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” by Hamermesh and Parker found that instructors who are viewed to be better looking receive higher instructional ratings.

Here, you 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, you will explore and visualize the data using the tidyverse suite of packages. The data can be found in the companion package for OpenIntro resources, openintro.

Let’s load the packages.

library(tidyverse)
library(openintro)
library(GGally)

This is the first time we’re using the GGally package. You 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. The result is a data frame where each row contains a different course and columns represent variables about the courses and professors. It’s called evals.

glimpse(evals)
## Rows: 463
## Columns: 23
## $ course_id     <int> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 1…
## $ prof_id       <int> 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5,…
## $ score         <dbl> 4.7, 4.1, 3.9, 4.8, 4.6, 4.3, 2.8, 4.1, 3.4, 4.5, 3.8, 4…
## $ rank          <fct> tenure track, tenure track, tenure track, tenure track, …
## $ ethnicity     <fct> minority, minority, minority, minority, not minority, no…
## $ gender        <fct> female, female, female, female, male, male, male, male, …
## $ language      <fct> english, english, english, english, english, english, en…
## $ age           <int> 36, 36, 36, 36, 59, 59, 59, 51, 51, 40, 40, 40, 40, 40, …
## $ cls_perc_eval <dbl> 55.81395, 68.80000, 60.80000, 62.60163, 85.00000, 87.500…
## $ cls_did_eval  <int> 24, 86, 76, 77, 17, 35, 39, 55, 111, 40, 24, 24, 17, 14,…
## $ cls_students  <int> 43, 125, 125, 123, 20, 40, 44, 55, 195, 46, 27, 25, 20, …
## $ cls_level     <fct> upper, upper, upper, upper, upper, upper, upper, upper, …
## $ cls_profs     <fct> single, single, single, single, multiple, multiple, mult…
## $ cls_credits   <fct> multi credit, multi credit, multi credit, multi credit, …
## $ bty_f1lower   <int> 5, 5, 5, 5, 4, 4, 4, 5, 5, 2, 2, 2, 2, 2, 2, 2, 2, 7, 7,…
## $ bty_f1upper   <int> 7, 7, 7, 7, 4, 4, 4, 2, 2, 5, 5, 5, 5, 5, 5, 5, 5, 9, 9,…
## $ bty_f2upper   <int> 6, 6, 6, 6, 2, 2, 2, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 9, 9,…
## $ bty_m1lower   <int> 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 7, 7,…
## $ bty_m1upper   <int> 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 6, 6,…
## $ bty_m2upper   <int> 6, 6, 6, 6, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 6, 6,…
## $ bty_avg       <dbl> 5.000, 5.000, 5.000, 5.000, 3.000, 3.000, 3.000, 3.333, …
## $ pic_outfit    <fct> not formal, not formal, not formal, not formal, not form…
## $ pic_color     <fct> color, color, color, color, color, color, color, color, …

We have observations on 21 different variables, some categorical and some numerical. The meaning of each variable can be found by bringing up the help file:

?evals

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.

Answer:

This is an observational study since there were no actual experiments setup involved. The observations were gathered by end of class evaluations and questionnaire answered by 6 students pertaining on the beauty assessment of the professors. As this is not an experimental study we cannot conclude casual relationship. We would need to rephrase the question as follows:

Rephrase: Does beauty correlates to professor’s evaluation or is the difference due to sampling variation?

  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? Answer:
hist(evals$score)

The distribution of ‘score’ is skewed to the left, with the majority of the observation with score of between 4 and 5. This is as expected, most students will provide feedback with positive evaluation (4 or 5).

  1. Excluding score, select two other variables and describe their relationship with each other using an appropriate visualization.

**Answer:*

We will consider the following variables: cls_perc_eval and cls_level. We will plot percentage of student that complete evaluation dependent on the class level using box plot.

ggplot(data = evals)+
  geom_boxplot(aes(x=cls_level, y=cls_perc_eval))+labs(x="Class Level", y="Class Percenatage Evaluation", title = "Class Level vs Percentage evaluation")+theme_bw()

We can see that the evaluation has less variability for upper level class (since the box plot is smaller) than for lower. There is no clear apparent difference for the median. There is a marked difference for the lower whiskers. For upper level class, the lower whisker ends higher with outliers. There are no outliers for lower class level.

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 you 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 geom_jitter as your layer. What was misleading about the initial scatterplot?

Answer:

ggplot(data = evals, aes(x = bty_avg, y = score)) +
  geom_jitter()+ylab("Score") + xlab("Beauty average")

May points have the same values for (x,y). So they could not be differentiated on the scatter plot. As a small amount of noise is added on the score variable(y),the points can be differentiated. So instead of 250 points, all 463 points are visible.

ggplot(data = evals, aes(x = bty_avg, y = score)) +
  geom_jitter()

  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. 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?

