Grading the professor

The data

load(url("http://www.openintro.org/stat/data/evals.RData"))

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.
  • It is an observational study.
  • How non-teaching related characteristics (mainly the physical appearance) influence on student’s evaluation of the course?
  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)

summary(evals$score)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   2.300   3.800   4.300   4.175   4.600   5.000
  • the distribution appears left skewed
  • the mean is 4.17473. The rates are more positive.
  • More information is needed to predict what to expect, but students’ ratings seem to be high and satisfactory for them.
  1. Excluding score, select two other variables and describe their relationship using an appropriate visualization (scatterplot, side-by-side boxplots, or mosaic plot).
mosaic(~ evals$ethnicity + evals$gender)

  • minority: the # female professors is greater than the male
  • non-minority: the # female professors is less than the male

Simple linear regression

plot(evals$score ~ evals$bty_avg)

  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?
qplot(x = evals$bty_avg, y = evals$score, geom = "point") + geom_jitter()

  • the points were more linear in the initial scatterplot
  • with jitter, the scatterplot displays points in a cloud-looking cluster of 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?
m_bty<-lm(evals$score ~ evals$bty_avg)
plot(jitter(evals$score) ~ jitter(evals$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
  • \(\hat{y} = 3.88034 + 0.06664 \times\) bty_avg
  • somewhat predictive with high residuals standard error of 53.48% and \(R^2_{adj}\) = 3.293%
  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(m_bty$residuals ~ evals$bty_avg)
abline(h = 0)

hist(m_bty$residuals)

qqnorm(m_bty$residuals)
qqline(m_bty$residuals)

  • Linearity: the residual plot is randomly dispersed, but satisfied, linearity is met
  • Normal residuals: left skewed histogram rather than normal distribution. The almost “S” shape indicates the non-normality of the residuals
  • Constant variability: it is fairly constant
  • Independence of observations: we assume the randomness and the indepence of the observations.

Multiple linear regression

plot(evals$bty_avg ~ evals$bty_f1lower)

cor(evals$bty_avg, evals$bty_f1lower)
## [1] 0.8439112
plot(evals[,13:19])

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
  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.
plot(m_bty_gen$residuals ~ evals$bty_avg)
abline(h = 0)

hist(m_bty_gen$residuals)

qqnorm(m_bty_gen$residuals)
qqline(m_bty_gen$residuals)

  • Linearity: the residual plot is randomly dispersed, but satisfied, linearity is met
  • Normal residuals: left skewed histogram rather than normal distribution. The almost “S” shape indicates the non-normality of the residuals
  • Constant variability: it is fairly constant
  • Independence of observations: we assume the randomness and the indepence of the observations.
  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?
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
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
  • bty_avg is still the significant predictor
  • gender addition didn’t effect much, it only helped \(R^2\) and \(R^2_{adj}\) to slightly improve
  • Residual standard error = 0.5287 didn’t change much

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?
  • gendermale = 1
  • \(\hat{y} = 3.74734 + 0.07416 \times\) bty_avg + 0.17239 = 3.91973 + 0.07416 \(\times\) bty_avg
  • \(\hat{y}_{male}\) > \(\hat{y}_{female}\)
  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
  • For n levels, R considers n-1 variables or n-1 categories.
  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.
variables<-c("rank" , "ethnicity" , "gender" , "language" , "age" , "cls_perc_eval" , "cls_students" , "cls_level" , "cls_profs" , "cls_credits" , "bty_avg" , "pic_outfit" , "pic_color")

