, not in your markdown document. Running the function also requires that the hfi dataset is loaded in your environment.

Think back to the way that we described the distribution of a single variable. Recall that we discussed characteristics such as center, spread, and shape. It’s also useful to be able to describe the relationship of two numerical variables, such as pf_expression_control and pf_score above.

Exercise 3 : Looking at your plot from the previous exercise, describe the relationship between these two variables. Make sure to discuss the form, direction, and strength of the relationship as well as any unusual observations.

The relationship between pf_expression_control and pf_score appears to be linear in form, meaning the data points roughly follow a straight-line pattern.

The direction of the relationship is positive: as pf_expression_control increases (i.e., media and expression are less controlled), the pf_score also increases, indicating higher levels of personal freedom.

The strength of the relationship is moderate to strong, as evidenced by the .79 correlation coefficient, the clear upward trend and relatively narrow band of points around the regression line, especially in the mid-to-high range of expression control.

There are a few unusual observations (potential outliers). For instance, some countries with high expression control (values near 0 or 1) still have moderately high personal freedom scores (around 6–7), which could indicate countries that restrict media but are otherwise free in personal domains.

Overall, the pattern suggests that greater freedom of expression is strongly associated with higher overall personal freedom, and a linear model is likely appropriate for modeling this relationship.

# This will only work interactively (i.e. will not show in the knitted document)
hfi <- hfi %>% filter(complete.cases(pf_expression_control, pf_score))
DATA606::plot_ss(x = hfi$pf_expression_control, y = hfi$pf_score)

After running this command, you’ll be prompted to click two points on the plot to define a line. Once you’ve done that, the line you specified will be shown in black and the residuals in blue. Note that there are 30 residuals, one for each of the 30 observations. Recall that the residuals are the difference between the observed values and the values predicted by the line:

ei=yi−y^i

The most common way to do linear regression is to select the line that minimizes the sum of squared residuals. To visualize the squared residuals, you can rerun the plot command and add the argument showSquares = TRUE.

DATA606::plot_ss(x = hfi$pf_expression_control, y = hfi$pf_score, showSquares = TRUE)

Note that the output from the plot_ss function provides you with the slope and intercept of your line as well as the sum of squares.

Exercise 4 : Using plot_ss, choose a line that does a good job of minimizing the sum of squares. Run the function several times. What was the smallest sum of squares that you got? How does it compare to your neighbors?

The smallest I can get is 952.153

m1 <- lm(pf_score ~ pf_expression_control, data = hfi)
summary(m1)
## 
## Call:
## lm(formula = pf_score ~ pf_expression_control, data = hfi)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -3.8467 -0.5704  0.1452  0.6066  3.2060 
## 
## Coefficients:
##                       Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.61707    0.05745   80.36   <2e-16 ***
## pf_expression_control  0.49143    0.01006   48.85   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.8318 on 1376 degrees of freedom
##   (80 observations deleted due to missingness)
## Multiple R-squared:  0.6342, Adjusted R-squared:  0.634 
## F-statistic:  2386 on 1 and 1376 DF,  p-value: < 2.2e-16

The linear model

It is rather cumbersome to try to get the correct least squares line, i.e. the line that minimizes the sum of squared residuals, through trial and error. Instead, you can use the lm function in R to fit the linear model (a.k.a. regression line).

m1 <- lm(pf_score ~ pf_expression_control, data = hfi) The first argument in the function lm is a formula that takes the form y ~ x. Here it can be read that we want to make a linear model of pf_score as a function of pf_expression_control. The second argument specifies that R should look in the hfi data frame to find the two variables.

The output of lm is an object that contains all of the information we need about the linear model that was just fit. We can access this information using the summary function.

summary(m1)
## 
## Call:
## lm(formula = pf_score ~ pf_expression_control, data = hfi)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -3.8467 -0.5704  0.1452  0.6066  3.2060 
## 
## Coefficients:
##                       Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.61707    0.05745   80.36   <2e-16 ***
## pf_expression_control  0.49143    0.01006   48.85   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.8318 on 1376 degrees of freedom
##   (80 observations deleted due to missingness)
## Multiple R-squared:  0.6342, Adjusted R-squared:  0.634 
## F-statistic:  2386 on 1 and 1376 DF,  p-value: < 2.2e-16

Let’s consider this output piece by piece. First, the formula used to describe the model is shown at the top. After the formula you find the five-number summary of the residuals. The “Coefficients” table shown next is key; its first column displays the linear model’s y-intercept and the coefficient of pf_expression_control. With this table, we can write down the least squares regression line for the linear model:

y^=4.61707+0.49143×pf_expression_control

One last piece of information we will discuss from the summary output is the Multiple R-squared, or more simply, R2 . The R2 value represents the proportion of variability in the response variable that is explained by the explanatory variable. For this model, 63.42% of the variability in runs is explained by at-bats.

