Grading the professor

Many college courses conclude by giving students the opportunity to evaluate the course and the instructor anonymously. However, the use of these student evaluations as an indicator of course quality and teaching effectiveness is often criticized because these measures may reflect the influence of non-teaching related characteristics, such as the physical appearance of the instructor. The article titled, “Beauty in the classroom: instructors’ pulchritude and putative pedagogical productivity” (Hamermesh and Parker, 2005) found that instructors who are viewed to be better looking receive higher instructional ratings. (Daniel S. Hamermesh, Amy Parker, Beauty in the classroom: instructors pulchritude and putative pedagogical productivity, Economics of Education Review, Volume 24, Issue 4, August 2005, Pages 369-376, ISSN 0272-7757, 10.1016/j.econedurev.2004.07.013. http://www.sciencedirect.com/science/article/pii/S0272775704001165.)

In this lab we will analyze the data from this study in order to learn what goes into a positive professor evaluation.

The data

The data were gathered from end of semester student evaluations for a large sample of professors from the University of Texas at Austin. In addition, six students rated the professors’ physical appearance. (This is aslightly modified version of the original data set that was released as part of the replication data for Data Analysis Using Regression and Multilevel/Hierarchical Models (Gelman and Hill, 2007).) The result is a data frame where each row contains a different course and columns represent variables about the courses and professors.

load("more/evals.RData")
variable description
score average professor evaluation score: (1) very unsatisfactory - (5) excellent.
rank rank of professor: teaching, tenure track, tenured.
ethnicity ethnicity of professor: not minority, minority.
gender gender of professor: female, male.
language language of school where professor received education: english or non-english.
age age of professor.
cls_perc_eval percent of students in class who completed evaluation.
cls_did_eval number of students in class who completed evaluation.
cls_students total number of students in class.
cls_level class level: lower, upper.
cls_profs number of professors teaching sections in course in sample: single, multiple.
cls_credits number of credits of class: one credit (lab, PE, etc.), multi credit.
bty_f1lower beauty rating of professor from lower level female: (1) lowest - (10) highest.
bty_f1upper beauty rating of professor from upper level female: (1) lowest - (10) highest.
bty_f2upper beauty rating of professor from second upper level female: (1) lowest - (10) highest.
bty_m1lower beauty rating of professor from lower level male: (1) lowest - (10) highest.
bty_m1upper beauty rating of professor from upper level male: (1) lowest - (10) highest.
bty_m2upper beauty rating of professor from second upper level male: (1) lowest - (10) highest.
bty_avg average beauty rating of professor.
pic_outfit outfit of professor in picture: not formal, formal.
pic_color color of professor’s picture: color, black & white.

Exploring the data

  1. 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 study is an observational study. This was a collection of surveys collected at the end of the semester, and the data was used to answer the question. It is not an experiment, and therefore, causation cannot be technically be determined from correlation. The better question would be: Does assigning beautiful/good-looking professors create differences in course evaluations? i.e. Will need to randomly assign students to professors who are unappealing vs. professors who are beautiful.

  1. 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?
hist(evals$score, main = "Histogram of Scores", xlab = "Scores", col = 'lightgreen', probability = TRUE)
x <- seq(from = 0, to = 5, by = 0.1)
y <- dnorm(x = x, mean = mean(evals$score), sd = sd(evals$score))
lines(x = x, y = y, col = "blue")

summary(evals$score)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   2.300   3.800   4.300   4.175   4.600   5.000

** This is skewed to the left. This suggests that there were multiple evaluations with much lower scores thus skewing the distribution to the left. Most students had rated the average at 4.175 (though the median is 4.3). However, there are very low scores (outliers, and likely from disgruntled students) that has caused this shift. I expect to see this as most students tend to rate their professor well, given that many professors tend to be staying and doing their job when they’re doing well. Of course, there are some students, whether because of their poor study habits, etc., may be upset from a low grade or perhaps genuinely feel that the professor was doing a good job would expectedly give this professor a poor score.**