cor_results<-c(cor(as.numeric(evals$score),as.numeric(evals$rank)),cor(as.numeric(evals$score),as.numeric(evals$ethnicity)),cor(as.numeric(evals$score),as.numeric(evals$gender)),cor(as.numeric(evals$score),as.numeric(evals$language)),cor(as.numeric(evals$score),as.numeric(evals$age)),cor(as.numeric(evals$score),as.numeric(evals$cls_perc_eval)),cor(as.numeric(evals$score),as.numeric(evals$cls_students)),cor(as.numeric(evals$score),as.numeric(evals$cls_level)),cor(as.numeric(evals$score),as.numeric(evals$cls_profs)),cor(as.numeric(evals$score),as.numeric(evals$cls_credits)),cor(as.numeric(evals$score),as.numeric(evals$bty_avg)),cor(as.numeric(evals$score),as.numeric(evals$pic_outfit)),cor(as.numeric(evals$score),as.numeric(evals$pic_color)))

for(i in cor_results){
  if(i>=-0.026 & i<=0.026){
      print(variables[which(cor_results == i)])
  }
}
## [1] "cls_students"
## [1] "cls_profs"
  • cls_profs is the closest to have no relationship (almost 0)
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_profssingle has the highest p-value of 0.77806 as we confirmed it previously by choosing cls_profs
  1. Interpret the coefficient associated with the ethnicity variable.
  • ethnicitynot minority variable has a coefficent of 0.1234929 that is a multiplier to itself
  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?
without_high_p_value <- 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(without_high_p_value)
## 
## 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
  • we notice minor changes in coefficients of the variables
  • we notice a decrease in the p-value
  • we notice an increase in the \(R^2_{adj}\)
  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.
backward_selection <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval + 
    cls_credits + bty_avg + pic_color, data = evals)
summary(backward_selection)
## 
## Call:
## lm(formula = score ~ ethnicity + gender + language + age + cls_perc_eval + 
##     cls_credits + bty_avg + pic_color, data = evals)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.85320 -0.32394  0.09984  0.37930  0.93610 
## 
## Coefficients:
##                        Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            3.771922   0.232053  16.255  < 2e-16 ***
## ethnicitynot minority  0.167872   0.075275   2.230  0.02623 *  
## gendermale             0.207112   0.050135   4.131 4.30e-05 ***
## languagenon-english   -0.206178   0.103639  -1.989  0.04726 *  
## age                   -0.006046   0.002612  -2.315  0.02108 *  
## cls_perc_eval          0.004656   0.001435   3.244  0.00127 ** 
## cls_creditsone credit  0.505306   0.104119   4.853 1.67e-06 ***
## bty_avg                0.051069   0.016934   3.016  0.00271 ** 
## pic_colorcolor        -0.190579   0.067351  -2.830  0.00487 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4992 on 454 degrees of freedom
## Multiple R-squared:  0.1722, Adjusted R-squared:  0.1576 
## F-statistic:  11.8 on 8 and 454 DF,  p-value: 2.58e-15
  • \(\hat{y}\) = 3.771922 + 0.167872 \(\times\) ethnicity_not_minority + 0.207112 \(\times\) gender_male - 0.206178 \(\times\) language_non_english - 0.006046 \(\times\) age + 0.004656 \(\times\) cls_perc_eval + 0.505306 \(\times\) cls_credits_one_credit + 0.051069 \(\times\) bty_avg - 0.190579 \(\times\) pic_color_color
  1. Verify that the conditions for this model are reasonable using diagnostic plots.
hist(backward_selection$residuals)

qqnorm(backward_selection$residuals)
qqline(backward_selection$residuals)

plot(backward_selection$residuals)
abline(h = 0)

  • Linearity: the residual plot is randomly dispersed, but satisfied, linearity is met
  • Normal residuals: left skewed histogram rather than normal distribution. The almost “S” shape indicates the non-normality of the residuals
  • Constant variability: it is fairly constant
  • Independence of observations: we assume the randomness and the indepence of the observations.
  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?
  • adding a course will impact on the conditions of linear regression
  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.

Characteristics would be:

  • a non-minority male
  • who speaks english
  • is younger
  • has a high percentage of students complete their evaluation
  • is considered attractive
  • has a more casual picture in color
  1. Would you be comfortable generalizing your conclusions to apply to professors generally (at any university)? Why or why not?

Better not generalize it because:

  • this is one observational study in one university
  • the definition of beauty may not be the case elsewhere