library(tidyverse)
library(openintro)
data('hfi', package='openintro')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.
The data
The data we’re working with is in the openintro package and it’s called hfi, short for Human Freedom Index.
- What are the dimensions of the dataset?
glimpse(hfi)Rows: 1,458
Columns: 123
$ year <dbl> 2016, 2016, 2016, 2016, 2016, 2016,…
$ ISO_code <chr> "ALB", "DZA", "AGO", "ARG", "ARM", …
$ countries <chr> "Albania", "Algeria", "Angola", "Ar…
$ region <chr> "Eastern Europe", "Middle East & No…
$ pf_rol_procedural <dbl> 6.661503, NA, NA, 7.098483, NA, 8.4…
$ pf_rol_civil <dbl> 4.547244, NA, NA, 5.791960, NA, 7.5…
$ pf_rol_criminal <dbl> 4.666508, NA, NA, 4.343930, NA, 7.3…
$ pf_rol <dbl> 5.291752, 3.819566, 3.451814, 5.744…
$ pf_ss_homicide <dbl> 8.920429, 9.456254, 8.060260, 7.622…
$ pf_ss_disappearances_disap <dbl> 10, 10, 5, 10, 10, 10, 10, 10, 10, …
$ pf_ss_disappearances_violent <dbl> 10.000000, 9.294030, 10.000000, 10.…
$ pf_ss_disappearances_organized <dbl> 10.0, 5.0, 7.5, 7.5, 7.5, 10.0, 10.…
$ pf_ss_disappearances_fatalities <dbl> 10.000000, 9.926119, 10.000000, 10.…
$ pf_ss_disappearances_injuries <dbl> 10.000000, 9.990149, 10.000000, 9.9…
$ pf_ss_disappearances <dbl> 10.000000, 8.842060, 8.500000, 9.49…
$ pf_ss_women_fgm <dbl> 10.0, 10.0, 10.0, 10.0, 10.0, 10.0,…
$ pf_ss_women_missing <dbl> 7.5, 7.5, 10.0, 10.0, 5.0, 10.0, 10…
$ pf_ss_women_inheritance_widows <dbl> 5, 0, 5, 10, 10, 10, 10, 5, NA, 0, …
$ pf_ss_women_inheritance_daughters <dbl> 5, 0, 5, 10, 10, 10, 10, 10, NA, 0,…
$ pf_ss_women_inheritance <dbl> 5.0, 0.0, 5.0, 10.0, 10.0, 10.0, 10…
$ pf_ss_women <dbl> 7.500000, 5.833333, 8.333333, 10.00…
$ pf_ss <dbl> 8.806810, 8.043882, 8.297865, 9.040…
$ pf_movement_domestic <dbl> 5, 5, 0, 10, 5, 10, 10, 5, 10, 10, …
$ pf_movement_foreign <dbl> 10, 5, 5, 10, 5, 10, 10, 5, 10, 5, …
$ pf_movement_women <dbl> 5, 5, 10, 10, 10, 10, 10, 5, NA, 5,…
$ pf_movement <dbl> 6.666667, 5.000000, 5.000000, 10.00…
$ pf_religion_estop_establish <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA,…
$ pf_religion_estop_operate <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA,…
$ pf_religion_estop <dbl> 10.0, 5.0, 10.0, 7.5, 5.0, 10.0, 10…
$ pf_religion_harassment <dbl> 9.566667, 6.873333, 8.904444, 9.037…
$ pf_religion_restrictions <dbl> 8.011111, 2.961111, 7.455556, 6.850…
$ pf_religion <dbl> 9.192593, 4.944815, 8.786667, 7.795…
$ pf_association_association <dbl> 10.0, 5.0, 2.5, 7.5, 7.5, 10.0, 10.…
$ pf_association_assembly <dbl> 10.0, 5.0, 2.5, 10.0, 7.5, 10.0, 10…
$ pf_association_political_establish <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA,…
$ pf_association_political_operate <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA,…
$ pf_association_political <dbl> 10.0, 5.0, 2.5, 5.0, 5.0, 10.0, 10.…
$ pf_association_prof_establish <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA,…