  1. Excluding score, select two other variables and describe their relationship using an appropriate visualization (scatterplot, side-by-side boxplots, or mosaic plot).
par(mfrow=c(3,1))
hist(evals$age, breaks = 20, main = "Age of the Student vs. Their Scores", xlab = "Age", col = "blue")
counts <- table(evals$language)
barplot(counts, main = "Languages", xlab = "English vs. Non-English", col = "pink")
# One more for extra luck (and practice)
plot(x = evals$bty_avg, y = evals$score, main = "Average Beauty vs. Score")

Simple linear regression

The fundamental phenomenon suggested by the study is that better looking teachers are evaluated more favorably. Let’s create a scatterplot to see if this appears to be the case:

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?

paste("Number of observations in data frame: ", length(evals$age))
## [1] "Number of observations in data frame:  463"

It appears that there are multiple points overlying on top of each other, which is why it looks less than the 463 that I had counted in R.

  1. 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?
x <- jitter(evals$bty_avg)
plot(evals$score ~ x)

There are a lot more points on this graph that the previous graph (which is the result of the jitter() function). With the jiter, the plots that were overlying on top of each other are now shifted slightly to allow better visualization of more points.

  1. 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?
plot(evals$score ~ evals$bty_avg)
m_bty <- lm(evals$score ~ evals$bty_avg)
abline(m_bty)

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

The equation is Score = 3.88034 + 0.6664 * B1 where B1 = Beauty Average Rating. And given that p value is way less than 0.05, we reject the null hypothesis and suggest that this is statistically significant.

  1. 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).
# https://www.r-bloggers.com/visualising-residuals/
library(psych)
par(mfrow = c(2,2))
plot(m_bty)

plot(m_bty$residuals ~evals$bty_avg)
abline(h = 0, lty = 3)
hist(m_bty$residuals)

The conditions for linear model to be assessed are: 1. Linearity, which according to the above plots, this model does. 2. Nearly normal residuals: which again, it appear it does. As of note, in the histogram, there appears to be a left skew. However, with a large sample size, we most likely will be able to discard the left skew. 3. Constant variability: there are some small outliers at the bottom of the plot, but for the most part, it appears fairly constant (this is questionable whether or not this is okay for the linear regression model.)

Multiple linear regression

The data set contains several variables on the beauty score of the professor: individual ratings from each of the six students who were asked to score the physical appearance of the professors and the average of these six scores. Let’s take a look at the relationship between one of these scores and the average beauty score.

plot(evals$bty_avg ~ evals$bty_f1lower)
cor(evals$bty_avg, evals$bty_f1lower)

As expected the relationship is quite strong - after all, the average score is calculated using the individual scores. We can actually take a look at the relationships between all beauty variables (columns 13 through 19) using the following command:

plot(evals[,13:19])

These variables are collinear (correlated), and adding more than one of these variables to the model would not add much value to the model. In this application and with these highly-correlated predictors, it is reasonable to use the average beauty score as the single representative of these variables.

In order to see if beauty is still a significant predictor of professor score after we’ve accounted for the gender of the professor, we can add the gender term into the model.

m_bty_gen <- lm(score ~ bty_avg + gender, data = evals)
summary(m_bty_gen)
  1. 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.
#par(mfrow=c(2,2))
#plot(m_bty_gen)

#hist(m_bty_gen$residuals)

# For some unusual reason, this chunk does not appear to knit, despite the fact that it runs without any hitches within R Studio. I had instead taken the liberty to post the images of the results in lieu.

Again, let’s go through the conditions. Above, in the multiple evals plots, there appears to be some form of linearity (as well as noted in the summary section as well.) The residuals appear to be left skew but again, with the large sample size, this could likely be disregarded. The residuals appear to be independent. The constant variability is debatable. While there is no clear cut pattern, most of the data appears to be in the negative residuals with what appears to be some negative outliers. It is debatable if this portion of the condition is met.

