Lecture 9 - Linear regression in R (2022)

Linear regression

Linear regression is just a more general form of ANOVA, which itself is a generalized t-test. In each case, we’re assessing if and how the mean of our outcome \(y\) varies with other variables. Unlike t-tests and ANOVA, which are restricted to the case where the factors of interest are all categorical, regression allows you to also model the effects of continuous variables.

linear regression is used to model linear relationship between an outcome variable, \(y\), and a set of covariates or predictor variables \(x_1, x_2, \ldots, x_p\).

For our first example we’ll look at a small data set in which we’re interested in predicting the crime rate per million population based on socio-economic and demographic information at the state level.

Let’s first import the data set and see what we’re working with.

# Import data setcrime <- read_delim("http://www.andrew.cmu.edu/user/achoulde/94842/data/crime_simple.txt", delim = "\t")
## Parsed with column specification:## cols(## R = col_double(),## Age = col_double(),## S = col_double(),## Ed = col_double(),## Ex0 = col_double(),## Ex1 = col_double(),## LF = col_double(),## M = col_double(),## N = col_double(),## NW = col_double(),## U1 = col_double(),## U2 = col_double(),## W = col_double(),## X = col_double()## )

The variable names that this data set comes with are very confusing, and even misleading.

R: Crime rate: # of offenses reported to police per million population

Age: The number of males of age 14-24 per 1000 population

S: Indicator variable for Southern states (0 = No, 1 = Yes)

Ed: Mean # of years of schooling x 10 for persons of age 25 or older

Ex0: 1960 per capita expenditure on police by state and local government

Ex1: 1959 per capita expenditure on police by state and local government

LF: Labor force participation rate per 1000 civilian urban males age 14-24

M: The number of males per 1000 females

N: State population size in hundred thousands

NW: The number of non-whites per 1000 population

U1: Unemployment rate of urban males per 1000 of age 14-24

U2: Unemployment rate of urban males per 1000 of age 35-39

W: Median value of transferable goods and assets or family income in tens of $

X: The number of families per 1000 earning below 1/2 the median income

We really need to give these variables better names

# Assign more meaningful variable names, also# Convert is.south to a factor# Divide average.ed by 10 so that the variable is actually average education# Convert median assets to 1000's of dollars instead of 10'scrime <- crime %>% rename(crime.per.million = R, young.males = Age, is.south = S, average.ed = Ed, exp.per.cap.1960 = Ex0, exp.per.cap.1959 = Ex1, labour.part = LF, male.per.fem = M, population = N, nonwhite = NW, unemp.youth = U1, unemp.adult = U2, median.assets = W, num.low.salary = X) %>% mutate(is.south = as.factor(is.south), average.ed = average.ed / 10, median.assets = median.assets / 100)# print summary of the datasummary(crime)
## crime.per.million young.males is.south average.ed ## Min. : 34.20 Min. :119.0 0:31 Min. : 8.70 ## 1st Qu.: 65.85 1st Qu.:130.0 1:16 1st Qu.: 9.75 ## Median : 83.10 Median :136.0 Median :10.80 ## Mean : 90.51 Mean :138.6 Mean :10.56 ## 3rd Qu.:105.75 3rd Qu.:146.0 3rd Qu.:11.45 ## Max. :199.30 Max. :177.0 Max. :12.20 ## exp.per.cap.1960 exp.per.cap.1959 labour.part male.per.fem ## Min. : 45.0 Min. : 41.00 Min. :480.0 Min. : 934.0 ## 1st Qu.: 62.5 1st Qu.: 58.50 1st Qu.:530.5 1st Qu.: 964.5 ## Median : 78.0 Median : 73.00 Median :560.0 Median : 977.0 ## Mean : 85.0 Mean : 80.23 Mean :561.2 Mean : 983.0 ## 3rd Qu.:104.5 3rd Qu.: 97.00 3rd Qu.:593.0 3rd Qu.: 992.0 ## Max. :166.0 Max. :157.00 Max. :641.0 Max. :1071.0 ## population nonwhite unemp.youth unemp.adult ## Min. : 3.00 Min. : 2.0 Min. : 70.00 Min. :20.00 ## 1st Qu.: 10.00 1st Qu.: 24.0 1st Qu.: 80.50 1st Qu.:27.50 ## Median : 25.00 Median : 76.0 Median : 92.00 Median :34.00 ## Mean : 36.62 Mean :101.1 Mean : 95.47 Mean :33.98 ## 3rd Qu.: 41.50 3rd Qu.:132.5 3rd Qu.:104.00 3rd Qu.:38.50 ## Max. :168.00 Max. :423.0 Max. :142.00 Max. :58.00 ## median.assets num.low.salary ## Min. :2.880 Min. :126.0 ## 1st Qu.:4.595 1st Qu.:165.5 ## Median :5.370 Median :176.0 ## Mean :5.254 Mean :194.0 ## 3rd Qu.:5.915 3rd Qu.:227.5 ## Max. :6.890 Max. :276.0

First step: some plotting and summary statistics

You can start by feeding everything into a regression, but it’s often a better idea to construct some simple plots (e.g., scatterplots and boxplots) and summary statistics to get some sense of how the data behaves.

