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)
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 There are 1458 observations/rows and 123 variables/columns in the dataset

dim(hfi)
## [1] 1458  123
  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 The type of plot I would use to display the relationship between the personal freedom score (pf_score) and the numerical variable pf_expression_control is a scatter plot. Scatter plots are commonly used to visualize the relationship between two numerical variables. The relationship appears to be linear.

Since the relationship appears to be linear in the scatter plot, I would be comfortable using a linear regression model to predict pf_score based on pf_expression_control. To make the prediction, I would check for linearity by calculating the \(R^2\), evaluating the residuals, and using statistical tests, to confirm the suitability of the linear model.

ggplot(hfi, aes(x=pf_expression_control, y=pf_score)) +
  geom_point() +
  labs(x = "pf_expression_control", y = "pf_score") +
  ggtitle("Scatter Plot of pf_score vs. pf_expression_control")

End of answer

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 ### Form of the Relationship: The relationship between pf_score and pf_expression_control appears to be roughly linear. In other words, as the value of pf_expression_control increases, there is a noticeable trend in the pf_score. The data points do not seem to follow a strict curve, but instead, they tend to follow a general linear trend.

Direction of the Relationship:

The relationship has a positive direction, which means that as the value of pf_expression_control increases, the pf_score tends to also increase. Conversely, as pf_expression_control decreases, the pf_score tends to decrease.

Strength of the Relationship:

The strength of the relationship seems to be moderately. While there is a noticeable trend, it’s not an extremely tight relationship. There is some variability in pf_score for any given value of pf_expression_control.

Unusual Observations:

I do not see any unusual or outlier observations in the scatter plot but the high variability in pf_score for any given value of pf_expression_control indicates that other factors may be having an influence. End of answer

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 plot_ss several times, the smallest sum of squares I got is 989.586 End of answer

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

The equation of the regression line is below: \[ \hat{y} = 5.153687 + 0.349862 \times pf\_expression\_control \]

The slope, 0.349862, tells us about the relationship between human freedom (hf_score) and the amount of political pressure on media content (pf_expression_control). Specifically:

Sign of the Slope: The slope is positive which means that as the amount of political pressure on media content increases, the total human freedom score (hf_score) tends to increase.

Magnitude of the Slope: Since the magnitude of the slope is a fraction, it indicates a weaker effect of political pressure on media content on hf_score.

Direction of the Relationship: The positive sign of slope implies a positive association. The positive slope suggests that as political pressure on media content increases, human freedom tends to increase. End of answer

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 The country’s personal freedom score pf_score can be predicted from: \[ \hat{y} = 4.61707 + 0.49143 \times pf\_expression\_control \]

predicted_pf_score <- 4.61707 + 0.49143*6.7
cat("predicted_pf_score = ", predicted_pf_score)
## predicted_pf_score =  7.909651

This is an underestimate by 0.8318 (Actual_pf_score = predicted_pf_score ).

actual_pf_score <- predicted_pf_score + 0.8318
cat("actual_pf_score = ", actual_pf_score)
## actual_pf_score =  8.741451

Since the residual is positive (Residual standard error = 0.8318), it means the prediction is an underestimate because the model predicted a lower pf_score than the actual score. In this case, the prediction is 0.8318 units lower than the actual score. End of answer

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 rename 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 There is no apparent pattern in the residuals plot

There is a random scatter of points around the horizontal line at 0. This suggests that the model is capturing the relationships between the variables adequately. Also, there is no clear trend or Pattern such as a curve, funnel shape, or any other consistent deviation from the horizontal line. There appear to to be an almost equal spread of residuals across the range of predicted values. The spread does not widen or narrow noticeably along the predicted values which in agreement with the key assumption of linear regression. There are a few but not significant outliers or extreme data points in the residuals plot ing data that the points conform to the linear relationship. End of answer

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

ggplot(data = m1, aes(x = .resid)) +
  geom_histogram() +
  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.

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

Insert your answer here The histogram shows a seemingly normal distribution of residuals. It is not completely bell shaped but there appear to be an equal distribution of the residuals about a mean of 0. There are no isolated bars or tails indicating outliers.

The points on the normal probability plot roughly align along a straight line indicating that the residuals closely follow a normal distribution. The curve does does not concave upwards or downward in any significant way to indicate skewness. Some points deviate from the expected straight line at the top end of the curve but the deviation is not significant. End of answer

Constant variability:

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

Insert your answer here In a residuals vs. fitted plot:

The spread of residuals is roughly uniform and shows no clear pattern as you move along the fitted values indicating that a constant variability condition is likely met.

Similarly the spread of residuals does not widen or narrow systematically as you move along the fitted values, suggesting constant variability.

In other words, the residuals vs. fitted plot shows a fairly consistent, random spread of residuals around zero as you move along the fitted values suggesting that a constant variability condition is met. * * *

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 My chosen variable is pf_ss_disappearances. Below is the scatter plot.

ggplot(hfi, aes(x=pf_ss_disappearances, y=pf_score)) +
  geom_point() +
  labs(x = "pf_ss_disappearances", y = "pf_score") +
  ggtitle("Scatter Plot of pf_score vs. pf_ss_disappearances")

At a glance:

Form of the Relationship:

The relationship between pf_score and pf_ss_disappearances appears to be weakly linear. In other words, as the value of pf_ss_disappearances increases, there is a weak positive trend in the pf_score. The data points does not seem to follow a strict curve.

Direction of the Relationship:

The relationship has a general positive direction, which means that as the value of pf_ss_disappearances increases, the pf_score tends to also increase. Conversely, as pf_ss_disappearances decreases, the pf_score tends to decrease.

Strength of the Relationship:

The strength of the relationship seems to be weak While there is a noticeable trend, it appears not be a tight relationship. There is a significant variability in pf_score for any given value of pf_ss_disappearances

Unusual Observations:

I do not see any unusual or outlier observations in the scatter plot but the high variability in pf_score for any given value of pf_ss_disappearances indicates that other factors may be having an influence. End of answer

  • 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

m3 <- lm(hf_score ~ pf_ss_disappearances, data = hfi)
summary(m3)
## 
## Call:
## lm(formula = hf_score ~ pf_ss_disappearances, data = hfi)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.82351 -0.58537 -0.02494  0.68647  2.08254 
## 
## Coefficients:
##                      Estimate Std. Error t value Pr(>|t|)    
## (Intercept)           3.59652    0.11699   30.74   <2e-16 ***
## pf_ss_disappearances  0.38551    0.01305   29.55   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.8027 on 1376 degrees of freedom
##   (80 observations deleted due to missingness)
## Multiple R-squared:  0.3882, Adjusted R-squared:  0.3877 
## F-statistic:   873 on 1 and 1376 DF,  p-value: < 2.2e-16

The \(R^2\) for the relationship between pf_score and pf_expression_control is 0.6342 while that between pf_score and pf_ss_disappearances is 0.3882. This indicates that there is a stronger relationship between pf_score and pf_expression_control than between pf_score and pf_ss_disappearances. The independent variable pf_expression_control predicts the response variable pf_score better than does pf_ss_disappearances because it has a higher \(R^2\) and thus a stronger relationship. End of answer

  • 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