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

Yes, bty_avg is still a significant predictor of score as noted with the p values < 0.05. The addition of gender to the model changed the parameter estimate from 0.06664 to 0.07416.

Note that the estimate for gender is now called gendermale. You’ll see this name change whenever you introduce a categorical variable. The reason is that R recodes gender from having the values of female and male to being an indicator variable called gendermale that takes a value of \(0\) for females and a value of \(1\) for males. (Such variables are often referred to as “dummy” variables.)

As a result, for females, the parameter estimate is multiplied by zero, leaving the intercept and slope form familiar from simple regression.

\[ \begin{aligned} \widehat{score} &= \hat{\beta}_0 + \hat{\beta}_1 \times bty\_avg + \hat{\beta}_2 \times (0) \\ &= \hat{\beta}_0 + \hat{\beta}_1 \times bty\_avg\end{aligned} \]

We can plot this line and the line corresponding to males with the following custom function.

multiLines(m_bty_gen)
  1. 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?

For males: Male.Score = 3.74734 + 0.07416 * (bty_avg) + 0.17239 * (gendermale = 1) –> Male.Score = 3.74734 + 0.07416 * (bty_avg) + 0.17239 * 1. For females: Female.Score = 3.74734 + 0.07416 * (bty_avg) + 0.17239 * (gendermale = 0) –> Female.Score = 3.74734 + 0.07416 * (bty_avg). Because of the formula where gendermale = 1, the male gender tends to have higher course evaluation scores.

The decision to call the indicator variable gendermale instead ofgenderfemale has no deeper meaning. R simply codes the category that comes first alphabetically as a \(0\). (You can change the reference level of a categorical variable, which is the level that is coded as a 0, using therelevel function. Use ?relevel to learn more.)

  1. 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

It appears that in R, they would assign a value of 1 or 0 for each categorical label. So if the professor was on ranktenure track, he would be assigned a 1 but a 0 for ranktenured. Likewise, if the profesor was ranktenured, he would be assigned a 1 but a 0 for ranktenure track. It it noted that one of the categorical labels is dropped. In this case, teaching has been dropped. However, it appears that if the professor is on the teaching track, he is technically NOT on ranktenured or ranktenure track, thus effectively making the teaching track 0 and 0 respectively.

The interpretation of the coefficients in multiple regression is slightly different from that of simple regression. The estimate for bty_avg reflects how much higher a group of professors is expected to score if they have a beauty rating that is one point higher while holding all other variables constant. In this case, that translates into considering only professors of the same rank with bty_avg scores that are one point apart.

The search for the best model

We will start with a full model that predicts professor score based on rank, ethnicity, gender, language of the university where they got their degree, age, proportion of students that filled out evaluations, class size, course level, number of professors, number of credits, average beauty rating, outfit, and picture color.

  1. 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 expect the number of credits to have the highest value as the professors rating should intuitively make any difference whether the class was 1 credit or 3 credits.

Let’s run the model…

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

It appears that the highest p value in this full model corresponds to cls_profssingle with a p value of 0.77806. Intersetingly ‘cls_creditsone credit’ is associated with a p value of 1.84e-05.

  1. Interpret the coefficient associated with the ethnicity variable.

When holding all other variables constant, if the professor is a not minority, then he is given an additional .1235 points (despite the fact that this coefficient is not statistically significant).

  1. 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_full.no_cls_profs <- 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_full.no_cls_profs)
## 
## 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 of other explanatory variables very slightly but not significantly (the p values did not change significant). However, the adjusted R-squared value increased slightly from 0.1617 to 0.1634 when the cls_profs was dropped.