# Scatter plot of outcome (crime.per.million) against average.edqplot(average.ed, crime.per.million, data = crime)

Lecture 9 - Linear regression in R (1)

# correlation between education and crimewith(crime, cor(average.ed, crime.per.million))
## [1] 0.3228349

This seems to suggest that higher levels of average education are associated with higher crime rates. Can you come up with an explanation for this phenomenon?

# Scatter plot of outcome (crime.per.million) against median.assetsqplot(median.assets, crime.per.million, data = crime)

Lecture 9 - Linear regression in R (2)

# correlation between education and crimewith(crime, cor(median.assets, crime.per.million))
## [1] 0.4413199

There also appears to be a positive association between median assets and crime rates.

# Boxplots showing crime rate broken down by southern vs non-southern stateqplot(is.south, crime.per.million, geom = "boxplot", data = crime)

Lecture 9 - Linear regression in R (3)

Constructing a regression model

To construct a linear regression model in R, we use the lm() function. You can specify the regression model in various ways. The simplest is often to use the formula specification.

The first model we fit is a regression of the outcome (crimes.per.million) against all the other variables in the data set. You can either write out all the variable names. or use the shorthand y ~ . to specify that you want to include all the variables in your regression.

crime.lm <- lm(crime.per.million ~ ., data = crime)# Summary of the linear regression modelcrime.lm
## ## Call:## lm(formula = crime.per.million ~ ., data = crime)## ## Coefficients:## (Intercept) young.males is.south1 average.ed ## -6.918e+02 1.040e+00 -8.308e+00 1.802e+01 ## exp.per.cap.1960 exp.per.cap.1959 labour.part male.per.fem ## 1.608e+00 -6.673e-01 -4.103e-02 1.648e-01 ## population nonwhite unemp.youth unemp.adult ## -4.128e-02 7.175e-03 -6.017e-01 1.792e+00 ## median.assets num.low.salary ## 1.374e+01 7.929e-01
summary(crime.lm)
## ## Call:## lm(formula = crime.per.million ~ ., data = crime)## ## Residuals:## Min 1Q Median 3Q Max ## -34.884 -11.923 -1.135 13.495 50.560 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -6.918e+02 1.559e+02 -4.438 9.56e-05 ***## young.males 1.040e+00 4.227e-01 2.460 0.01931 * ## is.south1 -8.308e+00 1.491e+01 -0.557 0.58117 ## average.ed 1.802e+01 6.497e+00 2.773 0.00906 ** ## exp.per.cap.1960 1.608e+00 1.059e+00 1.519 0.13836 ## exp.per.cap.1959 -6.673e-01 1.149e+00 -0.581 0.56529 ## labour.part -4.103e-02 1.535e-01 -0.267 0.79087 ## male.per.fem 1.648e-01 2.099e-01 0.785 0.43806 ## population -4.128e-02 1.295e-01 -0.319 0.75196 ## nonwhite 7.175e-03 6.387e-02 0.112 0.91124 ## unemp.youth -6.017e-01 4.372e-01 -1.376 0.17798 ## unemp.adult 1.792e+00 8.561e-01 2.093 0.04407 * ## median.assets 1.374e+01 1.058e+01 1.298 0.20332 ## num.low.salary 7.929e-01 2.351e-01 3.373 0.00191 ** ## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 21.94 on 33 degrees of freedom## Multiple R-squared: 0.7692, Adjusted R-squared: 0.6783 ## F-statistic: 8.462 on 13 and 33 DF, p-value: 3.686e-07

R’s default is to output values in scientific notation. This can make it hard to interpret the numbers. Here’s some code that can be used to force full printout of numbers.

options(scipen=4) # Set scipen = 0 to get back to default
summary(crime.lm)
## ## Call:## lm(formula = crime.per.million ~ ., data = crime)## ## Residuals:## Min 1Q Median 3Q Max ## -34.884 -11.923 -1.135 13.495 50.560 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -691.837588 155.887918 -4.438 0.0000956 ***## young.males 1.039810 0.422708 2.460 0.01931 * ## is.south1 -8.308313 14.911588 -0.557 0.58117 ## average.ed 18.016011 6.496504 2.773 0.00906 ** ## exp.per.cap.1960 1.607818 1.058667 1.519 0.13836 ## exp.per.cap.1959 -0.667258 1.148773 -0.581 0.56529 ## labour.part -0.041031 0.153477 -0.267 0.79087 ## male.per.fem 0.164795 0.209932 0.785 0.43806 ## population -0.041277 0.129516 -0.319 0.75196 ## nonwhite 0.007175 0.063867 0.112 0.91124 ## unemp.youth -0.601675 0.437154 -1.376 0.17798 ## unemp.adult 1.792263 0.856111 2.093 0.04407 * ## median.assets 13.735847 10.583028 1.298 0.20332 ## num.low.salary 0.792933 0.235085 3.373 0.00191 ** ## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 21.94 on 33 degrees of freedom## Multiple R-squared: 0.7692, Adjusted R-squared: 0.6783 ## F-statistic: 8.462 on 13 and 33 DF, p-value: 0.0000003686

Looking at the p-values, it looks like num.low.salary (number of families per 1000 earning below 1/2 the median income), unemp.adult (Unemployment rate of urban males per 1000 of age 35-39), average.ed (Mean # of years of schooling 25 or older), and young.males (number of males of age 14-24 per 1000 population) are all statistically significant predictors of crime rate.

