Multiple linear regression

load("more/evals.RData")

Exploring the data

Is this an observational study or an experiment? The original research question posed in the paper is whether beauty leads directly to the differences in course evaluations. Given the study design, is it possible to answer this question as it is phrased? If not, rephrase the question.

This is an observational study. The question should be rephrased as ’is the beauty of a professor a predictor of score in course evaluations?

Describe the distribution of score. Is the distribution skewed? What does that tell you about how students rate courses? Is this what you expected to see? Why, or why not?

library(tidyverse)
ggplot(evals,aes(x=score))+
  geom_histogram(aes(y=..count..,fill=..count..,alpha=.2))+
  scale_fill_gradient("Count", low="blue", high="red")+
  theme_classic()+
  theme(legend.position="none")

The distribution is left-skewed. Positive evaluations are more frequent than negative evaluations; a normal distribution was expected, wherein fewer positive evaluations were submitted.

Excluding score, select two other variables and describe their relationship using an appropriate visualization (scatterplot, side-by-side boxplots, or mosaic plot).

evals$agecat[evals$age<=40] <-1
evals$agecat[evals$age>=40 & evals$age<=50] <-2
evals$agecat[evals$age>=50 & evals$age<=60] <-3
evals$agecat[evals$age>=60] <-4
evals$agecat<-as.numeric(evals$agecat)

ggplot(evals,aes(y=evals$bty_avg,x=as.factor(evals$agecat)))+
  geom_boxplot(color="black",fill="#80bfff",alpha=0.2)+
  theme_classic()+
  theme(legend.position="none")+
  geom_text(aes(label = ..count.., y= ..prop..), stat= "count", vjust = -.5) +
  scale_x_discrete(labels=c("1"="40 or younger",
                              "2"="41 - 50",
                              "3"="51 - 60",
                              "4"="61 or older"))

There is a decline in median beauty as age increases when evaluations are categorized by age range.

Simple linear regression

plot(evals$score ~ evals$bty_avg)

Before we draw conclusions about the trend, compare the number of observations in the data frame with the approximate number of points on the scatterplot. Is anything awry?

The plotted points are superimposed; it appears that there are fewer plotted points than there should be.

Replot the scatterplot, but this time use the function jitter() on the $y$- or the $x$-coordinate. (Use ?jitter to learn more.) What was misleading about the initial scatterplot?

ggplot(evals, aes(bty_avg, score))+
  geom_point(position = position_jitter(w = 0.1, h = 0.1))+
  theme_classic()+
  ylab("score")+
  xlab("beauty")

Let’s see if the apparent trend in the plot is something more than natural variation. Fit a linear model called m_bty to predict average professor score by average beauty rating and add the line to your plot using abline(m_bty). Write out the equation for the linear model and interpret the slope. Is average beauty score a statistically significant predictor? Does it appear to be a practically significant predictor?

m_bty <- lm(evals$score ~ evals$bty_avg)
ggplot(evals, aes(bty_avg, score))+
  geom_point(position = position_jitter(w = 0.1, h = 0.1))+
  geom_abline(intercept = coef(m_bty)["(Intercept)"],
              slope = coef(m_bty)["evals$bty_avg"])+
  theme_classic()+
  ylab("score")+
  xlab("beauty")

The equation is $\hat{y}$=3.880338 + 0.066637(bty_avg)

summary(m_bty)

## 
## Call:
## lm(formula = evals$score ~ evals$bty_avg)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.9246 -0.3690  0.1420  0.3977  0.9309 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    3.88034    0.07614   50.96  < 2e-16 ***
## evals$bty_avg  0.06664    0.01629    4.09 5.08e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5348 on 461 degrees of freedom
## Multiple R-squared:  0.03502,    Adjusted R-squared:  0.03293 
## F-statistic: 16.73 on 1 and 461 DF,  p-value: 5.083e-05

bty_avg is indeed a predictor of evaluation score (p < 0.001), however the slope of the equation indicates that for every point increase in beauty there is a fractional increase in evaluation score (.06664).

