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.
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 we’re working with is in the openintro package and it’s
called hfi, short for Human Freedom Index.
We can see the dataframe fields and dimensions using R’s built-in
colnames and dim.data.frame functions.
colnames(hfi)## [1] "year" "ISO_code"
## [3] "countries" "region"
## [5] "pf_rol_procedural" "pf_rol_civil"
## [7] "pf_rol_criminal" "pf_rol"
## [9] "pf_ss_homicide" "pf_ss_disappearances_disap"
## [11] "pf_ss_disappearances_violent" "pf_ss_disappearances_organized"
## [13] "pf_ss_disappearances_fatalities" "pf_ss_disappearances_injuries"
## [15] "pf_ss_disappearances" "pf_ss_women_fgm"
## [17] "pf_ss_women_missing" "pf_ss_women_inheritance_widows"
## [19] "pf_ss_women_inheritance_daughters" "pf_ss_women_inheritance"
## [21] "pf_ss_women" "pf_ss"
## [23] "pf_movement_domestic" "pf_movement_foreign"
## [25] "pf_movement_women" "pf_movement"
## [27] "pf_religion_estop_establish" "pf_religion_estop_operate"
## [29] "pf_religion_estop" "pf_religion_harassment"
## [31] "pf_religion_restrictions" "pf_religion"
## [33] "pf_association_association" "pf_association_assembly"
## [35] "pf_association_political_establish" "pf_association_political_operate"
## [37] "pf_association_political" "pf_association_prof_establish"
## [39] "pf_association_prof_operate" "pf_association_prof"
## [41] "pf_association_sport_establish" "pf_association_sport_operate"
## [43] "pf_association_sport" "pf_association"
## [45] "pf_expression_killed" "pf_expression_jailed"
## [47] "pf_expression_influence" "pf_expression_control"
## [49] "pf_expression_cable" "pf_expression_newspapers"
## [51] "pf_expression_internet" "pf_expression"
## [53] "pf_identity_legal" "pf_identity_parental_marriage"
## [55] "pf_identity_parental_divorce" "pf_identity_parental"
## [57] "pf_identity_sex_male" "pf_identity_sex_female"
## [59] "pf_identity_sex" "pf_identity_divorce"
## [61] "pf_identity" "pf_score"
## [63] "pf_rank" "ef_government_consumption"
## [65] "ef_government_transfers" "ef_government_enterprises"
## [67] "ef_government_tax_income" "ef_government_tax_payroll"
## [69] "ef_government_tax" "ef_government"
## [71] "ef_legal_judicial" "ef_legal_courts"
## [73] "ef_legal_protection" "ef_legal_military"
## [75] "ef_legal_integrity" "ef_legal_enforcement"
## [77] "ef_legal_restrictions" "ef_legal_police"
## [79] "ef_legal_crime" "ef_legal_gender"
## [81] "ef_legal" "ef_money_growth"
## [83] "ef_money_sd" "ef_money_inflation"
## [85] "ef_money_currency" "ef_money"
## [87] "ef_trade_tariffs_revenue" "ef_trade_tariffs_mean"
## [89] "ef_trade_tariffs_sd" "ef_trade_tariffs"
## [91] "ef_trade_regulatory_nontariff" "ef_trade_regulatory_compliance"
## [93] "ef_trade_regulatory" "ef_trade_black"
## [95] "ef_trade_movement_foreign" "ef_trade_movement_capital"
## [97] "ef_trade_movement_visit" "ef_trade_movement"
## [99] "ef_trade" "ef_regulation_credit_ownership"
## [101] "ef_regulation_credit_private" "ef_regulation_credit_interest"
## [103] "ef_regulation_credit" "ef_regulation_labor_minwage"
## [105] "ef_regulation_labor_firing" "ef_regulation_labor_bargain"
## [107] "ef_regulation_labor_hours" "ef_regulation_labor_dismissal"
## [109] "ef_regulation_labor_conscription" "ef_regulation_labor"
## [111] "ef_regulation_business_adm" "ef_regulation_business_bureaucracy"
## [113] "ef_regulation_business_start" "ef_regulation_business_bribes"
## [115] "ef_regulation_business_licensing" "ef_regulation_business_compliance"
## [117] "ef_regulation_business" "ef_regulation"
## [119] "ef_score" "ef_rank"
## [121] "hf_score" "hf_rank"
## [123] "hf_quartile"
dim.data.frame(hfi)## [1] 1458 123
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 show a relationship between
pf_score and another numerical variable. First, we’ll need
to load the ggplot2 plotting librarylibrary(ggplot2)ggplot(hfi, aes(x=pf_expression_control, y=pf_score)) + geom_point()
Looking at our scatter plot, there does appear to be a rough linear
relationship between these two variables. In other words, as
pf_expression_control increases, pf_score also
increases linearly. I would be comfortable using a linear model to
predict a personal freedom scroe based on the political expression
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.
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.
There is a postiviely-increasing relationship between these two
variables. That is, as pf_expression_control increases, the
corresponding pf_score of a country also increases. While a
linear model could be used here, there is a fair amount of spread in
terms of pf_score for countries with a given
pf_expression_control score. Let’s take a look at the
largest spread for countries in this pf_expression_control
“band
hfi %>%
filter(pf_expression_control==5) %>%
summarise(min_pf = min(pf_score), max_pf = max(pf_score))## # A tibble: 1 × 2
## min_pf max_pf
## <dbl> <dbl>
## 1 4.99 9.33
We see a difference of over 5 points in pf_score between
two countries with the same expression score.
