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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.

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

Getting Started

Load packages

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)

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

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.
  1. Is this an observational study or an experiment?
    1. Observational study
    2. Experiment

    answer: Observational study

  2. The original research question posed in the paper is whether beauty leads directly to the differences in course evaluations. Given the study design, should the question be rephrased? If so, how?
    1. No, the question is worded accurately.
    2. Yes, revise wording to “Is there an association between beauty and course evaluations?”
    3. Yes, revise wording to “Does beauty score increase the professor’s course evaluations?”
    4. Yes, revise wording to “Does beauty score decrease the professor’s course evaluations?”

    answer: Yes, revise wording to “Is there an association between beauty and course evaluations?”

Exploring the data

  1. Which of the following statements is false about the distribution of score?
    1. The median of the distribution is 4.3.
    2. 25% of the students gave their professors a score of over 4.6.
    3. 11 of students gave a professor a score below 3.
    4. The left skewness of the data suggests that the students are less likely to rate the professors highly. 
# type your code for Question 3 here, and Knit
# median IS 4.3
s1 <- evals %>% summarise(med = median(score), n = n())
s2 <- evals %>% filter(score > 4.6) %>% summarise(n = n())
s3 <- evals %>% filter(score < 3) %>% summarise(n = n())
s1$med
## [1] 4.3
s2$n / s1$n
## [1] 0.2073434
s3$n
## [1] 11
ggplot(evals, aes(x = score)) +
  geom_histogram()
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

answer: The left skewness of the data suggests that the students are less likely to rate the professors highly.

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
ggplot(evals, aes(x = cls_level, y = bty_avg)) +
  geom_boxplot()

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:

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.

ggplot(data = evals, aes(x = bty_avg, y = score)) +
  geom_jitter() +
  geom_smooth(method = "lm")

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)

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
m_bty = lm(score ~ bty_avg, data = evals)
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

\[ y = 3.880 + .067 \times bty\_avg \] A one-unit increase in beauty score improves the prossessor rating by .067 points.

  1. Average beauty score is a statistically significant predictor of evaluation score.
    1. True
    2. False

    answer: true

  2. Use residual plots to evaluate whether the conditions of least squares regression are reasonable. Which of the following statements is an incorrect analysis of the residual plots and conditions?
    1. Linear association: The residuals plot shows a random scatter.
    2. Constant variance of residuals: No fan shape in residuals plot.
    3. Nearly normal residuals: Residuals are right skewed, but the sample size is large, so this may not be an important violation of conditions.
    4. Independent observations: Classes sampled randomly, no order effect. 
# type your code for the Question 5 here, and Knit
# A residuals plot of (e_i  ~ x_i ) with no pattern indicates a linear relationship because the linear model captures the trend and leaves behind only noise. 
plot(m_bty$residuals ~ evals$bty_avg)

# A residuals plot of (e_i~y_i ) should have random scatter in a band of constant width around 0 (no fan shape!).
plot(m_bty$residuals ~ m_bty$fitted)

# A histogram plot should be normal.  A normal probability plot should not have a bow-shaped pattern.  
hist(m_bty$residuals)

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

answer: Nearly normal residuals: Residuals are right skewed, but the sample size is large, so this may not be an important violation of conditions.

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.

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                   0.8439112

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:

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.

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. Using diagnostic plots, we can conclude that the conditions for this model are reasonable.
    1. True
    2. False 
# type your code for the Exercise here, and Knit
# Linearity: residuals plot has no pattern because pattern is already captured.
plot(m_bty_gen$residuals, evals$bty_avg)

boxplot(m_bty_gen$residuals, evals$gender)

plot(evals$score ~ m_bty_gen$fitted)

# Normal residuals:
hist(m_bty_gen$residuals)

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

# Homoscedasticity
plot(m_bty_gen$residuals ~ m_bty_gen$fitted)

# Autocorrelation
plot(m_bty_gen$residuals)

answer: True

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.

