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.

download.file("http://www.openintro.org/stat/data/evals.RData", destfile = "evals.RData")
load("evals.RData")

See lab tutorial instructions for a description of all the variables.

str(evals)
## 'data.frame':    463 obs. of  21 variables:
##  $ score        : num  4.7 4.1 3.9 4.8 4.6 4.3 2.8 4.1 3.4 4.5 ...
##  $ rank         : Factor w/ 3 levels "teaching","tenure track",..: 2 2 2 2 3 3 3 3 3 3 ...
##  $ ethnicity    : Factor w/ 2 levels "minority","not minority": 1 1 1 1 2 2 2 2 2 2 ...
##  $ gender       : Factor w/ 2 levels "female","male": 1 1 1 1 2 2 2 2 2 1 ...
##  $ language     : Factor w/ 2 levels "english","non-english": 1 1 1 1 1 1 1 1 1 1 ...
##  $ age          : int  36 36 36 36 59 59 59 51 51 40 ...
##  $ cls_perc_eval: num  55.8 68.8 60.8 62.6 85 ...
##  $ cls_did_eval : int  24 86 76 77 17 35 39 55 111 40 ...
##  $ cls_students : int  43 125 125 123 20 40 44 55 195 46 ...
##  $ cls_level    : Factor w/ 2 levels "lower","upper": 2 2 2 2 2 2 2 2 2 2 ...
##  $ cls_profs    : Factor w/ 2 levels "multiple","single": 2 2 2 2 1 1 1 2 2 2 ...
##  $ cls_credits  : Factor w/ 2 levels "multi credit",..: 1 1 1 1 1 1 1 1 1 1 ...
##  $ bty_f1lower  : int  5 5 5 5 4 4 4 5 5 2 ...
##  $ bty_f1upper  : int  7 7 7 7 4 4 4 2 2 5 ...
##  $ bty_f2upper  : int  6 6 6 6 2 2 2 5 5 4 ...
##  $ bty_m1lower  : int  2 2 2 2 2 2 2 2 2 3 ...
##  $ bty_m1upper  : int  4 4 4 4 3 3 3 3 3 3 ...
##  $ bty_m2upper  : int  6 6 6 6 3 3 3 3 3 2 ...
##  $ bty_avg      : num  5 5 5 5 3 ...
##  $ pic_outfit   : Factor w/ 2 levels "formal","not formal": 2 2 2 2 2 2 2 2 2 2 ...
##  $ pic_color    : Factor w/ 2 levels "black&white",..: 2 2 2 2 2 2 2 2 2 2 ...

Exploring the data

Exercise 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.

Observational study. No, not possible to answer the question as phrased. Cannot make a causal claim with observational study. Re-phrase question to: Is beauty associated with the course evaluation score?

Exercise 2

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 is left-skewed. Since the distribution is not normal, we can conclude that students tend to rate course evaluations more positively (give higher ratings than average).

My initial expectation was to see a normal distribution. But thinking about it now, I am not surprised to see the distribution lean toward higher ratings. I think it is human nature to give more positive feedback (unless something is truly terrible).

Exercise 3

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

Let’s see if there a correlation between the age of the professor and his/her beauty rating..

# Look at Age of professor vs Average Beauty Rating
plot(evals$age, evals$bty_avg, main = "Age of Professor vs Beauty Rating", xlab = "Age of Professor", ylab ="Avg Beauty Rating")

The points are everywhere, except for the two points for age > 70. There does not appear to be a clear association.

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?

There are 463 observations in the data frame. This plot shows fewer points.

Exercise 4

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?

# Use jitter() on x-coordinate
plot(jitter(evals$score) ~ evals$bty_avg)

This plots show there are overlapping points (those that appear darker).

The initial plot did not show where there are duplicate points.

Exercise 5

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, data = evals)
plot(jitter(evals$score) ~ evals$bty_avg)
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

Equation for Linear Model:

score = 3.8803 + 0.0666 * bty_avg

Slope is 0.0666. For every additional point in average beauty score, the course evaluation score improves by 0.067 points.

The p-value (.0000508) is very low. So average beauty score is a statistically significant predictor of evaluation score.

Is it a practically significant predictor?

No, because “beauty is in the eyes of the beholder.” The beauty scores were obtained from a very small sample (6 students). The results could be entirely different if a different sample of students rated the professors.

Exercise 6

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).

# Residuals Plot:  residuals vs. bty_avg
plot(m_bty$residuals ~ evals$bty_avg)
abline(h = 0, lty = 3)  # adds a horizontal dashed line at y = 0

The residuals do not appear quite random. The residuals above the zero line are more clustered together. The residuals below the zero line have more variability and a wider spread. This indicates some lack of normality.

Look at the histogram of residuals to confirm.

# Check normality of residuals on a histogram
hist(m_bty$residuals)

The histogram is not normal. It has a left skew.

F/U: Would this be considered “nearly normal”?

Check the qq plot too.

qqnorm(m_bty$residuals)  # create normal probability plot 
qqline(m_bty$residuals)  # adds diagonal line to the normal prob plot

The qq plot confirms the lack of normality. It shows a left skew on the bottom left.

F/U: What does the downward curve in the top right mean??

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)
## [1] 0.8439112

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)
## 
## 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

Exercise 7

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.