  1. 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.
# Looking at the previous question, we can see that the next highest p value is: cls_level.
m_backwards <- lm(formula = score ~ rank + ethnicity + gender + language + age + 
    cls_perc_eval + cls_students + cls_credits + 
    bty_avg + pic_outfit + pic_color, data = evals)
summary(m_backwards)
## 
## Call:
## lm(formula = score ~ rank + ethnicity + gender + language + age + 
##     cls_perc_eval + cls_students + cls_credits + bty_avg + pic_outfit + 
##     pic_color, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.7761 -0.3187  0.0875  0.3547  0.9367 
## 
## Coefficients:
##                         Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.0856255  0.2888881  14.143  < 2e-16 ***
## ranktenure track      -0.1420696  0.0818201  -1.736 0.083184 .  
## ranktenured           -0.0895940  0.0658566  -1.360 0.174372    
## ethnicitynot minority  0.1424342  0.0759800   1.875 0.061491 .  
## gendermale             0.2037722  0.0513416   3.969 8.40e-05 ***
## languagenon-english   -0.2093185  0.1096785  -1.908 0.056966 .  
## age                   -0.0087287  0.0031224  -2.795 0.005404 ** 
## cls_perc_eval          0.0053545  0.0015306   3.498 0.000515 ***
## cls_students           0.0003573  0.0003585   0.997 0.319451    
## cls_creditsone credit  0.4733728  0.1106549   4.278 2.31e-05 ***
## bty_avg                0.0410340  0.0174449   2.352 0.019092 *  
## pic_outfitnot formal  -0.1172152  0.0716857  -1.635 0.102722    
## pic_colorcolor        -0.1973196  0.0681052  -2.897 0.003948 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4975 on 450 degrees of freedom
## Multiple R-squared:  0.185,  Adjusted R-squared:  0.1632 
## F-statistic:  8.51 on 12 and 450 DF,  p-value: 1.275e-14

Noted is an adjusted R-squared of 0.1632, which is worse than the previous model. Therefore, we will stop from performing any further removal of coefficients. The final multiple regression formula is: score = 4.0872523 - 0.1476746 * ranktenuretrack - 0.0973829 * ranktenured + 0.1274458 * ethnicitynot minority + 0.2101231 * gendermale - 0.2282894 * languagenon-english - 0.0089992 * age + 0.0052888 * cls_perceval + 4.6871617 × 10−4 * cls_students + 0.0606374 * cls_slevelupper + 0.5061196 * cls_creditsone + 0.0398629 * bty_avg - 0.1083227 * pic_outfitnotformal + -0.2190527 * pic_colorcolor.

  1. Verify that the conditions for this model are reasonable using diagnostic plots.
summary(m_full.no_cls_profs)
## 
## 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
hist(m_full.no_cls_profs$residuals, main = "Residuals", xlab = "Residuals", col = "lightgreen", prob = TRUE)
x1 <- seq(-2,1, by=.1)
y1 <- dnorm(x = x1, mean = mean(m_full.no_cls_profs$residuals), sd = sd(m_full.no_cls_profs$residuals))
lines(x = x1, y = y1, col = "blue")

summary(m_full.no_cls_profs$residuals)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## -1.7840 -0.3257  0.0859  0.0000  0.3513  0.9551
plot(m_full.no_cls_profs$residuals ~ m_full.no_cls_profs$fitted.values)
abline(h = 0, lty = 3)

qqnorm(m_full.no_cls_profs$residuals)
qqline(m_full.no_cls_profs$residuals)

par(mfrow=c(2,2))
plot(m_full.no_cls_profs)

plot(evals)

The data appear to have a normal distribution (slight left skew, though can be ignored with a large sample size), however, there are some concerns for the constant variability and possibility with linearity.

  1. 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?

Any new data could certainly make an impact on the linear regression. New data can change whether or not there is a (degree of) positive or negative impact.

  1. 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.

Looking at the m_full.no_cls_profs summary, the professor at the University of Texas at Austin is on the teaching track, white, male, english speaking, and beautiful.

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

No because this is observational data. Observational data cannot be used to dictate causation. And in addition, this sample size was taken from one university, which is by no means sufficient as generalizable.

This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was written by Mine Çetinkaya-Rundel and Andrew Bray.