$ pf_association_prof_operate <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA,…
$ pf_association_prof <dbl> 10.0, 5.0, 5.0, 7.5, 5.0, 10.0, 10.…
$ pf_association_sport_establish <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA,…
$ pf_association_sport_operate <dbl> NA, NA, NA, NA, NA, NA, NA, NA, NA,…
$ pf_association_sport <dbl> 10.0, 5.0, 7.5, 7.5, 7.5, 10.0, 10.…
$ pf_association <dbl> 10.0, 5.0, 4.0, 7.5, 6.5, 10.0, 10.…
$ pf_expression_killed <dbl> 10.000000, 10.000000, 10.000000, 10…
$ pf_expression_jailed <dbl> 10.000000, 10.000000, 10.000000, 10…
$ pf_expression_influence <dbl> 5.0000000, 2.6666667, 2.6666667, 5.…
$ pf_expression_control <dbl> 5.25, 4.00, 2.50, 5.50, 4.25, 7.75,…
$ pf_expression_cable <dbl> 10.0, 10.0, 7.5, 10.0, 7.5, 10.0, 1…
$ pf_expression_newspapers <dbl> 10.0, 7.5, 5.0, 10.0, 7.5, 10.0, 10…
$ pf_expression_internet <dbl> 10.0, 7.5, 7.5, 10.0, 7.5, 10.0, 10…
$ pf_expression <dbl> 8.607143, 7.380952, 6.452381, 8.738…
$ pf_identity_legal <dbl> 0, NA, 10, 10, 7, 7, 10, 0, NA, NA,…
$ pf_identity_parental_marriage <dbl> 10, 0, 10, 10, 10, 10, 10, 10, 10, …
$ pf_identity_parental_divorce <dbl> 10, 5, 10, 10, 10, 10, 10, 10, 10, …
$ pf_identity_parental <dbl> 10.0, 2.5, 10.0, 10.0, 10.0, 10.0, …
$ pf_identity_sex_male <dbl> 10, 0, 0, 10, 10, 10, 10, 10, 10, 1…
$ pf_identity_sex_female <dbl> 10, 0, 0, 10, 10, 10, 10, 10, 10, 1…
$ pf_identity_sex <dbl> 10, 0, 0, 10, 10, 10, 10, 10, 10, 1…
$ pf_identity_divorce <dbl> 5, 0, 10, 10, 5, 10, 10, 5, NA, 0, …
$ pf_identity <dbl> 6.2500000, 0.8333333, 7.5000000, 10…
$ pf_score <dbl> 7.596281, 5.281772, 6.111324, 8.099…
$ pf_rank <dbl> 57, 147, 117, 42, 84, 11, 8, 131, 6…
$ ef_government_consumption <dbl> 8.232353, 2.150000, 7.600000, 5.335…
$ ef_government_transfers <dbl> 7.509902, 7.817129, 8.886739, 6.048…
$ ef_government_enterprises <dbl> 8, 0, 0, 6, 8, 10, 10, 0, 7, 10, 7,…
$ ef_government_tax_income <dbl> 9, 7, 10, 7, 5, 5, 4, 9, 10, 10, 8,…
$ ef_government_tax_payroll <dbl> 7, 2, 9, 1, 5, 5, 3, 4, 10, 10, 8, …
$ ef_government_tax <dbl> 8.0, 4.5, 9.5, 4.0, 5.0, 5.0, 3.5, …
$ ef_government <dbl> 7.935564, 3.616782, 6.496685, 5.346…
$ ef_legal_judicial <dbl> 2.6682218, 4.1867042, 1.8431292, 3.…
$ ef_legal_courts <dbl> 3.145462, 4.327113, 1.974566, 2.930…
$ ef_legal_protection <dbl> 4.512228, 4.689952, 2.512364, 4.255…
$ ef_legal_military <dbl> 8.333333, 4.166667, 3.333333, 7.500…
$ ef_legal_integrity <dbl> 4.166667, 5.000000, 4.166667, 3.333…
$ ef_legal_enforcement <dbl> 4.3874441, 4.5075380, 2.3022004, 3.…
$ ef_legal_restrictions <dbl> 6.485287, 6.626692, 5.455882, 6.857…
$ ef_legal_police <dbl> 6.933500, 6.136845, 3.016104, 3.385…
$ ef_legal_crime <dbl> 6.215401, 6.737383, 4.291197, 4.133…
$ ef_legal_gender <dbl> 0.9487179, 0.8205128, 0.8461538, 0.…
$ ef_legal <dbl> 5.071814, 4.690743, 2.963635, 3.904…
