HAZAL GUNDUZ
Introduction to linear regression
The Human Freedom Index is a report that attempts to summarize the idea of “freedom” through a bunch of different variables for many countries around the globe. It serves as a rough objective measure for the relationships between the different types of freedom - whether it’s political, religious, economical or personal freedom - and other social and economic circumstances. The Human Freedom Index is an annually co-published report by the Cato Institute, the Fraser Institute, and the Liberales Institut at the Friedrich Naumann Foundation for Freedom.
In this lab, you’ll be analyzing data from Human Freedom Index reports from 2008-2016. Your aim will be to summarize a few of the relationships within the data both graphically and numerically in order to find which variables can help tell a story about freedom.
Getting Started
Load packages
In this lab, you will explore and visualize the data using the tidyverse suite of packages. The data can be found in the companion package for OpenIntro resources, openintro.
Let’s load the packages.
library(tidyverse)## ── Attaching packages ─────────────────────────────────────── tidyverse 1.3.1 ──## ✓ ggplot2 3.3.5 ✓ purrr 0.3.4
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## x dplyr::filter() masks stats::filter()
## x dplyr::lag() masks stats::lag()library(openintro)## Loading required package: airports## Loading required package: cherryblossom## Loading required package: usdatalibrary(statsr)## Loading required package: BayesFactor## Loading required package: coda## Loading required package: Matrix##
## Attaching package: 'Matrix'## The following objects are masked from 'package:tidyr':
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## expand, pack, unpack## ************
## Welcome to BayesFactor 0.9.12-4.2. If you have questions, please contact Richard Morey (richarddmorey@gmail.com).
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## Type BFManual() to open the manual.
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## Attaching package: 'statsr'## The following objects are masked from 'package:openintro':
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## calc_streak, evals, nycflights, presentThe data
The data we’re working with is in the openintro package and it’s called hfi, short for Human Freedom Index.
Exercise 1. What are the dimensions of the dataset?
data('hfi', package='openintro')
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,…dim(hfi)## [1] 1458 123=> The data set has 1458 observations and 123 variables.
Exercise 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 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 could use a scatter plot to display the relationship between the personal freedom score, pf_score, and one of the other numerical variables.
plot(hfi$pf_score ~ hfi$pf_expression_control,
xlab = "Expression control", ylab = "Pf score", col = "green")
If the relationship looks linear, we can quantify the strength of the relationship with the correlation coefficient.
ggplot(hfi, aes(x=pf_expression_control, y=pf_score)) + geom_point() +
ggtitle("Human Freedom Index \nScore per expression control") +
xlab("Expression Control") + ylab("Score") +
theme(
plot.title = element_text(color="red", size=14, face="bold.italic"),
axis.title.x = element_text(color="green", size=12, face="bold"),
axis.title.y = element_text(color="#993333", size=12, face="bold"))## Warning: Removed 80 rows containing missing values (geom_point).
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 0.796Sum 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.
Exercise 3. 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 is an association between the two variables, the relationship between two variables is linear. It is a moderate strong positive correlation. As the pf_expression_control increases, the pf_score increases as well. We should also point that there are outliers or some points which predicted values would be far off from the observed values.
hfi1 <- hfi[c("pf_score", "pf_expression_control")]
hfi1 <- drop_na(hfi1)
row.names <- NULLplot_ss(x = pf_expression_control, y = pf_score, data = hfi1)
## Click two points to make a line.
## Call:
## lm(formula = y ~ x, data = pts)
##
## Coefficients:
## (Intercept) x
## 4.6171 0.4914
##
## Sum of Squares: 952.153plot_ss(x = pf_expression_control, y = pf_score, data = hfi1, 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: 952.153Exercise 4. 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?
plot_ss(x = pf_expression_control, y = pf_score, data = hfi1)
## Click two points to make a line.
## Call:
## lm(formula = y ~ x, data = pts)
##
## Coefficients:
## (Intercept) x
## 4.6171 0.4914
##
## Sum of Squares: 952.153=> The smallest sum of squares is 952.153.
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)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-16Exercise 5. Fit a new model that uses pf_expression_control to predict hf_score, or the total human freedom score. Using the estimates from the R output, write the equation of the regression line. What does the slope tell us in the context of the relationship between human freedom and the amount of political pressure on media content?
lm5 <- lm(hf_score ~ pf_expression_control, data = hfi)
summary(lm5)##
## 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.
Exercise 6. 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?
pf_exp_control <- 6.7
pf_score6 <- 4.61707 + 0.49143 * pf_exp_control
pf_score6## [1] 7.909651hfi %>%
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>, …residus <- 7.43 - 7.91
residus## [1] -0.48=> The prediction overestimated on 0.48
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 (ŷ ) 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.