Answer:

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

To check its statistical significance let;s call summary function on our model:

summary(model_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

As we can see that our model equation comes out to be:

\[ Y = 3.88+.06664* beauty average \] According to the equation of our model if we increase average beauty by 1 the score will increase by .06664. And the mentioned increase will be statically significant since p value is close to 0 but the since the increase in itself is not very high ther changes that it will bring to the score will not be that significant.

Add the line of the bet fit model to your plot using the following:

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 you about the variability you might expect in your 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)

  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).

Answer:

plot_ss <- function(x, y, showSquares = FALSE, leastSquares = FALSE){
  plot(y~x, asp = 1)# xlab = paste(substitute(x)), ylab = paste(substitute(y)))
  
  if(leastSquares){
    m1 <- lm(y~x)
    y.hat <- m1$fit
  } else{
    cat("Click two points to make a line.")
    pt1 <- locator(1)
    points(pt1$x, pt1$y, pch = 4)
    pt2 <- locator(1)
    points(pt2$x, pt2$y, pch = 4)
    pts <- data.frame("x" = c(pt1$x, pt2$x),"y" = c(pt1$y, pt2$y))
    m1 <- lm(y ~ x, data = pts)
    y.hat <- predict(m1, newdata = data.frame(x))
  }
  r <- y - y.hat
  abline(m1)

  oSide <- x - r
  LLim <- par()$usr[1]
  RLim <- par()$usr[2]
  oSide[oSide < LLim | oSide > RLim] <- c(x + r)[oSide < LLim | oSide > RLim] # move boxes to avoid margins

  n <- length(y.hat)
  for(i in 1:n){
    lines(rep(x[i], 2), c(y[i], y.hat[i]), lty = 2, col = "blue")
    if(showSquares){
    lines(rep(oSide[i], 2), c(y[i], y.hat[i]), lty = 3, col = "orange")
    lines(c(oSide[i], x[i]), rep(y.hat[i],2), lty = 3, col = "orange")
    lines(c(oSide[i], x[i]), rep(y[i],2), lty = 3, col = "orange")
    }
  }

  SS <- round(sum(r^2), 3)
  cat("\r                                ")
  print(m1)
  cat("Sum of Squares: ", SS)
}
plot_ss(x = evals$bty_avg, y = evals$score, showSquares = TRUE)

## Click two points to make a line.
                                
## Call:
## lm(formula = y ~ x, data = pts)
## 
## Coefficients:
## (Intercept)            x  
##     3.88034      0.06664  
## 
## Sum of Squares:  131.868
hist(model_bty$residuals)

# normal probability plot of the residuals
qqnorm(model_bty$residuals)
qqline(model_bty$residuals)

Conditions for the least squares line

Linearity: The data show a slightly linear and it is positive linearity.

Nearly Normal residuals: From the Histogram, the residuals show a slightly left skewed distribution. The normal probability plot of the residuals shows that the points do not follow the line for upper quadrilles.

Constant Variability: From the residual plot, we can observe that there seems to have constant variability.

Independent observations: We do not have much information on how the sample was taken. We can assume independence of the observations

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_point()

evals %>% 
  summarise(cor(bty_avg, bty_f1lower))
## # A tibble: 1 × 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. You can actually look at the relationships between all beauty variables (columns 13 through 19) using the following command:

evals %>%
  select(contains("bty")) %>%
  ggpairs()

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 you’ve accounted for the professor’s gender, you 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. Verify that the conditions for this model are reasonable using diagnostic plots.

Answer:

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

# residual plot against each predictor variable
plot(m_bty_gen$residuals ~ evals$bty_avg)
abline(h = 0, lty = 4)  # adds a horizontal dashed line at y = 0

plot(m_bty_gen$residuals ~ evals$gender)
abline(h = 0, lty = 4)  # adds a horizontal dashed line at y = 0

#Resiual vs Fitted, Normal Probability Plot, Scale-Location, Residual vs Leverage
plot(m_bty_gen)

hist(m_bty_gen$residuals)

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

plot(evals$score ~ evals$gender)

The histogram of residuals suggests that the residuals distribution is slightly skewed to the left.

The residuals do not follow the lines for upper quadriles in the Normal Probability Plot for residuals, .

Residuals vs Fitted, show that it appears to be constant variability for residuals. But as was established in the previous exercises, there is a linear relationship between beauty average and teaching evaluation score.

  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?

