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.

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.

Question 1

  1. What are the dimensions of the dataset?

1,458 rows of 123 columns where the rows are countries over a nine year period.

hfi %>% group_by(year)
## # A tibble: 1,458 × 123
## # Groups:   year [9]
##     year ISO_code countries region pf_rol_procedur… pf_rol_civil pf_rol_criminal
##    <dbl> <chr>    <chr>     <chr>             <dbl>        <dbl>           <dbl>
##  1  2016 ALB      Albania   Easte…             6.66         4.55            4.67
##  2  2016 DZA      Algeria   Middl…            NA           NA              NA   
##  3  2016 AGO      Angola    Sub-S…            NA           NA              NA   
##  4  2016 ARG      Argentina Latin…             7.10         5.79            4.34
##  5  2016 ARM      Armenia   Cauca…            NA           NA              NA   
##  6  2016 AUS      Australia Ocean…             8.44         7.53            7.36
##  7  2016 AUT      Austria   Weste…             8.97         7.87            7.67
##  8  2016 AZE      Azerbaij… Cauca…            NA           NA              NA   
##  9  2016 BHS      Bahamas   Latin…             6.93         6.01            6.26
## 10  2016 BHR      Bahrain   Middl…            NA           NA              NA   
## # … with 1,448 more rows, and 116 more variables: 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>,
## #   pf_ss_women_missing <dbl>, pf_ss_women_inheritance_widows <dbl>,
## #   pf_ss_women_inheritance_daughters <dbl>, pf_ss_women_inheritance <dbl>, …

Question 2

  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?

I would use a scatter plot to look at the relationship between two numberic values and I would set the predictor on the x-axis.

The relationship looks linear but I’d compare the linear relationship to the y ~ sqrt(x) relationship for fit. I would feel comfortable using a linear model but would want to try multiple variables to see which else correlate to personal freedom score.

ggplot(hfi, aes(x = pf_expression_control, y = pf_score, color = pf_score)) + 
        geom_point()
## Warning: Removed 80 rows containing missing values (geom_point).

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.

Question 3

  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.

The two variables are positively correlated, with a strength represented by a correlation coefficient of 80%. It looks like it could be a square root relationship, or a lesser root relationship like y^1.5 ~ x. It also seems like there’s room for another variable to account for the variation at any given level of Personal Freedom Expression Control.

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.

Question 4

  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?

Using set.seed(1264) I get a sum of squares of 952.153. Using a set.seed(2649) I get the same sum of squares of 952.153. I may be approaching this problem incorrectly.

I tried wrapping the y in a power like ‘y = (hfi$pf_score)^1.5’ but as I decreased the power it decreases the Sum of Squares so that seems to be affecting scale only and not fit.

I tried wrapping the x in a power like ’x = (hfi$pf_expression_control^2) and powers below 1 resulted in higher sum of squares. A power of two was higher as well but 1.1 was lower. Through iteration we could find the lowest sum of squares but I don’t think that’s what we’re going for in this exercise.

Iterating anyway meow and when you set ‘x’ to the power of 1.325 then we seem to get the lowest sum of squares at 942.02.

2.00 = 972
1.50 = 944.539
1.35 = 942.086
1.325 = 942.02
1.30 = 942.06
1.25 = 942.472
1.20 = 943.35
1.00 = 952.153

Correction: after I tried knitting the knit failed but now the function is working correctly and it’s letting me pick my own lines. I got various results between 1717.32 to 1074.879

DATA606::plot_ss(x = (hfi$pf_expression_control^1.325), y = (hfi$pf_score), showSquares = TRUE)

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.

Question 5

  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?

\[ \hat{y}_{hf\_score} = 5.153687 + 0.349862 \times pf\_expression\_control \]

As in you increase the amount of political pressure on media content by one unit you increase the human freedom score by a third of a unit.

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 values (stat_smooth).
## Warning: Removed 80 rows containing missing values (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.

Question 6

  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?