$ ef_money_growth <dbl> 8.986454, 6.955962, 9.385679, 5.233…
$ ef_money_sd <dbl> 9.484575, 8.339152, 4.986742, 5.224…
$ ef_money_inflation <dbl> 9.743600, 8.720460, 3.054000, 2.000…
$ ef_money_currency <dbl> 10, 5, 5, 10, 10, 10, 10, 5, 0, 10,…
$ ef_money <dbl> 9.553657, 7.253894, 5.606605, 5.614…
$ ef_trade_tariffs_revenue <dbl> 9.626667, 8.480000, 8.993333, 6.060…
$ ef_trade_tariffs_mean <dbl> 9.24, 6.22, 7.72, 7.26, 8.76, 9.50,…
$ ef_trade_tariffs_sd <dbl> 8.0240, 5.9176, 4.2544, 5.9448, 8.0…
$ ef_trade_tariffs <dbl> 8.963556, 6.872533, 6.989244, 6.421…
$ ef_trade_regulatory_nontariff <dbl> 5.574481, 4.962589, 3.132738, 4.466…
$ ef_trade_regulatory_compliance <dbl> 9.4053278, 0.0000000, 0.9171598, 5.…
$ ef_trade_regulatory <dbl> 7.489905, 2.481294, 2.024949, 4.811…
$ ef_trade_black <dbl> 10.00000, 5.56391, 10.00000, 0.0000…
$ ef_trade_movement_foreign <dbl> 6.306106, 3.664829, 2.946919, 5.358…
$ ef_trade_movement_capital <dbl> 4.6153846, 0.0000000, 3.0769231, 0.…
$ ef_trade_movement_visit <dbl> 8.2969231, 1.1062564, 0.1106256, 7.…
$ ef_trade_movement <dbl> 6.406138, 1.590362, 2.044823, 4.697…
$ ef_trade <dbl> 8.214900, 4.127025, 5.264754, 3.982…
$ ef_regulation_credit_ownership <dbl> 5, 0, 8, 5, 10, 10, 8, 5, 10, 10, 5…
$ ef_regulation_credit_private <dbl> 7.295687, 5.301526, 9.194715, 4.259…
$ ef_regulation_credit_interest <dbl> 9, 10, 4, 7, 10, 10, 10, 9, 10, 10,…
$ ef_regulation_credit <dbl> 7.098562, 5.100509, 7.064905, 5.419…
$ ef_regulation_labor_minwage <dbl> 5.566667, 5.566667, 8.900000, 2.766…
$ ef_regulation_labor_firing <dbl> 5.396399, 3.896912, 2.656198, 2.191…
$ ef_regulation_labor_bargain <dbl> 6.234861, 5.958321, 5.172987, 3.432…
$ ef_regulation_labor_hours <dbl> 8, 6, 4, 10, 10, 10, 6, 6, 8, 8, 10…
$ ef_regulation_labor_dismissal <dbl> 6.299741, 7.755176, 6.632764, 2.517…
$ ef_regulation_labor_conscription <dbl> 10, 1, 0, 10, 0, 10, 3, 1, 10, 10, …
$ ef_regulation_labor <dbl> 6.916278, 5.029513, 4.560325, 5.151…
$ ef_regulation_business_adm <dbl> 6.072172, 3.722341, 2.758428, 2.404…
$ ef_regulation_business_bureaucracy <dbl> 6.000000, 1.777778, 1.333333, 6.666…
$ ef_regulation_business_start <dbl> 9.713864, 9.243070, 8.664627, 9.122…
$ ef_regulation_business_bribes <dbl> 4.050196, 3.765515, 1.945540, 3.260…
$ ef_regulation_business_licensing <dbl> 7.324582, 8.523503, 8.096776, 5.253…
$ ef_regulation_business_compliance <dbl> 7.074366, 7.029528, 6.782923, 6.508…
$ ef_regulation_business <dbl> 6.705863, 5.676956, 4.930271, 5.535…
$ ef_regulation <dbl> 6.906901, 5.268992, 5.518500, 5.369…
$ ef_score <dbl> 7.54, 4.99, 5.17, 4.84, 7.57, 7.98,…
$ ef_rank <dbl> 34, 159, 155, 160, 29, 10, 27, 106,…
$ hf_score <dbl> 7.568140, 5.135886, 5.640662, 6.469…
$ hf_rank <dbl> 48, 155, 142, 107, 57, 4, 16, 130, …
$ hf_quartile <dbl> 2, 4, 4, 3, 2, 1, 1, 4, 2, 2, 4, 2,…The dataset has 1458 observations or cases and 123 columns or variables.