Use residual plots to evaluate whether the conditions of least squares regression are reasonable. Provide plots and comments for each one (see the Simple Regression Lab for a reminder of how to make these).

ggplot(evals, aes(y=m_bty$residuals, x=evals$bty_avg))+ geom_point()+ geom_abline(intercept = 0, slope = 0)+ theme_classic()

ggplot(evals,aes(x=m_bty$residuals))+ geom_histogram(aes(y=..count..,fill=..count..,alpha=.2))+ theme_classic()+ theme(legend.position="none")

ggplot(evals,aes(sample=m_bty$residuals))+ stat_qq()+ stat_qq_line()+ theme_classic()

While the residuals appear to confirm constant variability, the histogram and the normal probability plot both suggest a distribution that deviates from a normal distribution.

Multiple linear regression

m_bty_gen <- lm(score ~ bty_avg + gender, data = evals) summary(m_bty_gen)

## ## Call: ## lm(formula = score ~ bty_avg + gender, data = evals) ## ## Residuals: ## Min 1Q Median 3Q Max ## -1.8305 -0.3625 0.1055 0.4213 0.9314 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 3.74734 0.08466 44.266 < 2e-16 *** ## bty_avg 0.07416 0.01625 4.563 6.48e-06 *** ## gendermale 0.17239 0.05022 3.433 0.000652 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.5287 on 460 degrees of freedom ## Multiple R-squared: 0.05912, Adjusted R-squared: 0.05503 ## F-statistic: 14.45 on 2 and 460 DF, p-value: 8.177e-07

P-values and parameter estimates should only be trusted if the conditions for the regression are reasonable. Verify that the conditions for this model are reasonable using diagnostic plots.

ggplot(evals, aes(y=m_bty_gen$residuals, x=m_bty_gen$fitted.values))+ geom_point()+ geom_abline(intercept = 0, slope = 0)+ theme_classic()

Plot of residuals against fitted values indicates constant variance.

ggplot(evals, aes(y=m_bty_gen$residuals, x=evals$bty_avg))+ geom_point()+ geom_abline(intercept = 0, slope = 0)+ theme_classic()

Variance appears constant.

ggplot(evals,aes(sample=m_bty_gen$residuals))+ stat_qq()+ stat_qq_line()+ theme_classic()

The distribution is approximiately normal. Conditions for the regression are reasonable.

Is bty_avg still a significant predictor of score? Has the addition of gender to the model changed the parameter estimate for bty_avg?

The inclusion of the variable improved the model; bty_avg is still a sgnificant predictor of score.

multiLines(m_bty_gen)

What is the equation of the line corresponding to males? (Hint: For males, the parameter estimate is multiplied by 1.) For two professors who received the same beauty rating, which gender tends to have the higher course evaluation score?

\[ \begin{aligned} \widehat{score} &= \hat{\beta}_0 + \hat{\beta}_1 \times bty\_avg + \hat{\beta}_2 \times (1) \\ &= \hat{\beta}_0 + \hat{\beta}_1 \times bty\_avg + \hat{\beta}_2 \end{aligned} \]

The additional term in the equation implies that males would have the higher course evaluation if beauty rating is held constant.

Create a new model called m_bty_rank with gender removed and rank added in. How does R appear to handle categorical variables that have more than two levels? Note that the rank variable has three levels: teaching, tenure track, tenured.

m_bty_rank <- lm(score ~ bty_avg + rank, data = evals) summary(m_bty_rank)

## ## Call: ## lm(formula = score ~ bty_avg + rank, data = evals) ## ## Residuals: ## Min 1Q Median 3Q Max ## -1.8713 -0.3642 0.1489 0.4103 0.9525 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 3.98155 0.09078 43.860 < 2e-16 *** ## bty_avg 0.06783 0.01655 4.098 4.92e-05 *** ## ranktenure track -0.16070 0.07395 -2.173 0.0303 * ## ranktenured -0.12623 0.06266 -2.014 0.0445 * ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.5328 on 459 degrees of freedom ## Multiple R-squared: 0.04652, Adjusted R-squared: 0.04029 ## F-statistic: 7.465 on 3 and 459 DF, p-value: 6.88e-05

multiLines(m_bty_rank)

R creates two variables, ranktenure track and ranktenured.

