Week 12 Discussion Board
Gabriel Santos
2022-11-08
Using R, build a multiple regression model for data that interests you. Include in this model at least one quadratic term, one dichotomous term, and one dichotomous vs. quantitative interaction term. Interpret all coefficients. Conduct residual analysis. Was the linear model appropriate? Why or why not?
Data Set: the salary data used in Weisberg’s book, consisting of 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
library(foreign)
salary <- read.csv("https://raw.githubusercontent.com/GabrielSantos33/DATA605_HW12/main/salary.csv")dim(salary)## [1] 52 6summary(salary)## sx rk yr dg
## Min. :0.0000 Min. :1.000 Min. : 0.000 Min. :0.0000
## 1st Qu.:0.0000 1st Qu.:1.000 1st Qu.: 3.000 1st Qu.:0.0000
## Median :0.0000 Median :2.000 Median : 7.000 Median :1.0000
## Mean :0.2692 Mean :2.038 Mean : 7.481 Mean :0.6538
## 3rd Qu.:1.0000 3rd Qu.:3.000 3rd Qu.:11.000 3rd Qu.:1.0000
## Max. :1.0000 Max. :3.000 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. :38045head(salary)## sx rk yr dg yd sl
## 1 0 3 25 1 35 36350
## 2 0 3 13 1 22 35350
## 3 0 3 10 1 23 28200
## 4 1 3 7 1 27 26775
## 5 0 3 19 0 30 33696
## 6 0 3 16 1 21 28516Convert 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)Linear Model:
The purpose of this analysis is building that predicts professor’s salary.
Dichotomous variable - Sex (sx).
Quadratic variable - square of rank (rk2).
# 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-16Regressive Removal:
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-16salary_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-16salary_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-16salary_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-16salary_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-16plot(salary_lm$fitted.values, salary_lm$residuals, xlab="Fitted Values", ylab="Residuals", main="Residuals vs. Fitted", col = "blue")
abline(h=0)
qqnorm(salary_lm$residuals, col = "blue")
qqline(salary_lm$residuals, col = "darkblue")
Conclusion:
Answer: According to the data and graphs, the variability is constant, although there are a few outliers.The distribution of the residuals is almost normal. It is correct to apply the linear model.
The equation for Number of years since highest degree was earned is:
\[ Salary\_lm = 11336.67 + 376.50 \times yd\ (Number\ of\ years\ since\ highest\ degree\ was\ earned) \]
Based on the adjusted R-squared value, the model explains that of variability in the data is 83.73%.