- 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 variablepf_expression_controlas the predictor. Does the relationship look linear? If you knew a country’spf_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?
The type of plot I would use to display the relationship between the personal freedom score, pf_score, and any other numerical variable is a scatter plot. When I look at the relationship betweenthe variable pf_score and pf_expression_control, with pf_expression_control as the predictor, the relationship does look linear. Yes, if I 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, I think it makes sense to use a linear model to predict the personal freedom score.
hfi |>
ggplot(aes(x = pf_expression_control, y = pf_score)) +
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 0.796Here, 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.
- 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 relationship between pf_expression_control and pf_score is linear and positive. We can observe a positive slope: the two variables are positively correlated. This means that as pf_expression_control increases, so too does pf_score. We quantified the strength of the relationship with a correlation coefficient of 0.796. This indicates that there is a strong positive linear relationship between the two variables. We can observe a handful of outliers (“unusual observations”).
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 - _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.
- 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?
The smallest sum of squares I got was 971.235. The largest sum of squares I got was 1494.809.
First I will remove NAs with complete.cases. This makes sure that the dataframe is ready for analysis without missing data in the pf_expression_control and pf_score columns.
hfi_q4 <- hfi |>
filter(complete.cases(pf_expression_control, pf_score))Then I will try and choose a line that does a good job of minimizing the sum of squares.
DATA606::plot_ss(x = hfi_q4$pf_expression_control, y = hfi_q4$pf_score)
Click two points to make a line.
Call:
lm(formula = y ~ x, data = pts)
Coefficients:
(Intercept) x
4.6171 0.4914
Sum of Squares: 952.153The first sum of squares I got was 1494.809. The second sum of squares I got was 1029.723. The third sum of squares I got was 1169.127. The fourth and smallest sum of squares I got was 971.235, which had a x=0.5322 an an intercept of 4.3279.
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-16Let’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:
[ = 4.61707 + 0.49143 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 pf_score is explained by pf_expression_control.
- Fit a new model that uses
pf_expression_controlto predicthf_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?
hfi_q5 <- lm(hf_score ~ pf_expression_control, data = hfi)
summary(hfi_q5)
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-16The equation is [ = 5.153687 + 0.349862 pf_expression_control ]
The y-intercept is 5.153687. With a positive slope of 0.349862 we can interpret that for each additional increase of political pressure on media content, we would expect to see an increase of the human freedom score using an increase in relation to the slope coefficient. Our adjusted R-squared value is 0.5772 which tells us the proportion of variability in the response variable that’s explained by the explanatory variable: in this model, 57.72% of the variability in hf_score is explained by pf_expression_control.
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.
- 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?
hfi_q6 <- lm(pf_score ~ pf_expression_control, data = hfi)
summary(hfi_q6)
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-16We can see that the equation for this is y-hat for pf_score = 4.61707 + 0.49143 * pf_expression_control.
estimate_q6 <- 4.61707 + 0.49143 * 6.7
estimate_q6[1] 7.909651The estimate would be 7.909651. We will see below that this is an overestimate.
filter_q6 <- hfi |>
select(pf_score, pf_expression_control) |>
filter(pf_expression_control >= 6.6) |>
arrange(pf_expression_control)
filter_q6# A tibble: 437 × 2
pf_score pf_expression_control
<dbl> <dbl>
1 7.43 6.75
2 8.22 6.75
3 8.77 6.75
4 7.87 6.75
5 7.39 6.75
6 7.25 6.75
7 7.79 6.75
8 8.27 6.75
9 7.19 6.75
10 7.75 6.75
# ℹ 427 more rowsWe don’t see any actual data points with a pf_expression_control of 6.7 exactly, but we do see some with a pf_expression_control of 6.75 which is nearby. Arranging by pf_expression_control ascending order we see the results that have the first pf_score listed as 7.43 so we will use that to calculate our residual.
residual_q6 <- 7.43 - 7.909651
residual_q6[1] -0.479651The residual is -0.479651 for this 7.909651 prediction for a country’s personal freedom with a 6.7 rating for pf_expression_control.