The search for the best model

Which variable would you expect to have the highest p-value in this model? Why? Hint: Think about which variable would you expect to not have any association with the professor score.

I would expect cls_profs to have the least association with evaluation score.

m_full <- lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval + cls_students + cls_level + cls_profs + cls_credits + bty_avg + pic_outfit + pic_color, data = evals) summary(m_full)

Check your suspicions from the previous exercise. Include the model output in your response.

cls_profs has the greatest p-value, with cls_profssingle p-value at 0.77806.

ggplot(evals,aes(y=evals$score,x=as.factor(evals$cls_profs)))+ geom_boxplot(color="black",fill="#80bfff",alpha=0.2)+ theme_classic()+ theme(legend.position="none")+ geom_text(aes(label = ..count.., y= ..prop..), stat= "count", vjust = -.5)

Interpret the coefficient associated with the ethnicity variable.

The coefficient indicates that professors not identified as a minority have higher evaluation scores, but the p-value indicates negligible significance.

Drop the variable with the highest p-value and re-fit the model. Did the coefficients and significance of the other explanatory variables change? (One of the things that makes multiple regression interesting is that coefficient estimates depend on the other variables that are included in the model.) If not, what does this say about whether or not the dropped variable was collinear with the other explanatory variables?

m_full2 <- lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval + cls_students + cls_level + cls_credits + bty_avg + pic_outfit + pic_color, data = evals) summary(m_full2)

## ## Call: ## lm(formula = score ~ rank + ethnicity + gender + language + age + ## cls_perc_eval + cls_students + cls_level + cls_credits + ## bty_avg + pic_outfit + pic_color, data = evals) ## ## Residuals: ## Min 1Q Median 3Q Max ## -1.7836 -0.3257 0.0859 0.3513 0.9551 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 4.0872523 0.2888562 14.150 < 2e-16 *** ## ranktenure track -0.1476746 0.0819824 -1.801 0.072327 . ## ranktenured -0.0973829 0.0662614 -1.470 0.142349 ## ethnicitynot minority 0.1274458 0.0772887 1.649 0.099856 . ## gendermale 0.2101231 0.0516873 4.065 5.66e-05 *** ## languagenon-english -0.2282894 0.1111305 -2.054 0.040530 * ## age -0.0089992 0.0031326 -2.873 0.004262 ** ## cls_perc_eval 0.0052888 0.0015317 3.453 0.000607 *** ## cls_students 0.0004687 0.0003737 1.254 0.210384 ## cls_levelupper 0.0606374 0.0575010 1.055 0.292200 ## cls_creditsone credit 0.5061196 0.1149163 4.404 1.33e-05 *** ## bty_avg 0.0398629 0.0174780 2.281 0.023032 * ## pic_outfitnot formal -0.1083227 0.0721711 -1.501 0.134080 ## pic_colorcolor -0.2190527 0.0711469 -3.079 0.002205 ** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.4974 on 449 degrees of freedom ## Multiple R-squared: 0.187, Adjusted R-squared: 0.1634 ## F-statistic: 7.943 on 13 and 449 DF, p-value: 2.336e-14

The coefficients and significance changed; each variable is more significantly associated.

Using backward-selection and p-value as the selection criterion, determine the best model. You do not need to show all steps in your answer, just the output for the final model. Also, write out the linear model for predicting score based on the final model you settle on.

m_full_final <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval + cls_credits + bty_avg + pic_color, data = evals) summary(m_full_final)

## ## Call: ## lm(formula = score ~ ethnicity + gender + language + age + cls_perc_eval + ## cls_credits + bty_avg + pic_color, data = evals) ## ## Residuals: ## Min 1Q Median 3Q Max ## -1.85320 -0.32394 0.09984 0.37930 0.93610 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 3.771922 0.232053 16.255 < 2e-16 *** ## ethnicitynot minority 0.167872 0.075275 2.230 0.02623 * ## gendermale 0.207112 0.050135 4.131 4.30e-05 *** ## languagenon-english -0.206178 0.103639 -1.989 0.04726 * ## age -0.006046 0.002612 -2.315 0.02108 * ## cls_perc_eval 0.004656 0.001435 3.244 0.00127 ** ## cls_creditsone credit 0.505306 0.104119 4.853 1.67e-06 *** ## bty_avg 0.051069 0.016934 3.016 0.00271 ** ## pic_colorcolor -0.190579 0.067351 -2.830 0.00487 ** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 0.4992 on 454 degrees of freedom ## Multiple R-squared: 0.1722, Adjusted R-squared: 0.1576 ## F-statistic: 11.8 on 8 and 454 DF, p-value: 2.58e-15

