Introduction to linear regression
Leticia Salazar
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.
- What are the dimensions of the dataset?
The data set has 1458 observation with 123 variables.
# glimpse of data
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,…- 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?
A scatter plot would be best to show the relationship between
the pf_score and other numerical variables. Being that the relationship
appears linear and if we new a country’s
pf_expression_control, or it’s score out of 10, with 0
being the most, of political pressures and control media content, I
would be comfortable using a linear model to predict the personal
freedom score pf_score.
# scatter plot
hfi %>%
ggplot(aes(x = pf_expression_control, y = pf_score)) +
geom_point() +
xlab("Expression Control") +
ylab("Pf Score") +
ggtitle("")
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 the variables is linear with a
positive upward correlation meaning that as the
pf_expression_control increases so does the
pf_score. There are a couple of outliers but majority of
the points are closer together.
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.
- 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 is 953.209.
set.seed(75)
hfi_1 <- select(hfi, pf_expression_control, pf_score)
# Check for missing values
sum(is.na(hfi_1))## [1] 160# Drop NAs
hfi_2 <- na.omit(hfi_1)
# Plot
DATA606::plot_ss(x = hfi_2$pf_expression_control, y = hfi_2$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: 952.153set.seed(78)
DATA606::plot_ss(x = hfi_2$pf_expression_control, y = hfi_2$pf_score, showSquares = TRUE)
## 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 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 at_bats. 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.
- 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?
**The equation: hf_score = 5.153687 + 0.349862*(pf_expression_control), the intercept: 5.153687, and the slope: 0.349862. **
m2 <- lm(pf_score ~ pf_expression_control, data = hfi)
summary(m2)##
## 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-16Prediction 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?
The prediction is overestimated by 0.48.
pf_expression_cont <- 6.7
pf_score_6 <- 4.61707 + 0.49143 * pf_expression_cont
pf_score_6## [1] 7.909651# Checking for observed value for pf_score 6.7 rating
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>, …# Check residual
residual <- 7.43 - 7.91
residual## [1] -0.48Model 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.
- Is there any apparent pattern in the residuals plot? What does this indicate about the linearity of the relationship between the two variables?
There doesn’t seem to be a pattern in the residual plot therefore being a relationship between the variables.
Nearly normal residuals: To check this condition, we can look at a histogram
ggplot(data = m1, aes(x = .resid)) +
geom_histogram(binwidth = 1) +
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 histogram and normal probability plot of the residuals show a distribution that appears to be normal. I am assuming that the condition are met due to the appearance.
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 a constant variability, therefore, the condition apppears 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?
At a glance it looks like there’s a linear relationtion with a downward negative slope, as pf_association increases the pf_rank decreases.
# Using new variables pf_association and pf_rank
hfi %>%
ggplot(aes( x = pf_association, y = pf_rank)) +
geom_point() +
geom_smooth(method = "lm", se = FALSE) +
xlab("Pf Association") +
ylab("Pf Rank")
- 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?
For Model 1 the r-squared is 57.75% and for Model 2 it’s 52.52%. The independent variable does not seem to predict the dependent variable due to the r-squared of Model 1 being lower than that of Model 2.
# Model 1
lm_11 <- lm(hf_score ~ pf_expression_control, data = hfi)
summary(lm_11)##
## 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# Model 2
lm_12 <- lm(hfi$pf_rank ~ hfi$pf_ss)
summary(lm_12)##
## Call:
## lm(formula = hfi$pf_rank ~ hfi$pf_ss)
##
## Residuals:
## Min 1Q Median 3Q Max
## -90.288 -23.554 -5.786 22.067 77.687
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 269.9610 5.0094 53.89 <2e-16 ***
## hfi$pf_ss -23.5592 0.6039 -39.01 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 30.67 on 1376 degrees of freedom
## (80 observations deleted due to missingness)
## Multiple R-squared: 0.5252, Adjusted R-squared: 0.5248
## F-statistic: 1522 on 1 and 1376 DF, p-value: < 2.2e-16- What’s one freedom relationship you were most surprised about and why? Display the model diagnostics for the regression model analyzing this relationship.
There seems to be little to no relationship between ef_legal_integrity and ef_government.
model_2 <- lm(ef_legal_integrity ~ ef_money_growth, data = hfi)
summary(model_2)##
## Call:
## lm(formula = ef_legal_integrity ~ ef_money_growth, data = hfi)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.2785 -1.6567 0.0264 1.7994 5.6234
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.97216 0.48134 4.097 4.47e-05 ***
## ef_money_growth 0.48825 0.05545 8.804 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.092 on 1176 degrees of freedom
## (280 observations deleted due to missingness)
## Multiple R-squared: 0.06184, Adjusted R-squared: 0.06104
## F-statistic: 77.52 on 1 and 1176 DF, p-value: < 2.2e-16# Plot
hfi %>%
ggplot(aes(x = ef_money_growth, y = ef_legal_integrity)) +
geom_point() +
xlab("Ef_Money_Growth") +
ylab("Ef_Legal_Integrity")