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.
- Is there any apparent pattern in the residuals plot? What does this indicate about the linearity of the relationship between the two variables?
There is no apparent pattern in the residuals plot, which indicates there is likely 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.
- Based on the histogram and the normal probability plot, does the nearly normal residuals condition appear to be met?
The normal probability plot and the histogram indicate that the distribution is / appears nearly normal, so the nearly normal residuals condition appears to be met.
Constant variability:
- Based on the residuals vs. fitted plot, does the constant variability condition appear to be met?
Based on the residuals vs. fitted plot, the points are scattered around 0 and there is constant variability, so the constant variability condition appears 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?
I am going to look at women’s security pf_ss_women and women’s movement pf_movement_women. At a glance, yes, there does seem to be somewhat of a linear relationship, but not nearly as strong as I was expecting to see. There is some variability, clearly.
womenfreedom <- lm(pf_ss_women ~ pf_movement_women, data = hfi)
summary(womenfreedom)
Call:
lm(formula = pf_ss_women ~ pf_movement_women, data = hfi)
Residuals:
Min 1Q Median 3Q Max
-5.562 -0.662 0.838 1.005 2.687
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.64628 0.11185 50.48 <2e-16 ***
pf_movement_women 0.33490 0.01299 25.79 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.449 on 1315 degrees of freedom
(141 observations deleted due to missingness)
Multiple R-squared: 0.3358, Adjusted R-squared: 0.3353
F-statistic: 664.9 on 1 and 1315 DF, p-value: < 2.2e-16ggplot(hfi, aes(x = pf_movement_women, y = pf_ss_women)) +
geom_point() +
geom_smooth(method = "lm") +
labs(
title = "Scatterplot of Women's Security vs Women's Freedom of Movement",
x = "Women's Movement (`pf_movement_women`)",
y = "Women's Security (`pf_ss_women`)"
)
- How does this relationship compare to the relationship between
pf_expression_controlandpf_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?
The model using pf_movement_women to predict pf_ss_women R-squared is 0.3358, which means that 33.58% of the variability in pf_ss_women is explained. That is a lot lower of an R-squared value compared to the model using pf_expression_control to predict pf_score, which had a R-squared of 0.6342, indicating that it accounts for 63.42% of the variability in pf_score. The independent variable I used does not seem to predict the dependent variable better based on this difference – the model does not do as good of a job as having an independent variable that prediects a dependent variable compared with the model using pf_expression_control to predict pf_score. In some ways this does make sense because of the varied definitionof what ‘security’ and ‘movement’ might mean in terms of the way countries are scored for these variables impacting women.
- What’s one freedom relationship you were most surprised about and why? Display the model diagnostics for the regression model analyzing this relationship.
After looking into the above model using pf_movement_women to predict pf_ss_women with a R-squared of 0.3358, I wanted to see if a model using pf_movement_women to predict pf_score would lead to a greater R-squared value, perhaps one closer to the R-squared of 0.6342 we saw with the the model using pf_expression_control to predict pf_score. While the below model using pf_movement_women to predict pf_score does have a higher R-squared value than using pf_movement_women to predict pf_ss_women, it is much lower than the R-squared using pf_movement_women to predict pf_score. The R-squared is 0.4123 for the model using pf_movement_women to predict pf_score, which is still a linear positive relationship, albeit much less strong of a positive linear relationship than expected. I really thought that a higher score for women’s ability to move freely (which I think is what pf_movement_women is measuring, though I could be misinterpreting) would be more strongly correlated with the pf_score for personal freedom than it appears to be based on the model.
womenmovement <- lm(pf_score ~ pf_movement_women, data = hfi)
summary(womenmovement)
Call:
lm(formula = pf_score ~ pf_movement_women, data = hfi)
Residuals:
Min 1Q Median 3Q Max
-2.76339 -0.74093 -0.01009 0.95994 1.94920
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.851785 0.082209 59.02 <2e-16 ***
pf_movement_women 0.289935 0.009546 30.37 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.065 on 1315 degrees of freedom
(141 observations deleted due to missingness)
Multiple R-squared: 0.4123, Adjusted R-squared: 0.4118
F-statistic: 922.5 on 1 and 1315 DF, p-value: < 2.2e-16