The coefficients for these predictors are all positive, so crime rates are positively associated with wealth inequality, adult unemployment rates, average education levels, and high rates of young males in the population.

Exploring the lm object

What kind of output do we get when we run a linear model (lm) in R?

# List all attributes of the linear modelattributes(crime.lm)
## $names## [1] "coefficients" "residuals" "effects" "rank" ## [5] "fitted.values" "assign" "qr" "df.residual" ## [9] "contrasts" "xlevels" "call" "terms" ## [13] "model" ## ## $class## [1] "lm"
# coefficientscrime.lm$coef
## (Intercept) young.males is.south1 average.ed ## -691.837587905 1.039809653 -8.308312889 18.016010601 ## exp.per.cap.1960 exp.per.cap.1959 labour.part male.per.fem ## 1.607818377 -0.667258285 -0.041031047 0.164794968 ## population nonwhite unemp.youth unemp.adult ## -0.041276887 0.007174688 -0.601675298 1.792262901 ## median.assets num.low.salary ## 13.735847285 0.792932786

None of the attributes seem to give you p-values. Here’s what you can do to get a table that allows you to extract p-values.

# Pull coefficients element from summary(lm) objectround(summary(crime.lm)$coef, 3)
## Estimate Std. Error t value Pr(>|t|)## (Intercept) -691.838 155.888 -4.438 0.000## young.males 1.040 0.423 2.460 0.019## is.south1 -8.308 14.912 -0.557 0.581## average.ed 18.016 6.497 2.773 0.009## exp.per.cap.1960 1.608 1.059 1.519 0.138## exp.per.cap.1959 -0.667 1.149 -0.581 0.565## labour.part -0.041 0.153 -0.267 0.791## male.per.fem 0.165 0.210 0.785 0.438## population -0.041 0.130 -0.319 0.752## nonwhite 0.007 0.064 0.112 0.911## unemp.youth -0.602 0.437 -1.376 0.178## unemp.adult 1.792 0.856 2.093 0.044## median.assets 13.736 10.583 1.298 0.203## num.low.salary 0.793 0.235 3.373 0.002

If you want a particular p-value, you can get it by doing the following

# Pull the coefficients table from summary(lm)crime.lm.coef <- round(summary(crime.lm)$coef, 3)# See what this givesclass(crime.lm.coef)
## [1] "matrix"
attributes(crime.lm.coef)
## $dim## [1] 14 4## ## $dimnames## $dimnames[[1]]## [1] "(Intercept)" "young.males" "is.south1" ## [4] "average.ed" "exp.per.cap.1960" "exp.per.cap.1959"## [7] "labour.part" "male.per.fem" "population" ## [10] "nonwhite" "unemp.youth" "unemp.adult" ## [13] "median.assets" "num.low.salary" ## ## $dimnames[[2]]## [1] "Estimate" "Std. Error" "t value" "Pr(>|t|)"
crime.lm.coef["average.ed", "Pr(>|t|)"]
## [1] 0.009

The coefficients table is a matrix with named rows and columns. You can therefore access particular cells either by numeric index, or by name (as in the example above).

Plotting the lm object
plot(crime.lm)

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These four plots are important diagnostic tools in assessing whether the linear model is appropriate. The first two plots are the most important, but the last two can also help with identifying outliers and non-linearities.

Residuals vs.Fitted When a linear model is appropriate, we expect

  1. the residuals will have constant variance when plotted against fitted values; and

  2. the residuals and fitted values will be uncorrelated.

If there are clear trends in the residual plot, or the plot looks like a funnel, these are clear indicators that the given linear model is inappropriate.

Normal QQ plot You can use a linear model for prediction even if the underlying normality assumptions don’t hold. However, in order for the p-values to be believable, the residuals from the regression must look approximately normally distributed.

Scale-location plot This is another version of the residuals vs fitted plot. There should be no discernible trends in this plot.

Residuals vs Leverage. Leverage is a measure of how much an observation influenced the model fit. It’s a one-number summary of how different the model fit would be if the given observation was excluded, compared to the model fit where the observation is included. Points with high residual (poorly described by the model) and high leverage (high influence on model fit) are outliers. They’re skewing the model fit away from the rest of the data, and don’t really seem to fit with the rest of the data.

The residual vs fitted and scale-location diagnostic plots for the crime data aren’t especially insightful, largely due to the very small sample size. Below we look at the diamonds data to see what a more typical anaylsis of linear model diagnostic plots might reveal.

Diagnostic plots for diamonds data.
diamonds.lm <- lm(price ~ carat + cut + clarity + color, data = diamonds)plot(diamonds.lm)

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