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.
In this lab we will analyze the data from this study in order to learn what goes into a positive professor evaluation.
In this lab we will explore the data using the dplyr
package and visualize it using the ggplot2
package for data visualization. The data can be found in the companion package for this course, statsr
.
Let’s load the packages.
library(statsr)
library(dplyr)
library(ggplot2)
library(GGally)
library(plotly)
This is the first time we’re using the GGally
package. We will be using the ggpairs
function from this package later in the lab.
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 a slightly 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.
Let’s load the data:
data(evals)
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. |
score
?
# type your code for Question 3 here, and Knit
<- plot_ly(data=evals, x=~score, type="histogram")
fig fig
summary(evals$score)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 2.300 3.800 4.300 4.175 4.600 5.000
Exercise: Excluding score
, select two other variables and describe their relationship using an appropriate visualization (scatterplot, side-by-side boxplots, or mosaic plot).
# type your code for the Exercise here, and Knit
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:
ggplot(data = evals, aes(x = bty_avg, y = score)) +
geom_point()
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?
Exercise: Replot the scatterplot, but this time replace the geom_point()
layer with a geom_jitter()
layer. (Use ?geom_jitter
to learn more.) What was misleading about the initial scatterplot?
# type your code for the Exercise here, and Knit
ggplot(data = evals, aes(x = bty_avg, y = score)) +
geom_jitter()
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 the following. If you do not remember how to do this, refer to the previous lab.
<- lm(score ~ bty_avg,data = evals) m_bty
ggplot(data = evals, aes(x = bty_avg, y = score)) +
geom_jitter() +
geom_smooth(method = "lm")
## `geom_smooth()` using formula 'y ~ x'
The blue line is the model. The shaded gray area around the line tells us about the variability we might expect in our predictions. To turn that off, use se = FALSE
.
ggplot(data = evals, aes(x = bty_avg, y = score)) +
geom_jitter() +
geom_smooth(method = "lm", se = FALSE)
## `geom_smooth()` using formula 'y ~ x'
Exercise: Print a summary of the linear model, write out the equation, and interpret the slope.
# type your code for the Exercise here, and Knit
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
# type your code for the Question 5 here, and Knit
par(mfrow = c(2, 2)) # Split the plotting panel into a 2 x 2 grid
plot(m_bty)
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.
ggplot(data = evals, aes(x = bty_f1lower, y = bty_avg)) +
geom_jitter()
%>%
evals summarise(cor(bty_avg, bty_f1lower))
## # A tibble: 1 x 1
## `cor(bty_avg, bty_f1lower)`
## <dbl>
## 1 0.844
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:
ggpairs(evals, columns = 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.
<- lm(score ~ bty_avg + gender, data = evals)
m_bty_gen 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
# type your code for the Exercise here, and Knit
Exercise: Print a summary of the multiple linear regression model. Is bty_avg
still a significant predictor of score
? Has the addition of gender
to the model changed the parameter estimate for bty_avg
?
# type your code for the Exercise here, and Knit
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
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.
\[ \begin{aligned} \widehat{score} &= \hat{\beta}_0 + \hat{\beta}_1 \times bty\_avg + \hat{\beta}_2 \times (0) \\ &= \hat{\beta}_0 + \hat{\beta}_1 \times bty\_avg\end{aligned} \]
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.)
Exercise: 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.
# type your code for the Exercise here, and Knit
<- lm(score ~ bty_avg + rank, data = evals)
m_bty_rank 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
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.
Suppose we want to use the model we created earlier, m_bty_gen
to predict the evaluation score for a professor, Dr. Hypo Thetical, who is a male tenure track professor with an average beauty of 3.
If we wanted to do this by hand, we would simply plug in these values into the linear model.
We can also calculate the predicted value in R.
First, we need to create a new data frame for this professor.
<- data.frame(gender = "male", bty_avg = 3) newprof
Note that I didn’t need to add rank = "tenure track"
to this data frame since this variable is not used in our model.
