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.

library(tidyverse)
library(openintro)
library(ggplot2)
data('hfi', package='openintro')

The data

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

  1. What are the dimensions of the dataset?

Insert your answer here

dim(hfi)
## [1] 1458  123
The dataset is 123 columns with 1458 observations
  1. 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?

Insert your answer here

plot(hfi$pf_score ~ hfi$pf_expression_control, 
     xlab = "Expression control", ylab = "Personal Freedom score") +
  abline(lm(hfi$pf_score ~ hfi$pf_expression_control))

## integer(0)
# Alternatively, with ggplot
hfi %>%  ggplot(aes(x = pf_expression_control, y = pf_score)) + geom_point(col="black", na.rm = TRUE)  +
  geom_smooth(method = 'lm', se = TRUE, col="red", na.rm = TRUE) +
labs(x="Expression Control", y="Personal Freedom Score")

To display the relationship between pf_score and some numeric variable, I’d use a scatterplot. The data itself is rather scattered overall, but there is a slight upward trend which makes sense- from a logical perspective, it makes sense to see a higher expression control as the personal freedom score increases. If i knew the score, I’d be comfortable using a linear model to predict the pf score.

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"))
## # A tibble: 1 × 1
##   `cor(pf_expression_control, pf_score, use = "complete.obs")`
##                                                          <dbl>
## 1                                                        0.796

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.

  1. 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.

Insert your answer here

Based on the correlation coefficient of 0.796 , there is a strong correlation for a linear relation between pf_expression_control and pf_score. The pf_expression_control variable increases as pf_score increases, and there are some expected outliers that are far from the best-fit line.

Just as you’ve used the mean and standard deviation to summarize a single variable, you can summarize the relationship between these two variables by finding the line that best follows their association. Use the following interactive function to select the line that you think does the best job of going through the cloud of points.

# 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:

\[ e_i = y_i - \hat{y}_i \]

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.

  1. 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?

Insert your answer here

After running the above function several times, for the sum of squares I got the smallest value of 965.632

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:

\[ \hat{y} = 4.61707 + 0.49143 \times pf\_expression\_control \]

One last piece of information we will discuss from the summary output is the Multiple R-squared, or more simply, \(R^2\). The \(R^2\) 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.

  1. 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?

Insert your answer here

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
hf_score = 0.349862*(pf_expression_control) + 5.153687. With regards to the relationship between human freedom and political pressure, the slope tells us that as human freedom increases, the greater the amount of political pressure on media content. The human freedom score increases by 0.35 for each increase of 1 in political pressure on media content.

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)

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.

  1. 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?

Insert your answer here

# m1 Intercept = 4.61707    
# m1 Slope = 0.49143

predicted <- 4.61707 + (0.49143 * 6.7)
predicted
## [1] 7.909651
# As there are multiple values for pf_expression_control = 6.7, we should take the average
hfi6.7 <- hfi %>%
  filter(pf_expression_control >= 6.7 & pf_expression_control <6.8)
Observed <- mean(hfi6.7$pf_score)

# See the residual
Observed - predicted
## [1] 0.09666408
If someone saw the least squares regression line and not the data, for pf_expression_control = 6.7, it makes logical sense that they’d simply choose the value of pf_score that corresponds to 6.7, which looks to be approximately 8. Residual value = Observed value - Predicted value, and the residual of those 2 values is actually rather small, so a minor underestimate.

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")

Notice here that m1 can also serve as a data set because stored within it are the fitted values (\(\hat{y}\)) 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.

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

Insert your answer here

The residuals plot shows no easily discernible pattern. The relationship between the two variables seems to be linear, as there is no variance in the error in the predicted values.

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

ggplot(data = m1, aes(x = .resid)) +
  geom_histogram(binwidth = 2.5) +
  xlab("Residuals")

# The 25 is likely a typo, as I see no data. I've fixed it to 2.5, which actually shows the histogram of 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.

  1. Based on the histogram and the normal probability plot, does the nearly normal residuals condition appear to be met?

Insert your answer here

Yes, it does appear to be met, as the majority of residuals have value of 0 (per the histogram), and the qq plot appears rather normal.

