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.

Since this is an observational study, it is not recommended to make causal conclusions. A safer way to phrase this question - Is there an association between beauty and course evaluation scores.

  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)

The distribution of ‘score’ is left skewed. It means that there are fewer student giving low scores. Instead, most of them are generous and give higher evaluation scores.

  1. Excluding score, select two other variables and describe their relationship using an appropriate visualization (scatterplot, side-by-side boxplots, or mosaic plot).
scatter.smooth(evals$age, evals$bty_avg)

The variables of my choosing are age and bty_avg. The data show the average beauty score decreases with professors age.

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?

  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?
plot(evals$score ~ jitter(evals$bty_avg))

Fewer points appear on the initial plot because there are multiple cases with the same ratings which causes them to overlay.

  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?
m_bty <- lm(score ~ bty_avg, evals)
plot(score ~ bty_avg, evals)
abline(m_bty)

summary(m_bty)
## 
## Call:
## lm(formula = score ~ bty_avg, data = evals)
## 
## 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 ***
## 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 linear formula is y = 3.88034 + 0.06664 * bty_avg

The p-value, 0.00005083, is smaller than 0.05 which means we can reject the null hypothesis (no impact on evaluation)

  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).
plot(residuals(m_bty))
abline(h = 0, lty = 5)

hist(residuals(m_bty))

Linearity - TRUE

Nearly normal residuals - TRUE (slight left skeweness)

Constant variability - TRUE

Independent observations - TRUE

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

Residuals vs Fitted confirms linearity and homoscedasticity as the line does not significantly diverge from the 0. The normality is represented with the QQ plot. Even though the upper portion curves away from the line, the sample size is large enough to assume this as a nearly normal distribution. Scale-location shows any patterns that would confirm or reject the homoscedasticity assumption. The last plot shows cooks distance - there are no points greater than 1.

  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?

The pvalue of 0.0000008177 is < 0.05 which allows us to reject the H0 (beauty has no impact) which means it is a statisically significant predictor. The gender variable has slightly increased the parameter estimate for bty_agv.

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?

From the summary function: 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 ***

ev_score <- 3.74734 + 0.07416 * bty_avg + 0.17239

When the bty_avg is the same for two professors of different genders, the male professor will have around 0.17 higher score.

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
multiLines(m_bty_rank)

There are two new variables added - ‘ranktenure track’ and ‘ranktenured.’ The function goes through all variables and changes all to 0 except the one that is currently being calculated.

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.

My highest p-value assertion goes to variable cls_students.

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.
cls_students <- lm(score ~ cls_students, data = evals)
summary(cls_students)
## 
## Call:
## lm(formula = score ~ cls_students, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.8666 -0.3677  0.1281  0.4300  0.8336 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  4.1643491  0.0314034 132.608   <2e-16 ***
## cls_students 0.0001881  0.0003373   0.558    0.577    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5443 on 461 degrees of freedom
## Multiple R-squared:  0.0006744,  Adjusted R-squared:  -0.001493 
## F-statistic: 0.3111 on 1 and 461 DF,  p-value: 0.5773

It turns out that age does not have a high p-value.

  1. Interpret the coefficient associated with the ethnicity variable.
ethnicity <- lm(score ~ ethnicity, data = evals)
summary(ethnicity)
## 
## Call:
## lm(formula = score ~ ethnicity, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.8912 -0.3816  0.1088  0.4088  0.9281 
## 
## Coefficients:
##                       Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.07188    0.06786  60.003   <2e-16 ***
## ethnicitynot minority  0.11935    0.07310   1.633    0.103    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5429 on 461 degrees of freedom
## Multiple R-squared:  0.005749,   Adjusted R-squared:  0.003593 
## F-statistic: 2.666 on 1 and 461 DF,  p-value: 0.1032

Using the ethnicity variable we can determine that professors that are not minority are evaluated more generously.

  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?
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(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

After removing cl_profs the p-value did not much. It means the coefficient estimates did not depend on the other variables that were included.

  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.

Dropping all variables with p-values and leaving only the ones with values that are less than 0.1

best_model <- lm(score ~ rank + gender + age + cls_perc_eval 
             + cls_level  + cls_credits + bty_avg 
             + pic_color, data = evals)
summary(best_model)
## 
## Call:
## lm(formula = score ~ rank + gender + age + cls_perc_eval + cls_level + 
##     cls_credits + bty_avg + pic_color, data = evals)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.73093 -0.33454  0.08301  0.38724  0.93323 
## 
## Coefficients:
##                        Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.204554   0.243666  17.255  < 2e-16 ***
## ranktenure track      -0.193269   0.078838  -2.451  0.01460 *  
## ranktenured           -0.078612   0.064140  -1.226  0.22097    
## gendermale             0.223373   0.051509   4.337 1.78e-05 ***
## age                   -0.009288   0.003088  -3.008  0.00278 ** 
## cls_perc_eval          0.003981   0.001450   2.746  0.00627 ** 
## cls_levelupper         0.040931   0.054250   0.754  0.45095    
## cls_creditsone credit  0.444285   0.112858   3.937 9.56e-05 ***
## bty_avg                0.047702   0.017135   2.784  0.00560 ** 
## pic_colorcolor        -0.218653   0.068364  -3.198  0.00148 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5035 on 453 degrees of freedom
## Multiple R-squared:  0.1597, Adjusted R-squared:  0.143 
## F-statistic: 9.564 on 9 and 453 DF,  p-value: 2.124e-13
best_model <- lm(score ~ rank, data = evals)
summary(best_model)
## 
## Call:
## lm(formula = score ~ rank, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.8546 -0.3391  0.1157  0.4305  0.8609 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       4.28431    0.05365  79.853   <2e-16 ***
## ranktenure track -0.12968    0.07482  -1.733   0.0837 .  
## ranktenured      -0.14518    0.06355  -2.284   0.0228 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5419 on 460 degrees of freedom
## Multiple R-squared:  0.01163,    Adjusted R-squared:  0.007332 
## F-statistic: 2.706 on 2 and 460 DF,  p-value: 0.06786
  1. Verify that the conditions for this model are reasonable using diagnostic plots.
par(mfrow=c(2,2))
plot(best_model)

All conditions seem to be satisfied.

  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?

It would be impactful to the results because the same profesors could be counted multiple times breaking the independency of the samples.

  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.

The highest evaluated professor are young and attractive males.

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

I would be comfortable generalizing this conclusion because it simply is highly believable. However, due to the observational nature of this study I would recommed to conduct an experiment in order to confirm and publicize the findings.

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.