Equation: $\hat{score}$ = 3.7719215 + 0.1678723(ethnicitynot minority) + 0.2071121(gendermale) + -0.2061781(languagenon-english) + -0.0060459(age) + 0.0046559(cls_perc_eval) + 0.5053062(cls_creditsone credit) + 0.0510693(bty_avg) + -0.1905788(pic_colorcolor)

Verify that the conditions for this model are reasonable using diagnostic plots.

ggplot(evals,aes(sample=m_full_final$residuals))+ stat_qq()+ stat_qq_line()+ theme_classic()

ggplot(evals, aes(y=m_full_final$residuals, x=m_full_final$fitted.values))+ geom_point()+ theme_classic()

ggplot(evals,aes(y=evals$score,x=as.factor(evals$ethnicity)))+ geom_boxplot(color="black",fill="#80bfff",alpha=0.2)+ theme_classic()+ theme(legend.position="none")+ geom_text(aes(label = ..count.., y= ..prop..), stat= "count", vjust = -.5)

ggplot(evals,aes(y=evals$score,x=as.factor(evals$gender)))+ geom_boxplot(color="black",fill="#80baaf",alpha=0.2)+ theme_classic()+ theme(legend.position="none")+ geom_text(aes(label = ..count.., y= ..prop..), stat= "count", vjust = -.5)

ggplot(evals,aes(y=evals$score,x=as.factor(evals$language)))+ geom_boxplot(color="black",fill="#60eaaf",alpha=0.2)+ theme_classic()+ theme(legend.position="none")+ geom_text(aes(label = ..count.., y= ..prop..), stat= "count", vjust = -.5)

ggplot(evals,aes(y=evals$score,x=evals$age))+ geom_point(shape=19,color="blue",fill="#efccff")+ theme_classic()+ theme(legend.position="none")+ scale_x_continuous(name="cls_perc_eval", breaks=c(0,20,40,60,80,100))

ggplot(evals,aes(y=evals$score,x=evals$cls_perc_eval))+ geom_point(shape=19,color="red",fill="#efccff")+ theme_classic()+ theme(legend.position="none")

ggplot(evals,aes(y=evals$score,x=as.factor(evals$cls_credits)))+ geom_boxplot(color="black",fill="#60eaaf",alpha=0.2)+ theme_classic()+ theme(legend.position="none")+ geom_text(aes(label = ..count.., y= ..prop..), stat= "count", vjust = -.5)

ggplot(evals,aes(y=evals$score,x=evals$bty_avg))+ geom_point(shape=19,color="gray",fill="#efccff")+ theme_classic()+ theme(legend.position="none")

ggplot(evals,aes(y=evals$score,x=as.factor(evals$pic_color)))+ geom_boxplot(color="black",fill="#60ffff",alpha=0.2)+ theme_classic()+ theme(legend.position="none")+ geom_text(aes(label = ..count.., y= ..prop..), stat= "count", vjust = -.5)

The original paper describes how these data were gathered by taking a sample of professors from the University of Texas at Austin and including all courses that they have taught. Considering that each row represents a course, could this new information have an impact on any of the conditions of linear regression?

This information should not change any of the conditions of linear regression.

Based on your final model, describe the characteristics of a professor and course at University of Texas at Austin that would be associated with a high evaluation score.

The highest scores are associated with younger, attractive (high rank in beauty score) males who do not identify as a minority and teach a course for one credit.

Would you be comfortable generalizing your conclusions to apply to professors generally (at any university)? Why or why not?

These conclusions would not apply to all universities and professors. University demographics would likely greatly interfere with our model.