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?

Answer:

We can check the dimension of datset using dim() function as shown below:

dim(hfi)
## [1] 1458  123

The data set has 1458 Observations with 123 variable with a dimension of \(1458 X 123\)

  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?

Answer:

For all the relational variables I would use a scatter or line plot. In this particular case I will use Scatter plot so that I’m able to see all the points in its exact spots.

Plotting the graph:

ggplot()+
  geom_point(data= hfi, mapping = aes(x=pf_expression_control, y=pf_score, alpha = .5 ),shape = 10)+labs(x= "Expression Control", y = "Score")+theme_bw()+theme(legend.position = 'none')

Relationship between pf_score and pf_expression_control does look somewhat linear because with the increase in expression control the score also increases. By looking at the graph I would be rather comfortable to use Linear model.

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.

Answer:

There is a strong association between the two variable which was confirmed by the correlation function which gave us the value in moderately strong positive correlation value. Similarly we did see that as pf_expression_controlwas increasing the pf_score was also increasing.

But we have to bear in mind that even the though there was a strong positive correlation but still there were some outliers or point which can effect the results and the model might give us a result which can be far off from the actual 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?

Answer:

The least sum of squares that I got was 955.065. Below are some info about intercept and x axis. Coefficients:

(Intercept) x
4.6079 0.5009

Sum of Squares: 955.065

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?

Answer:

lm_new <- lm(hf_score ~ pf_expression_control, data = hfi)
summary(lm_new)
## 
## 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 comes out to be: 5.1536+.349862 X pf_expression_control
  • The Slope which is .3499 over here tells us that For each additional amount of political pressure on media content score, we would expect the human freedom score to increase by 0.3499
  • If there is no political pressure the total human freedom comes out to be 5.1536 which is the intercept in the equation.

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?

Answer: If some only saw the least square regression line then they could predict the values using the equation of line. Since we did get the equation earlier while applying the linear regression model so we will use that to predict the pf_score

# Since pf_expression_control is equal to 6.7
exp_control <- 6.7

score_cal <- 4.61707 + 0.49143 * exp_control
score_cal
## [1] 7.909651

Our freedom score came out to be 7.91 if expression control is 6.7.

Now in order to find out the actual freedom score for expression control of 6.7 we will go back to the data frame.

hfi %>%
  group_by(pf_score) %>%
  filter(pf_expression_control == 6.7)
## # A tibble: 0 × 123
## # Groups:   pf_score [0]
## # … with 123 variables: year <dbl>, ISO_code <chr>, countries <chr>,
## #   region <chr>, pf_rol_procedural <dbl>, pf_rol_civil <dbl>,
## #   pf_rol_criminal <dbl>, pf_rol <dbl>, pf_ss_homicide <dbl>,
## #   pf_ss_disappearances_disap <dbl>, pf_ss_disappearances_violent <dbl>,
## #   pf_ss_disappearances_organized <dbl>,
## #   pf_ss_disappearances_fatalities <dbl>, pf_ss_disappearances_injuries <dbl>,
## #   pf_ss_disappearances <dbl>, pf_ss_women_fgm <dbl>, …

There is no observed value of pf_score with 6.7 rating for pf_expression_score.I would consider the closest one

residual <- 7.43 - 7.91
residual
## [1] -0.48

The prediction overestimated by 0.48.

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?

Answer:

There is no apparent pattern in the residuals plot and this indicates there is a linear relationship between the two variables.


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.

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

Answer:

Both the histogram and the normal probability plot show that the distribution of these data are nearly normal. Thus, the nearly normal residuals condition appears to be met.


Constant variability:

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

Answer:

The points residuals vs. fitted plot show that points are scattered around 0, there is a constant variability.Thus, the constant variability condition appear to be 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?

Answer:

ggplot(data = hfi, aes(y = hf_score, x = ef_regulation)) +
  geom_point() +
  stat_smooth(method = "lm", se = FALSE)

At a glance, the relationship do seem linear between the two variable and both have a positive association, meaning that the freedom score increases with the increase in freedom based on economical regulations.

  • 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?

Answer:

lm_new <- lm(hf_score ~ pf_expression_control, data = hfi)
summary(lm_new)
## 
## 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
lm_10 <- lm(hf_score~ef_regulation, data = hfi)
summary(lm_10)
## 
## Call:
## lm(formula = hf_score ~ ef_regulation, data = hfi)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.3716 -0.5796  0.0428  0.6234  1.9150 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    2.30349    0.14185   16.24   <2e-16 ***
## ef_regulation  0.66811    0.01999   33.41   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.7625 on 1376 degrees of freedom
##   (80 observations deleted due to missingness)
## Multiple R-squared:  0.4479, Adjusted R-squared:  0.4475 
## F-statistic:  1117 on 1 and 1376 DF,  p-value: < 2.2e-16

From the r. square values of both models, we have this:

pf_expression_control and hf_score model: 57.75% of the variability in pf_score can be explained by pf_expression_control.

hf_score and ef_regulation model: 44.79% of the variability in hf_rank can be explained by pf_religion

My independent variable does not seem to predict my dependent variable better because my r square (as explained above) is lower than r square of pf_expression_control and pf_score` model, it counts less variation.

  • What’s one freedom relationship you were most surprised about and why? Display the model diagnostics for the regression model analyzing this relationship.

Answer:

hf_score and pf_religion relationship is the one which surprise me the most because I was expecting that there will be very strong relation with more variability of property, but not in this case. The model has a r square lower than the previous models. See summary and the model diagnostics for the regression model analyzing this relationship.

ggplot(data = hfi, aes(y = hf_score, x = pf_religion)) +
  geom_point() +
  stat_smooth(method = "lm", se = FALSE)

lm_12 <- lm(hf_score~pf_religion, data = hfi)
summary(lm_12)
## 
## Call:
## lm(formula = hf_score ~ pf_religion, data = hfi)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -3.10229 -0.58501 -0.04865  0.77466  2.00693 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   4.7081     0.1502   31.34   <2e-16 ***
## pf_religion   0.2917     0.0188   15.51   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.9377 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

Model diagnostics:

Linearity and constant variability:

ggplot(data = lm_12, aes(x = .fitted, y = .resid)) +
  geom_point() +
  geom_hline(yintercept = 0, linetype = "dashed") +
  xlab("Fitted values") +
  ylab("Residuals")

There is not a apparent pattern in the residuals plot and this indicates there is a linear relationship between the two variables.The points residuals vs. fitted plot show that points are somewhat scattered around 0, there is a certain constant variability.

Normality of residuals:

To check this condition, we can look at a histogram

ggplot(data = lm_12, aes(x = .resid)) +
  geom_histogram(binwidth = .25) +
  xlab("Residuals")

or a normal probability plot of the residuals.

ggplot(data = lm_12, aes(sample = .resid)) +
  stat_qq()

From the histogram and the normal probability plot, we can say that the distribution of these data are nearly normal.