Data
There’s this interesting dataset at the site for Princeton’s course in Generalized Linear Models (http://data.princeton.edu/wws509/datasets/#salary).
I am using salary data that include observations on six variables for 52 tenure-track professors in a small college. The variables are:
sx: Sex, coded 1 for female and 0 for male
rk: Rank, coded 1 for assistant professor, 2 for associate professor, and 3 for full professor
yr: Number of years in current rank
dg: Highest degree, coded 1 if doctorate, 0 if masters
yd: Number of years since highest degree was earned
sl: Academic year salary, in dollars.
# Data import
library(foreign)
salary <- read.dta("http://data.princeton.edu/wws509/datasets/salary.dta")
# Data size
dim(salary)
## [1] 52 6
# General data description
summary(salary)
## sx rk yr dg
## Male :38 Assistant:18 Min. : 0.000 Min. :0.0000
## Female:14 Associate:14 1st Qu.: 3.000 1st Qu.:0.0000
## Full :20 Median : 7.000 Median :1.0000
## Mean : 7.481 Mean :0.6538
## 3rd Qu.:11.000 3rd Qu.:1.0000
## Max. :25.000 Max. :1.0000
## yd sl
## Min. : 1.00 Min. :15000
## 1st Qu.: 6.75 1st Qu.:18247
## Median :15.50 Median :23719
## Mean :16.12 Mean :23798
## 3rd Qu.:23.25 3rd Qu.:27259
## Max. :35.00 Max. :38045
# Data sample
knitr::kable(head(salary))
| sx | rk | yr | dg | yd | sl |
|---|---|---|---|---|---|
| Male | Full | 25 | 1 | 35 | 36350 |
| Male | Full | 13 | 1 | 22 | 35350 |
| Male | Full | 10 | 1 | 23 | 28200 |
| Female | Full | 7 | 1 | 27 | 26775 |
| Male | Full | 19 | 0 | 30 | 33696 |
| Male | Full | 16 | 1 | 21 | 28516 |
Convert sex and rank into numerical representation.
salary$sx <- as.character(salary$sx)
salary$sx[salary$sx == "Male"] <- 0
salary$sx[salary$sx == "Female"] <- 1
salary$sx <- as.integer(salary$sx)
salary$rk <- as.character(salary$rk)
salary$rk[salary$rk == "Assistant"] <- 1
salary$rk[salary$rk == "Associate"] <- 2
salary$rk[salary$rk == "Full"] <- 3
salary$rk <- as.integer(salary$rk)
The model will predict professor’s salary. All variables seem like they may be relevant.
Sex (sx) will be my dichotomous variable.
It may make sense that increase in rank does not have a strictly linear relationship with the target variable. Higher rank may warrant higher salary increases, so my model will use square of rank (rk2) as a quadratic variable.
Finally, I wanted to see if the model is influenced by interaction between sex and number of years since highest degree was earned (sx×yd).
# Quadratic variable
rk2 <- salary$rk^2
# Dichotomous vs. quantative interaction
sx_yd <- salary$sx * salary$yd
# Initial model
salary_lm <- lm(sl ~ sx + rk + rk2 + yr + dg + yd + sx_yd, data=salary)
summary(salary_lm)
##
## Call:
## lm(formula = sl ~ sx + rk + rk2 + yr + dg + yd + sx_yd, data = salary)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3827.5 -1180.3 -288.7 844.7 8709.7
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 13045.79 3302.80 3.950 0.000279 ***
## sx 127.83 1359.23 0.094 0.925502
## rk 4230.31 3530.16 1.198 0.237202
## rk2 340.06 845.99 0.402 0.689655
## yr 523.83 105.21 4.979 1.03e-05 ***
## dg -1514.35 1024.89 -1.478 0.146645
## yd -174.42 90.99 -1.917 0.061751 .
## sx_yd 80.00 76.74 1.042 0.302882
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2396 on 44 degrees of freedom
## Multiple R-squared: 0.8585, Adjusted R-squared: 0.836
## F-statistic: 38.15 on 7 and 44 DF, p-value: < 2.2e-16
Perform backwards elimination - removing one variable (the one with highest p-value) at a time. Removing sex.