Constant variability:

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

Insert your answer here

Yes, there appears to be no pattern in the plot, which means the assumption of constant variance/variability is met when the points are scattered around 0.

More Practice

  • 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?

Insert your answer here

# Let's use the freedom of movement variable 
hfi %>%  ggplot(aes(x = pf_movement, y = pf_score)) + geom_point(col="black", na.rm = TRUE)  +
  geom_smooth(method = 'lm', se = TRUE, col="red", na.rm = TRUE) +
labs(x="Freedom of Movement", y="Personal Freedom Score")

At a glance, yes there does appear to be a linar relationship. When people have freedom of movement, they have more overall freedom and vice versa.
  • How does this relationship compare to the relationship between pf_expression_control and pf_score? Use the \(R^2\) values from the two model summaries to compare. Does your independent variable seem to predict your dependent one better? Why or why not?

Insert your answer here

m1 <- lm(pf_score ~ pf_expression_control, data = hfi)
m_movement <- lm(pf_score ~ pf_movement, 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
summary(m_movement)
## 
## Call:
## lm(formula = pf_score ~ pf_movement, data = hfi)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.4604 -0.6565  0.0227  0.6880  3.1706 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 4.015554   0.072434   55.44   <2e-16 ***
## pf_movement 0.407197   0.008774   46.41   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.8587 on 1376 degrees of freedom
##   (80 observations deleted due to missingness)
## Multiple R-squared:  0.6102, Adjusted R-squared:  0.6099 
## F-statistic:  2154 on 1 and 1376 DF,  p-value: < 2.2e-16
cor(hfi$pf_score, hfi$pf_expression_control)
## [1] NA
cor(hfi$pf_score, hfi$pf_movement)
## [1] NA
This relationship actually resembles the relationship between pf_expression_control and pf_score very closely, as they both view how (pf_movement) and (pf_expression_control) correspond to the personal freedom score. When personal freedom goes up, it makes sense these do as well. The R^2 value is 0.6342 for pf_expression_control, and 0.6102 for pf_movement, which means that the variable I chose is not predicted as accurately as pf_expression_control. This is likely because there are larger variations in my data, thus larger residuals. This is also confirmed by the fact that my chosen variable has a lower correlation.
  • What’s one freedom relationship you were most surprised about and why? Display the model diagnostics for the regression model analyzing this relationship.

Insert your answer here

# Now let's look at the relationship between movement and disappearances

hfi %>%  ggplot(aes(x = pf_movement, y = pf_ss_disappearances)) + geom_point(col="black", na.rm = TRUE)  +
  geom_smooth(method = 'lm', se = TRUE, col="red", na.rm = TRUE) +
labs(x="Freedom of Movement", y="Dissapearances")

#-------------------------------------------------------------------------------
m_movement_dissaperances <- lm(pf_ss_disappearances ~ pf_movement, data = hfi)
summary(m_movement_dissaperances)
## 
## Call:
## lm(formula = pf_ss_disappearances ~ pf_movement, data = hfi)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -8.0558 -0.4082  0.4404  0.5204  3.2183 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  6.23220    0.11910   52.33   <2e-16 ***
## pf_movement  0.32969    0.01443   22.86   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.412 on 1376 degrees of freedom
##   (80 observations deleted due to missingness)
## Multiple R-squared:  0.2752, Adjusted R-squared:  0.2746 
## F-statistic: 522.3 on 1 and 1376 DF,  p-value: < 2.2e-16
#-------------------------------------------------------------------------------
cor(hfi$pf_ss_disappearances, hfi$pf_movement)
## [1] NA
#-------------------------------------------------------------------------------
ggplot(data = m_movement_dissaperances, aes(x = .resid)) +
  geom_histogram(binwidth = 1) +
  xlab("Residuals")

#-------------------------------------------------------------------------------
ggplot(data = m_movement_dissaperances, aes(sample = .resid)) +
  stat_qq()