Exercise 7. Is there any apparent pattern in the residuals plot? What does this indicate about the linearity of the 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 = 2) +
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.
Exercise 8. Based on the histogram and the normal probability plot, does the nearly normal residuals condition appear to be met?
=> Both the histogram and the normal probability plot show that the distribution of these data are nearly normal. The nearly normal residuals condition appear to be met.
Exercise 9. Based on the residuals vs. fitted plot, does the constant variability condition appear to be met?
=> The points residuals vs. fitted plot show that points are scattered around 0, there is a constant variability.Thus, the constant variability condition appear to be met.
More Practice
1. 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?
ggplot(data = hfi, aes(x = pf_ss, y = hf_rank)) +
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).
=> At a glance, the relationship between these two variables seems linear. It is a negative association: As pf_ss increases, hf_rank decreases.
2. How does this relationship compare to the relationship between pf_expression_control and pf_score? Use the R2 values from the two model summaries to compare. Does your independent variable seem to predict your dependent one better? Why or why not?
lm5 <- lm(hf_score ~ pf_expression_control, data = hfi)
summary(lm5)##
## 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-16lma <- lm(hfi$hf_rank ~ hfi$pf_ss)
summary(lma)##
## Call:
## lm(formula = hfi$hf_rank ~ hfi$pf_ss)
##
## Residuals:
## Min 1Q Median 3Q Max
## -100.628 -23.567 -3.302 23.189 77.679
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 266.3430 5.1084 52.14 <2e-16 ***
## hfi$pf_ss -23.1430 0.6159 -37.58 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 31.28 on 1376 degrees of freedom
## (80 observations deleted due to missingness)
## Multiple R-squared: 0.5065, Adjusted R-squared: 0.5061
## F-statistic: 1412 on 1 and 1376 DF, p-value: < 2.2e-16=> From the r. square values of both models, we have this: pf_expression_controlandpf_score` model: 57.75% of the variability in pf_score can be explained by pf_expression_control. pf_ss and hf_rank model: 50.65% of the variability in hf_rank can be explained by pf_ss
My independent variable does not seem to predict my dependent variable better because my r square (as explained above) is lower than r square of pf_expression_controlandpf_score` model, it counts less variation.
3 What’s one freedom relationship you were most surprised about and why? Display the model diagnostics for the regression model analyzing this relationship.
lmb <- lm(hfi$ef_legal_protection ~ hfi$ef_legal_integrity)
summary(lmb)##
## Call:
## lm(formula = hfi$ef_legal_protection ~ hfi$ef_legal_integrity)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.3674 -0.8148 0.1089 0.9094 3.8036
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 2.09412 0.11409 18.36 <2e-16 ***
## hfi$ef_legal_integrity 0.56976 0.01719 33.15 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.226 on 1109 degrees of freedom
## (347 observations deleted due to missingness)
## Multiple R-squared: 0.4978, Adjusted R-squared: 0.4973
## F-statistic: 1099 on 1 and 1109 DF, p-value: < 2.2e-16hfi %>%
summarise(cor(ef_legal_protection, ef_legal_integrity, use = "complete.obs"))## # A tibble: 1 × 1
## `cor(ef_legal_protection, ef_legal_integrity, use = "complete.obs")`
## <dbl>
## 1 0.706Linearity: We should also verify this condition with a plot of the residuals vs. fitted (predicted) values.
ggplot(data = lmb, aes(x = .fitted, y = .resid)) +
geom_point() +
geom_hline(yintercept = 0, linetype = "dashed") +
xlab("Fitted values") +
ylab("Residuals")
Nearly normal residuals: To check this condition, we can look at a histogram.
ggplot(data = lmb, aes(x = .resid)) +
geom_histogram(binwidth = 2) +
xlab("Residuals")
ggplot(data = lmb, aes(sample = .resid)) +
stat_qq()
Constant variability:We will look again on residuals vs. fitted plot.
Rpubs => https://rpubs.com/gunduzhazal/830881