# Version 2
salary_lm <- update(salary_lm, .~. -sx)
summary(salary_lm)
##
## Call:
## lm(formula = sl ~ rk + rk2 + yr + dg + yd + sx_yd, data = salary)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3822.3 -1186.7 -284.7 851.5 8710.6
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 13159.28 3040.37 4.328 8.28e-05 ***
## rk 4142.04 3365.41 1.231 0.2248
## rk2 358.78 813.13 0.441 0.6612
## yr 523.94 104.04 5.036 8.16e-06 ***
## dg -1506.10 1009.82 -1.491 0.1428
## yd -175.51 89.24 -1.967 0.0554 .
## sx_yd 85.29 51.63 1.652 0.1055
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2370 on 45 degrees of freedom
## Multiple R-squared: 0.8585, Adjusted R-squared: 0.8396
## F-statistic: 45.51 on 6 and 45 DF, p-value: < 2.2e-16
Removing square of rank.
# Version 3
salary_lm <- update(salary_lm, .~. -rk2)
summary(salary_lm)
##
## Call:
## lm(formula = sl ~ rk + yr + dg + yd + sx_yd, data = salary)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3635.0 -1330.9 -218.3 615.3 8730.4
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11898.01 1026.60 11.590 3.04e-15 ***
## rk 5598.87 645.84 8.669 3.11e-11 ***
## yr 531.62 101.67 5.229 4.06e-06 ***
## dg -1411.88 978.31 -1.443 0.1557
## yd -180.92 87.61 -2.065 0.0446 *
## sx_yd 88.38 50.70 1.743 0.0880 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2349 on 46 degrees of freedom
## Multiple R-squared: 0.8579, Adjusted R-squared: 0.8424
## F-statistic: 55.54 on 5 and 46 DF, p-value: < 2.2e-16
Removing highest degree.
# Version 4
salary_lm <- update(salary_lm, .~. -dg)
summary(salary_lm)
##
## Call:
## lm(formula = sl ~ rk + yr + yd + sx_yd, data = salary)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3545.3 -1585.0 -432.7 884.0 8520.8
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11087.98 869.42 12.753 < 2e-16 ***
## rk 5090.45 547.49 9.298 3.18e-12 ***
## yr 480.09 96.28 4.986 8.81e-06 ***
## yd -94.26 64.53 -1.461 0.151
## sx_yd 66.10 48.85 1.353 0.182
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2376 on 47 degrees of freedom
## Multiple R-squared: 0.8515, Adjusted R-squared: 0.8388
## F-statistic: 67.35 on 4 and 47 DF, p-value: < 2.2e-16
Removing interaction between sex and number of years since highest degree was earned.
# Version 5
salary_lm <- update(salary_lm, .~. -sx_yd)
summary(salary_lm)
##
## Call:
## lm(formula = sl ~ rk + yr + yd, data = salary)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3329.7 -1135.6 -377.9 801.5 9576.6
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11282.90 864.79 13.047 < 2e-16 ***
## rk 4973.64 545.30 9.121 4.71e-12 ***
## yr 405.67 79.71 5.089 5.94e-06 ***
## yd -40.86 51.50 -0.794 0.431
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2396 on 48 degrees of freedom
## Multiple R-squared: 0.8457, Adjusted R-squared: 0.836
## F-statistic: 87.68 on 3 and 48 DF, p-value: < 2.2e-16
Removing number of years since highest degree was earned.
# Version 6
salary_lm <- update(salary_lm, .~. -yd)
summary(salary_lm)
##
## Call:
## lm(formula = sl ~ rk + yr, data = salary)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3339.4 -1451.0 -323.3 821.3 9502.6
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11336.67 858.87 13.200 < 2e-16 ***
## rk 4731.26 450.01 10.514 3.72e-14 ***
## yr 376.50 70.46 5.344 2.36e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2387 on 49 degrees of freedom
## Multiple R-squared: 0.8436, Adjusted R-squared: 0.8373
## F-statistic: 132.2 on 2 and 49 DF, p-value: < 2.2e-16
The final model has two variables - rank and number of years in current rank - that can be used to predict the target variable.
The median of residuals is a little off zero and they are a little uneven (the maximum value is noticeably farther away than the minimum value). Standard error for the yr variable could be smaller.
Two coefficients imply that for every increase in rank the salary increases by $4,731.26 and with every year in the current rank the salary increases by $376.50.
Based on the Residuals vs. Fitted plot below there are some outliers in the data, but overall variability is fairly consistent. Based on the Q-Q plot, distribution of residuals is close to normal.
Based on R2 value, the model explains 84.36% of variability in the data.
plot(salary_lm$fitted.values, salary_lm$residuals, xlab="Fitted Values", ylab="Residuals", main="Residuals vs. Fitted")
abline(h=0)
qqnorm(salary_lm$residuals)
qqline(salary_lm$residuals)