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

  1. For two professors (one male and one female) who received the same beauty rating, the male professor is predicted to have the higher course evaluation score than the female.
    1. True
    2. False

answer: True

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
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
  1. Which of the following is the correct order of the three levels of rank if we were to order them from lowest predicted course evaluation score to highest predicted course evaluation score?
    1. Teaching, Tenure Track, Tenured
    2. Tenure track, Tenured
    3. Tenure Track, Tenured, Teaching
    4. Teaching, Tenured, Tenure Track

answer: Tenure Track, Tenured, Teaching

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.

Prediction

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.

newprof <- data.frame(gender = "male", bty_avg = 3)

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.

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.

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…

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

  1. Which of the following is the correct intrepetation of the coefficient associated with the ethnicity variable.
    Non-minority professors are expected on average to score …
    1. 0.12 points lower than minority professors, all else held constant.
    2. 0.12 points higher than minority professors, all else held constant.
    3. 0.02 points lower than minority professors, all else held constant.
    4. 0.02 points higher than minority professors, all else held constant.

answer: 0.12 points higher than minority professors, all else held constant.

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
m_full <- 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)
## 
## 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.

m1 <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval 
             + 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):

m2 = lm(score ~ rank + gender + language + age + cls_perc_eval + 
    cls_students + cls_level + cls_profs + cls_credits + bty_avg, data = evals)
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
# remove cls_profs
m1 <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval 
             + cls_students + cls_level + cls_credits + bty_avg, data = evals)
summary(m1)$adj.r.squared
## [1] 0.1436001
  1. Elimination of which variable from the full model yielded the highest adjusted R-squared?
    1. bty_avg
    2. cls_profs
    3. cls_students
    4. rank

answer: cls_profs

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?

answer: yes, because some professors may teach more classes than others. So the observations are not independent.

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.

m1 <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval 
             + cls_students + cls_credits + bty_avg, data = evals)
summary(m1)$adj.r.squared
## [1] 0.1453655
summary(m1)
## 
## Call:
## lm(formula = score ~ ethnicity + gender + language + age + cls_perc_eval + 
##     cls_students + cls_credits + bty_avg, data = evals)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.89519 -0.31227  0.08596  0.37022  1.09853 
## 
## Coefficients:
##                         Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            3.3863086  0.2094164  16.170  < 2e-16 ***
## ethnicitynot minority  0.2044482  0.0746764   2.738 0.006428 ** 
## gendermale             0.1768250  0.0503142   3.514 0.000485 ***
## languagenon-english   -0.1511723  0.1035293  -1.460 0.144930    
## age                   -0.0048725  0.0026073  -1.869 0.062298 .  
## cls_perc_eval          0.0057538  0.0015405   3.735 0.000212 ***
## cls_students           0.0004073  0.0003428   1.188 0.235355    
## cls_creditsone credit  0.5230953  0.1050306   4.980 9.03e-07 ***
## bty_avg                0.0618985  0.0165267   3.745 0.000203 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5028 on 454 degrees of freedom
## Multiple R-squared:  0.1602, Adjusted R-squared:  0.1454 
## F-statistic: 10.82 on 8 and 454 DF,  p-value: 5.463e-14

answer: The best scores are for non-minorty, male, English Speaking, younger, garnering a higher percentage evaluations, having a larger class, teaching a single credit course, and having more beauty.

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

answer: no, because the observations were not random. At best, they could be generalized only to the university.

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.

References

Gelman, Andrew, and Jennifer Hill. 2007. Data Analysis Using Regression and Multilevel/Hierarchical Models. 1st ed. Cambridge University Press.

Hamermesh, Daniel S., and Amy Parker. 2005. “Beauty in the Classroom - Instructors’ Pulchritude and Putative Pedagogical Productivity” 24 (4). Economics of Education Review: 369–76. doi:10.1016/j.econedurev.2004.07.013.