Exercise 5 : Fit a new model that uses pf_expression_control to predict hf_score, or the total human freedom score. Using the estimates from the R output, write the equation of the regression line. What does the slope tell us in the context of the relationship between human freedom and the amount of political pressure on media content?

# Fit model to predict hf_score from pf_expression_control
m2 <- lm(hf_score ~ pf_expression_control, data = hfi)
summary(m2)
## 
## Call:
## lm(formula = hf_score ~ pf_expression_control, data = hfi)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.6198 -0.4908  0.1031  0.4703  2.2933 
## 
## Coefficients:
##                       Estimate Std. Error t value Pr(>|t|)    
## (Intercept)           5.153687   0.046070  111.87   <2e-16 ***
## pf_expression_control 0.349862   0.008067   43.37   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.667 on 1376 degrees of freedom
##   (80 observations deleted due to missingness)
## Multiple R-squared:  0.5775, Adjusted R-squared:  0.5772 
## F-statistic:  1881 on 1 and 1376 DF,  p-value: < 2.2e-16

Prediction and prediction errors

Let’s create a scatterplot with the least squares line for m1 laid on top.

ggplot(data = hfi, aes(x = pf_expression_control, y = pf_score)) +
  geom_point() +
  stat_smooth(method = "lm", se = FALSE)
## `geom_smooth()` using formula = 'y ~ x'
## Warning: Removed 80 rows containing non-finite outside the scale range
## (`stat_smooth()`).
## Warning: Removed 80 rows containing missing values or values outside the scale range
## (`geom_point()`).

Here, we are literally adding a layer on top of our plot. geom_smooth creates the line by fitting a linear model. It can also show us the standard error se associated with our line, but we’ll suppress that for now.

This line can be used to predict y at any value of x . When predictions are made for values of x that are beyond the range of the observed data, it is referred to as extrapolation and is not usually recommended. However, predictions made within the range of the data are more reliable. They’re also used to compute the residuals.

Exercise 6 : If someone saw the least squares regression line and not the actual data, how would they predict a country’s personal freedom school for one with a 6.7 rating for pf_expression_control? Is this an overestimate or an underestimate, and by how much? In other words, what is the residual for this prediction?

Ran into an issue with this. Initially I looked for a score of exactly 6.7 and there wasn’t one, so ended up using slice_min to find the countries that most closely matched 6.7.

predicted <- predict(m1, newdata = data.frame(pf_expression_control = 6.7))

# Find the actual observation closest to 6.7
closest <- hfi %>%
  filter(!is.na(pf_expression_control), !is.na(pf_score)) %>%
  mutate(diff = abs(pf_expression_control - 6.7)) %>%
  slice_min(order_by = diff, n = 1)

actual <- closest$pf_score
residual <- actual - predicted

data.frame(
  pf_expression_control = closest$pf_expression_control,
  actual = actual,
  predicted = predicted,
  residual = residual
)
## Warning in data.frame(pf_expression_control = closest$pf_expression_control, :
## row names were found from a short variable and have been discarded

Model diagnostics

To assess whether the linear model is reliable, we need to check for (1) linearity, (2) nearly normal residuals, and (3) constant variability.

Linearity: You already checked if the relationship between pf_score and `pf_expression_control’ is linear using a scatterplot. We should also verify this condition with a plot of the residuals vs. fitted (predicted) values.

ggplot(data = m1, aes(x = .fitted, y = .resid)) +
  geom_point() +
  geom_hline(yintercept = 0, linetype = "dashed") +
  xlab("Fitted values") +
  ylab("Residuals")
## Warning: `fortify(<lm>)` was deprecated in ggplot2 4.0.0.
## ℹ Please use `broom::augment(<lm>)` instead.
## ℹ The deprecated feature was likely used in the ggplot2 package.
##   Please report the issue at <https://github.com/tidyverse/ggplot2/issues>.
## This warning is displayed once per session.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was
## generated.