Using the formula copied below I would estimate the country’s personal freedom score at 7.909651. The residual for this prediction is zero because it’s directly on the line. It’s not a real data point. We would need the country’s actual personal freedom score to determine if our result is an over or underestimate.

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

4.61707 + 0.49143 * 6.7
## [1] 7.909651

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.

Question 7

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

I would compare this pattern to a rectangle of -2 to 2 on the y-axis across the x-axis and note that there are negative outliers for low and high values of x, positive outliers for medium low values of x and a hard line over which no residuals are higher than heading toward zero from the middle of the x-axis going right. The hard line might be due to maximum assignable scores. The outliers could indicate a slight curvature to the line but looking at the qq-plot below it seems that difference is normal for the extremes of our predictor data and this is fairly linear.


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

ggplot(data = m1, aes(x = .resid)) +
  geom_histogram(binwidth = .5) +
  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.

Question 8

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

The nearly normal residuals condition appears to be met. The histogram’s bin width was initially set at 25 and all data points fit in one bin. A bin width of 1 makes it look like a normal distribution. A bin width of 0.5 makes it look left-skewed.


Constant variability:

Question 9

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

This seems similar to Question 7 and yes the data points seem randomly scattered around zero though we detected that slight left or negative-skewness in the histogram for Question 8, which is supported by the number of dots below the line and there’s that hard line on positive residuals for higher predictor values.

More Practice

Question 10

  • 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?

I picked judicial independence (ef_legal_judicial) would predict integrity of the legal system (ef_legal_integrity). However likely legal integrity is a formula using judicial independence and other factors as inputs so yes there will be a correlation but comparing raw observations instead of calculated observations might be more interesting. We could find strong correlations by doing a nested for loop to capture all of the correlation coefficients and then setting the identity correlations to n/a.

At a glance there appears to be a linear relationship.

ggplot(hfi, aes(x = ef_legal_judicial, y = ef_legal_integrity, color = ef_legal_integrity)) + 
        geom_point()
## Warning: Removed 347 rows containing missing values (geom_point).

m3 <- lm(ef_legal_integrity ~ ef_legal_judicial, data = hfi)
summary(m3)
## 
## Call:
## lm(formula = ef_legal_integrity ~ ef_legal_judicial, data = hfi)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -4.9474 -0.9580  0.1382  1.1485  4.4820 
## 
## Coefficients:
##                   Estimate Std. Error t value Pr(>|t|)    
## (Intercept)        3.01798    0.11688   25.82   <2e-16 ***
## ef_legal_judicial  0.65499    0.02143   30.56   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.579 on 1109 degrees of freedom
##   (347 observations deleted due to missingness)
## Multiple R-squared:  0.4572, Adjusted R-squared:  0.4567 
## F-statistic: 934.1 on 1 and 1109 DF,  p-value: < 2.2e-16

Question 11

  • 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?

My independent variable doesn’t seem to predict my dependent one better. I assume that the formula used to calculate integrity of the legal system has more variables or assigns a lower weight to the judicial independence variable. This is supported by the lower correlation coefficient of 0.676 compared to 0.796.

The Multiple R-squared for my legal model is 0.4572, versus the multiple R-squared value for the main, personal freedom, model is 0.6342. From Google’s top search result, “R-squared is a goodness-of-fit measure for linear regression models. This statistic indicates the percentage of the variance in the dependent variable that the independent variables explain collectively.” This means that judicial independence explains a lower percentage of the variance in integrity of the legal system than political pressures and controls on media content does on the personal freedom score.

hfi %>%
  summarise(cor(ef_legal_judicial, ef_legal_integrity, use = "complete.obs"))
## # A tibble: 1 × 1
##   `cor(ef_legal_judicial, ef_legal_integrity, use = "complete.obs")`
##                                                                <dbl>
## 1                                                              0.676

Question 12

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

I used a for loop to check on the correlation coefficient between all of the variables to the human freedom score and ignoring the correlation coefficient with personal freedom of 0.943, the highest was Procedural justice (pf_rol_procedural) at 0.875. Also high were Rule of law (pf_rol) at 0.823 and Legal system and property rights (ef_legal) at 0.810. For comparison Freedom of expression (pf_expression) was at 0.727. This tells me how important the legal system is for protecting and advancing human freedom and the danger of a compromised legal system.