Then, I can do the prediction using the predict
function:
predict(m_bty_gen, newprof)
## 1
## 4.142194
We can also construct a prediction interval around this prediction, which will provide a measure of uncertainty around the prediction.
predict(m_bty_gen, newprof, interval = "prediction", level = 0.95)
## fit lwr upr
## 1 4.142194 3.100559 5.183829
Hence, the model predicts, with 95% confidence, that a male professor with an average beauty score of 3 is expected to have an evaluation score between 3.1 and 5.18.
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.
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.
Let’s run the model…
<- lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval
m_full + cls_students + cls_level + cls_profs + cls_credits + bty_avg
+ pic_outfit + pic_color, data = evals)
summary(m_full)
##
## Call:
## lm(formula = 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)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.77397 -0.32432 0.09067 0.35183 0.95036
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.0952141 0.2905277 14.096 < 2e-16 ***
## ranktenure track -0.1475932 0.0820671 -1.798 0.07278 .
## ranktenured -0.0973378 0.0663296 -1.467 0.14295
## ethnicitynot minority 0.1234929 0.0786273 1.571 0.11698
## gendermale 0.2109481 0.0518230 4.071 5.54e-05 ***
## languagenon-english -0.2298112 0.1113754 -2.063 0.03965 *
## age -0.0090072 0.0031359 -2.872 0.00427 **
## cls_perc_eval 0.0053272 0.0015393 3.461 0.00059 ***
## cls_students 0.0004546 0.0003774 1.205 0.22896
## cls_levelupper 0.0605140 0.0575617 1.051 0.29369
## cls_profssingle -0.0146619 0.0519885 -0.282 0.77806
## cls_creditsone credit 0.5020432 0.1159388 4.330 1.84e-05 ***
## bty_avg 0.0400333 0.0175064 2.287 0.02267 *
## pic_outfitnot formal -0.1126817 0.0738800 -1.525 0.12792
## pic_colorcolor -0.2172630 0.0715021 -3.039 0.00252 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.498 on 448 degrees of freedom
## Multiple R-squared: 0.1871, Adjusted R-squared: 0.1617
## F-statistic: 7.366 on 14 and 448 DF, p-value: 6.552e-14
Exercise: Check your suspicions from the previous exercise. Include the model output in your response.
Exercise: 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?
# type your code for the Exercise here, and Knit
<- lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval
m_full_no_cls_profs + 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
Now we try a different model selection method: adjusted \(R^2\). Create a new model, m1
, where you remove rank
from the list of explanatory variables. Check out the adjusted \(R^2\) of this new model and compare it to the adjusted \(R^2\) of the full model.
<- lm(score ~ ethnicity + gender + language + age + cls_perc_eval
m1 + cls_students + cls_level + cls_profs + cls_credits + bty_avg, data = evals)
summary(m1)$adj.r.squared
Then, try dropping the next variable from the full model (ethnicity
):
= lm(score ~ rank + gender + language + age + cls_perc_eval +
m2 + cls_level + cls_profs + cls_credits + bty_avg, data = evals)
cls_students summary(m2)$adj.r.squared
Exercise: Repeat this process until you have tried removing each variable from the full model at a time, and determine removal of which variable yields the highest improvement in the adjusted \(R^2\).
# type your code for the Exercise here, and Knit
= lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval +
m3 + cls_level + cls_credits + bty_avg, data = evals)
cls_students summary(m3)$adj.r.squared
## [1] 0.1430683
# type your code for the Exercise here, and Knit
= lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval +
m4 + cls_profs + cls_credits + bty_avg, data = evals)
cls_level summary(m4)$adj.r.squared
## [1] 0.1401804
= lm(score ~ ethnicity + gender + language + age + cls_perc_eval +
m5 + cls_students + cls_level + cls_profs + cls_credits + bty_avg, data = evals)
summary(m5)$adj.r.squared
## [1] 0.1417823
= lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval +
m6 + cls_students + cls_level + cls_profs + cls_credits, data = evals)
summary(m6)$adj.r.squared
## [1] 0.1174189
bty_avg
cls_profs
cls_students
rank
To complete the model selection we would continue removing variables one at a time until removal of another variable did not increase adjusted \(R^2\).
Exercise: 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?
Exercise: 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.
Exercise: Would you be comfortable generalizing your conclusions to apply to professors generally (at any university)? Why or why not?
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.