library(tidyverse)
library(openintro)
library(devtools)
install_github('jbryer/DATA606')
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? #1,458 Rows, 123 Columns
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? #Answer #2 I used a regular scatterplot. The relationship looks fairly linear / there seems to be a a somewhat strong correlation between the two variables. I would be comfortable using a linear regression here as the relationship looks fairly linear, however the relationship seems to breakdown as the pf score nears 0 ( or to be more accurate 0-4.)
ggplot(data = hfi, mapping = aes(x = pf_score , y = pf_expression_control)) +
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 0.796Here, 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 plot shows that the relationship looks positively linear. And, after calculating the correlation we know there is a faily strong correlation between the 2 variables. Although, it is important to note that there is obviously some residuals in the relationship.
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?
#Answer 4 : The sum of squares I got was 952.
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:
\[ \hat{y} = 4.61707 + 0.49143 \times pf\_expression\_control \]
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?
#Answer 5 - # Y intercept is 5.153687 #Coefficient 0.349862 # Y hat = 5.153687 * (0.349862 * pf_expression_control) #The slope tells us there is a roughly 35% connection/correlation between pf_expression_Control and hf_score
example_5 <- lm(hf_score ~ pf_expression_control , data = hfi)
summary(example_5)##
## 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-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)## `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.
- If someone saw the least squares regression line and not the actual
data, how would they predict a country’s personal freedom school for one
with a 6.7 rating for
pf_expression_control? Is this an overestimate or an underestimate, and by how much? In other words, what is the residual for this prediction? #Answer 6 - one would expect a freedom score of roughly 7.9. The range of outcomes could be as low 6.75 and as high as 9.
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.
- Is there any apparent pattern in the residuals plot? What does this
indicate about the linearity of the relationship between the two
variables?
#Based on the residuals plot , there is not pattern, and the variance of the error predicted value is constant across the x-axis 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? #ANSWER #8 Yes, the residuals seem to be normally distributed, so our normal residuals condition is met.
Constant variability:
- Based on the residuals vs. fitted plot, does the constant variability condition appear to be met? #ANSWER #9 Yes, the constant variability condition is met. The variance is constant in the positive and negative directions across the x-axis. * * *
More Practice
- Choose another freedom variable and a variable you think would strongly correlate with it.. Produce a scatterplot of the two variables and fit a linear model. At a glance, does there seem to be a linear relationship?
#ANSWER - PF Religion restrictions / PF religion harassment - there does seem to be a linear relationship ( somewhat ), there is a lot of scattered observations in the upper right hand corner of the plot
view(hfi)
hfi %>% ggplot(aes(x =pf_religion_restrictions , y = pf_religion_harassment)) +
geom_point()## Warning: Removed 94 rows containing missing values (geom_point).
hfi_1 <- drop_na(hfi[c('pf_religion_restrictions','pf_religion')])
DATA606::plot_ss(x = hfi_1$pf_religion_restrictions, y = hfi_1$pf_religion)
## Click two points to make a line.
## Call:
## lm(formula = y ~ x, data = pts)
##
## Coefficients:
## (Intercept) x
## 3.6265 0.5847
##
## Sum of Squares: 931.005- 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?
#Answer - The R^2 value in my new model is 0.5941 compared to 0.634 in the original model. Since our previous model has an R-squared higher it is a better predicter.
m3 <- lm(pf_religion_restrictions ~ pf_religion, data = hfi_1)
summary(m3)##
## Call:
## lm(formula = pf_religion_restrictions ~ pf_religion, data = hfi_1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.4408 -0.6859 0.0137 0.6571 3.9790
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.72688 0.18202 -3.993 6.86e-05 ***
## pf_religion 1.01659 0.02275 44.680 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.09 on 1362 degrees of freedom
## Multiple R-squared: 0.5944, Adjusted R-squared: 0.5941
## F-statistic: 1996 on 1 and 1362 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.