“Procedural justice refers to the idea of fairness in the processes that resolve disputes and allocate resources. It is a concept that, when embraced, promotes positive organizational change and bolsters better relationships.” (Top Google search result)

ggplot(hfi, aes(x = pf_rol_procedural, y = hf_score, color = pf_score)) + 
        geom_point()
## Warning: Removed 578 rows containing missing values (geom_point).

m4 <- lm(hf_score ~ pf_rol_procedural, data = hfi)
summary(m4)
## 
## Call:
## lm(formula = hf_score ~ pf_rol_procedural, data = hfi)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.31411 -0.25359  0.01229  0.29636  1.54322 
## 
## Coefficients:
##                   Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       4.923822   0.044559  110.50   <2e-16 ***
## pf_rol_procedural 0.399389   0.007472   53.45   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.461 on 878 degrees of freedom
##   (578 observations deleted due to missingness)
## Multiple R-squared:  0.7649, Adjusted R-squared:  0.7647 
## F-statistic:  2857 on 1 and 878 DF,  p-value: < 2.2e-16
ggplot(data = m4, aes(x = .fitted, y = .resid)) +
  geom_point() +
  geom_hline(yintercept = 0, linetype = "dashed") +
  xlab("Fitted values") +
  ylab("Residuals")

ggplot(data = m4, aes(x = .resid)) +
  geom_histogram(binwidth = .2) +
  xlab("Residuals")

ggplot(data = m4, aes(sample = .resid)) +
  stat_qq()

cc_hfscore <- c(5:118)
for (x in c(5:118)){
  cc_hfscore[x] <- pull(hfi %>% summarise(cor(hfi$hf_score, hfi[x], use = "complete.obs")))
}
cc_hfscore
##   [1]  5.00000000  6.00000000  7.00000000  8.00000000  0.87461264  0.72110803
##   [7]  0.73103698  0.82321978  0.27228399  0.50708451  0.35081479  0.69123185
##  [13]  0.36868807  0.31707731  0.62302505  0.32556647  0.32570512  0.65053725
##  [19]  0.66370298  0.65179332  0.65099361  0.72764758  0.57078911  0.54337133
##  [25]  0.56836493  0.71820995  0.46059130  0.49279725  0.52115199  0.12523173
##  [31]  0.12447809  0.38697111  0.61210803  0.62426114  0.48207496  0.56370659
##  [37]  0.58141626  0.45684595  0.61686838  0.58113333  0.49363451  0.57926813
##  [43]  0.58647860  0.66923167  0.21105551  0.19445719  0.74100382  0.75993721
##  [49]  0.45308438  0.57378879  0.47936248  0.72665240  0.32722524  0.61643736
##  [55]  0.54686828  0.61080562  0.50624594  0.48227781  0.53234051  0.54128287
##  [61]  0.67775420  0.94277284 -0.93200811 -0.26491046 -0.55974323  0.53309173
##  [67] -0.09882326 -0.16126147 -0.14454681 -0.06111807  0.57833632  0.48106520
##  [73]  0.61614992  0.71305791  0.57393004  0.48105660  0.36658226  0.58968649
##  [79]  0.44856797  0.61897710  0.81014723  0.28089971  0.56438581  0.43325356
##  [85]  0.49625992  0.71630355  0.45583311  0.58538590  0.07281405  0.45552979
##  [91]  0.56199591  0.66667059  0.71344682  0.28743252  0.59411016  0.47694172
##  [97]  0.51309482  0.68121618  0.79414926  0.53679392  0.21454951  0.34341498
## [103]  0.55657588  0.11632172  0.09120504  0.02089074  0.08562953  0.31076802
## [109]  0.25923642  0.32530348  0.17530779  0.66684815  0.47156658  0.62123954
## [115]  0.28827906  0.41122282  0.64229944  0.66928742