Notice here that m1 can also serve as a data set because stored within it are the fitted values (y^) and the residuals. Also note that we’re getting fancy with the code here. After creating the scatterplot on the first layer (first line of code), we overlay a horizontal dashed line at y=0 (to help us check whether residuals are distributed around 0), and we also reanme the axis labels to be more informative.

Exercise 7 :Is there any apparent pattern in the residuals plot? What does this indicate about the linearity of the relationship between the two variables?

The residuals plot does not show any clear curved or funnel-shaped pattern, and the residuals appear to be randomly scattered around the horizontal line at zero.

This suggests that the linearity assumption is reasonable. The relationship between pf_expression_control and pf_score appears to be appropriately modeled with a linear function.

If the residuals had shown a systematic curve or increasing/decreasing spread, that would indicate non-linearity. But in this case, we don’t see those issues, so a linear model is justified.

Nearly normal residuals: To check this condition, we can look at a histogram

ggplot(data = m1, aes(x = .resid)) +
  geom_histogram(binwidth = 25) +
  xlab("Residuals")

or a normal probability plot of the residuals.

ggplot(data = m1, aes(sample = .resid)) +
  stat_qq()

Note that the syntax for making a normal probability plot is a bit different than what you’re used to seeing: we set sample equal to the residuals instead of x, and we set a statistical method qq, which stands for “quantile-quantile”, another name commonly used for normal probability plots.

Exercise 8 :Based on the histogram and the normal probability plot, does the nearly normal residuals condition appear to be met?

Yes, the nearly normal residuals condition appears to be met.

While the histogram isn’t informative due to bin size (the entire thing is one big block), the Q-Q plot shows that the residuals follow a roughly straight line, indicating approximate normality. There is some deviation at the extremes, but the center of the distribution aligns well with the expected normal distribution. Using a linear model with these residuals is appropriate.

Constant variability:

Exercise 9 :Based on the residuals vs. fitted plot, does the constant variability condition appear to be met?

The residuals vs. fitted values plot shows that the spread of the residuals remains fairly constant across the range of predicted values. There is no clear fan shape or increasing/decreasing pattern in the variance.

So, the constant variability condition appears to be met, supporting the use of a linear regression model for this relationship.

More Practice

  1. Choose another freedom variable and a variable you think would strongly correlate with it.. Produce a scatterplot of the two variables and fit a linear model. At a glance, does there seem to be a linear relationship?

For this, we are going to look at the religion freedom variable: pf_religion and the predictor variable will be the total human freedom score, hf_score

hfi_clean <- hfi %>%
  filter(!is.na(pf_religion), !is.na(hf_score))

ggplot(hfi_clean, aes(x = hf_score, y = pf_religion)) +
  geom_point(alpha = 0.6) +
  geom_smooth(method = "lm", se = FALSE, color = "darkgreen") +
  labs(
    title = "Freedom of Religion vs Human Freedom Score",
    x = "Human Freedom Score",
    y = "Freedom of Religion"
  )
## `geom_smooth()` using formula = 'y ~ x'

  1. How does this relationship compare to the relationship between pf_expression_control and pf_score? Use the R2 values from the two model summaries to compare. Does your independent variable seem to predict your dependent one better? Why or why not?
m3 <- lm(pf_religion ~ hf_score, data = hfi)
summary(m3)
## 
## Call:
## lm(formula = pf_religion ~ hf_score, data = hfi)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -7.2548 -0.6808  0.2069  0.8671  2.8577 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   4.2787     0.2343   18.26   <2e-16 ***
## hf_score      0.5134     0.0331   15.51   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.244 on 1366 degrees of freedom
##   (90 observations deleted due to missingness)
## Multiple R-squared:  0.1497, Adjusted R-squared:  0.1491 
## F-statistic: 240.6 on 1 and 1366 DF,  p-value: < 2.2e-16
  1. What’s one freedom relationship you were most surprised about and why? Display the model diagnostics for the regression model analyzing this relationship.

I was most surprised that the relationship between hf_score (overall human freedom) and pf_religion (freedom of religion) was relatively weak. I expected a much stronger correlation between overall freedom and religious freedom.

Below are the model diagnostics for this regression model:

Overall, the model fits acceptably, but the low explanatory power was unexpected.

par(mfrow = c(2, 2))
plot(m3)