Answer:

Yes it is. In fact, gender made beauty average even more significant as the p-value computed is even smaller (6.48e-06 < 5.08e-05) now compared to a model where beauty average was the only variable.

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 male and female to being an indicator variable called gendermale that takes a value of \(0\) for female professors and a value of \(1\) for male professors. (Such variables are often referred to as “dummy” variables.)

As a result, for female professors, the parameter estimate is multiplied by zero, leaving the intercept and slope form familiar from simple regression.

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

ggplot(data = evals, aes(x = bty_avg, y = score, color = pic_color)) +
 geom_smooth(method = "lm", formula = y ~ x, se = FALSE)

  1. What is the equation of the line corresponding to those with color pictures? (Hint: For those with color pictures, the parameter estimate is multiplied by 1.) For two professors who received the same beauty rating, which color picture tends to have the higher course evaluation score?

Answer:

\[ Score = 3.74 .07416 * Beauty Average + 017239 * gender male \] For gender = Male, we will evaluate the equation with gender_male = 1. In case, of female gender, we will substitute a 0

Let’s put gender male equals 1

\[ Score = 3.74 .07416 * Beauty Average + 017239 \] Male professor will have a evaluation score higher by 0.17239 all other things being equal.

The decision to call the indicator variable gendermale instead of genderfemale 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.

Answer:

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 variable with more than 2 levels, it appears to handle it considering them 2 different variables.

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, gender, ethnicity, 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.

Answer:

The “professor single” (cls_profssingle) as the variable to have the least association with the professor’s evaluation score.

Let’s run the model…

m_full <- lm(score ~ rank + gender + ethnicity + 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 + gender + ethnicity + 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    
## gendermale             0.2109481  0.0518230   4.071 5.54e-05 ***
## ethnicitynot minority  0.1234929  0.0786273   1.571  0.11698    
## 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
  1. Check your suspicions from the previous exercise. Include the model output in your response.

Answer:

The “professor single” (cls_profssingle) as the variable to have the least association with the professor’s evaluation score. That has the maximum p-value(0.77806)

  1. Interpret the coefficient associated with the ethnicity variable.

Answer:

Evaluation for professor that is not minority tends to be 0.1234929 higher but The ethnicity p-value of about 0.11 means that it has a weak relationship to scores and may be dropped as part of the model

  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?

Answer:

model_back <- 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(model_back)
## 
## 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

Yes. There was a slight change in the coefficients and significance of the other explanatory variables when cls_profssingle was removed. All the values are now slightly lower - meaning they are more significant now to the level than before.

  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.

Answer:

model_best <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval + 
    cls_credits + bty_avg + pic_color, data = evals)
summary(model_best)
## 
## 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

\[ Score = B_0 + B_1 * ethnicity.not.minority +B_2 * gender_male+B_3 * language.non.english + B_4 * age + B_5 * class.perc.eval + B_6 * class.credits.one + B_7 * bty.avg + B_8 * picture.color.colored \]

  1. Verify that the conditions for this model are reasonable using diagnostic plots.

Answer: 1. the residuals of the model are nearly normal

qqnorm(model_best$residuals)
qqline(model_best$residuals)

The residuals of the model is not normal as residual values for the the higher and lower quantiles are less than what a normal distribution would predict

  1. Absolute values of residuals against fitted values. (the variability of the residuals is nearly constant)
plot(abs(model_best$residuals) ~ model_best$fitted.values)

There some outliers although overall, most of the residual values are close to the fitted values.

  1. the residuals are independent
plot(model_best$residuals ~ c(1:nrow(evals)))

Yes, this condition is met. The residuals based on the sequence when it was gathered shows that they were randomly gathered.

  1. each variable is linearly related to the outcome.
plot(evals$score ~ evals$ethnicity)

plot(evals$score ~ evals$gender)

plot(evals$score ~ evals$language)

plot(evals$score ~ evals$age)

plot(evals$score ~ evals$cls_perc_eval)

plot(evals$score ~ evals$cls_credits)

plot(evals$score ~ evals$bty_avg)

plot(evals$score ~ evals$pic_color)

The variables above are linearly related to the score - some more so than others.

  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?

Answer:

No. Class courses are independent of each other so evaluation scores from one course is independent of the other even if the course is being taught by the same professor.

  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.

Answer:

The professor is not a minority and male, must have graduated from an American (or English speaking) school and teaches a one credit course. He must also have a high beauty average score from the students and the professor’s class photo should be in black and white. He must also be relatively young. And a good percentage of his class must have completed the evaluation

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

Answer:

No. The sample size of 6 is too small. Also, some of the predictor variables are subjective and may vary with culture. Beauty, for one, is in the eye of the beholder.Picture preferences may also be culturally biased.