Verifying conditions for the model accounting for GENDER of professor:

# 1. Plot the m_bty_gen regression line on the scatterplot
m_bty_gen <- lm(score ~ bty_avg + gender, data = evals)
plot(jitter(evals$score) ~ evals$bty_avg)
abline(m_bty_gen)
## Warning in abline(m_bty_gen): only using the first two of 3 regression
## coefficients

# 2. Create residuals plot:  m_bty_gen residuals vs. bty_avg
plot(m_bty_gen$residuals ~ evals$bty_avg)
abline(h = 0, lty = 3)  # adds a horizontal dashed line at y = 0

The residuals plot looks fairly random except for greater variability below the zero line.

# 3. Plot histogram of m_bty_gen residuals to check for outliers
hist(m_bty_gen$residuals)

# 4. Plot normal probability plot
qqnorm(m_bty_gen$residuals)
qqline(m_bty_gen$residuals)  # adds diagonal line to the normal prob plot

This histogram is also left-skewed, even more skewed than the model without accounting for gender. But there are no extreme outliers that might cause concern. So although the histogram is skewed, the condition for normality can be relaxed.

The qq plot is fairly normal except the upper right hand corner (not sure what this represents).

The 3 conditions for the linear model are met.

  1. Linearity? - Sort of. In the scatterplot, points do not closely fit along the regression line.

  2. Nearly normal residuals? - Acceptable.

  3. Constant variablity? - Yes. Variability of points around the regression line are roughly constant as x changes.

The p-values and parameter estimates can be trusted.

Exercise 8

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

# Obtain statistics on the m_bty_gen model
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

The p-value (.000652) is still low with the addition of gender to the model. So bty_avg is still a significant predictor of evaluation score.

score = 3.7473 + 0.07416 * bty_avg

The addition of gender has changed the parameter estimate for bty_avg. The parameter estimate increased from 0.066 to 0.074.

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.

scoreˆ=β̂ 0+β̂ 1×bty_avg+β̂ 2×(0)=β̂ 0+β̂ 1×bty_avg

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

multiLines(m_bty_gen)

Exercise 9

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?

b1 is parameter estimate for makes (= 1) b2 is parameter estimate for females (= 0)

score = b0 + b1 * bty_avg

score = 3.747 + 0.172 * bty_avg

Males tend to have a higher course evaluation 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.)

Exercise 10

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

The output above shows only 2 of the 3 levels for rank (tenure track + tenured). The level for teaching is missing. This is because teaching is the reference level, the default level that other levels are measured against.

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.

Exercise 11

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.

F/U: Why does the variable with the least association have the highest p-value?

I expect the number of credits to have the highest p-value. The number of credits should have no association with a professor’s evalution score.

Let’s run the model…

# This is the full model (all variables included)
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)
## 
## 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 12

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

# Run the model without cls_credits
m_drop_credits <- lm(score ~ rank + ethnicity + gender + language + age +         
                       cls_perc_eval + cls_students + cls_level + cls_profs + 
                       bty_avg + pic_outfit + pic_color, data = evals)
summary(m_drop_credits)
## 
## Call:
## lm(formula = score ~ rank + ethnicity + gender + language + age + 
##     cls_perc_eval + cls_students + cls_level + cls_profs + bty_avg + 
##     pic_outfit + pic_color, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.7498 -0.3200  0.1056  0.3679  0.9200 
## 
## Coefficients:
##                         Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.3098194  0.2918733  14.766  < 2e-16 ***
## ranktenure track      -0.1957586  0.0829015  -2.361 0.018635 *  
## ranktenured           -0.1809000  0.0647027  -2.796 0.005398 ** 
## ethnicitynot minority  0.0429967  0.0778938   0.552 0.581229    
## gendermale             0.2366593  0.0524895   4.509 8.33e-06 ***
## languagenon-english   -0.2589399  0.1133484  -2.284 0.022810 *  
## age                   -0.0090463  0.0031973  -2.829 0.004873 ** 
## cls_perc_eval          0.0059006  0.0015636   3.774 0.000182 ***
## cls_students           0.0002954  0.0003829   0.771 0.440863    
## cls_levelupper        -0.0065495  0.0565243  -0.116 0.907807    
## cls_profssingle       -0.0427280  0.0525927  -0.812 0.416974    
## bty_avg                0.0315543  0.0177371   1.779 0.075917 .  
## pic_outfitnot formal  -0.1362125  0.0751223  -1.813 0.070467 .  
## pic_colorcolor        -0.2091633  0.0728769  -2.870 0.004297 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5077 on 449 degrees of freedom
## Multiple R-squared:  0.1531, Adjusted R-squared:  0.1286 
## F-statistic: 6.243 on 13 and 449 DF,  p-value: 7.671e-11

To evaluate which model is better, compare the adjusted R-squared value. Choose the one with the larger R-squared. Adjusted R-squared describes the strength of the model fit.

Removing cls_credits variable yields a smaller adjusted R-squared value (0.1286) than the first model (0.1617). So the first model (keeping cls_credits) is better.

Exercise 13

Interpret the coefficient associated with the ethnicity variable.

The ethnicity parameter has a B coefficient of 0.1235 (in the first model). The coefficient represents a 0.1234 point increase in evaluation score we would expect if the ethnicity changed from 0 (minority) to 1 (not minority), all other factors held constant.

Exercise 14

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?

The variable with the highest p-value is cls_profs (number of professors teaching sections in course in sample: single, multiple.)

# Drop the cls_profs variable from the model
m_drop_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_drop_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 barely changed. The significance of the other explanatory variables changed a little. This means these variables are collinear because dropping cls_prof does not add much value to the model.