In addition, there are several points towards the bottom left of this
plot that do not follow the general trend of the plot. They have a lower
pf_score than we would typically expect for their
corresponding pf_expression_control score.
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.
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? My smallest sum of squares was 971.021. It’s in a similar
range to my other trials as I attempted to select points that would form
a line through the middle of the “cloud”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.
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?First, let’s generate our linear model:
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
Using our output model from m2, we can write the
equation of our linear model between hf_score and
pf_expressionc_control as:
\[ \hat{y} = 5.153687 + 0.349862 \times pf\_expression\_control \]
The slope of our model (0.349862) tells us the increase we should
expect in our hf_score given an increase in
pf_expression_control. In other words, as political
pressure exerted on a country’s media is lessened, and the expression
score increases, the human freedom score associated with that country
will increase as well.
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.
pf_expression_control? Is this an
overestimate or an underestimate, and by how much? In other words, what
is the residual for this prediction?They would input an expression score of 6.7 into the linear model
equation above to determine the y-value (pf_score)
y_pred <- 4.61707 + (0.49143 * 6.7)
y_pred## [1] 7.909651
So we would expect a country with a pf_expression_score
to have a pf_score of 7.909651. The residual for this
prediction would be 0, as it lies along the model line and would have no
distance (residual) between the prediction and model.
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.
There does not seem to be a trend/relationship in the residual plot (i.e., a slope of 0 or nearly 0 between our residuals and fitted values). This indicates that our predicted values are in line with the data. In addition, the residuals appear to be centered around 0, so our model is not “offset” in comparison to the data.
Nearly normal residuals: To check this condition, we can look at a histogram
ggplot(data = m1, aes(x = .resid)) +
geom_histogram(binwidth = 0.25) +
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.
Constant variability:
I’d like to take a look at the relationship between a country’s
hf_score and the freedom of movement score pf_movement.
Let’s first construct a linear model and plot. There does seem to be a
linear relationship between the two variables.
m3 = lm(hf_score ~ pf_movement, data = hfi)
ggplot(hfi, aes(x = pf_movement, y = hf_score)) +
geom_point() +
stat_smooth(method = "lm", se = FALSE)summary(m3)##
## Call:
## lm(formula = hf_score ~ pf_movement, data = hfi)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.93512 -0.48806 0.01532 0.55176 2.37263
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.808176 0.060229 79.83 <2e-16 ***
## pf_movement 0.279320 0.007295 38.29 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.714 on 1376 degrees of freedom
## (80 observations deleted due to missingness)
## Multiple R-squared: 0.5158, Adjusted R-squared: 0.5155
## F-statistic: 1466 on 1 and 1376 DF, p-value: < 2.2e-16
Our linear model between these two variables has a coefficient of 0.279320 and intercept of 4.8082.
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?Our relationship between freedom of movement scores and human freedom scores has an \(R^2 = 0.5158\). This represents a weaker correlation between variables than the one from above. If we switch the independent and dependent variable we get the results below:
m3_inverted = lm(pf_movement ~ hf_score, data = hfi)
summary(m3_inverted)##
## Call:
## lm(formula = pf_movement ~ hf_score, data = hfi)
##
## Residuals:
## Min 1Q Median 3Q Max
## -7.3626 -1.1174 0.0016 1.2185 5.8322
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -5.09140 0.34092 -14.93 <2e-16 ***
## hf_score 1.84672 0.04823 38.29 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.836 on 1376 degrees of freedom
## (80 observations deleted due to missingness)
## Multiple R-squared: 0.5158, Adjusted R-squared: 0.5155
## F-statistic: 1466 on 1 and 1376 DF, p-value: < 2.2e-16
Which has the same correlation as the ones above, but different
intercepts and residuals. This makes sense as the correlation
coefficient only represents the strength of a relationship between two
variables. In this case, our pf_movement variable is a
better predictor of our hf_score variable than vice
versa.
I’m interested in looking at the relationship between the number of
violent conflicts, captured in the
pf_ss_disappearances_violent field and a country’s access
to the internet (pf_expression_internet):
m4 <- lm(pf_ss_disappearances_violent ~ pf_expression_internet, data = hfi)
summary(m4)##
## Call:
## lm(formula = pf_ss_disappearances_violent ~ pf_expression_internet,
## data = hfi)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.6461 0.3539 0.3539 0.5296 1.0569
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 8.94313 0.18177 49.200 < 2e-16 ***
## pf_expression_internet 0.07030 0.02048 3.433 0.000619 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.625 on 1127 degrees of freedom
## (329 observations deleted due to missingness)
## Multiple R-squared: 0.01035, Adjusted R-squared: 0.00947
## F-statistic: 11.78 on 1 and 1127 DF, p-value: 0.0006191
We have a correlation coefficient \(R^2 = 0.01035\) which does not show a strong relationship. Let’s create a scatter plot of these two variables to confirm.
ggplot(hfi, aes(x = pf_expression_internet, y = pf_ss)) +
geom_point() +
stat_smooth(method = "lm", se = FALSE)