Multiple linear regression

Submitted by: Preetha Rajan

Email: praj016@aucklanduni.ac.nz

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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)

## Warning: package 'GGally' was built under R version 3.5.2

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: This is an observational study. An observational study draws inferences from a sample to a population where the independent variable is not under the control of the researcher because of ethical concerns or logistical constraints.

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: Observational studies are descriptive in nature and often lead to the determination of associations. Such studies are often important to generate the hypotheses, for subsequent interventional studies. These studies typically do not have well-defined mechanistic hypotheses, but rather have a stated goal to obtain data or determine an association. Because observational studies are not randomized, they cannot control for all the other un-measurable and confounding factors that may actually be driving the results. Thus, any “link” between cause and effect in observational studies is speculative at best. Hence, it is appropriate to change the original research question by altering its wording to: “Is there an association between beauty and course evaluations?”.

Exploring the data

3. 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.
   Answer: This question can be answered by means of a histogram

hist(evals$score)

The output of the histogram plot indicates that the data is indeed left skewed. The distribution of the variable ‘score’ having a long left tail, indicates that students are highly likely to rate the professors highly. This fact is also demonstrated by the summary statistis table. The distribution is centered at an average evaluation score that is at the higher end of the scale. The majority of the distribution is at the right end.

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

Answer: Let us explore the relationship between the average beauty score and ethnicity

boxplot(evals$bty_avg~evals$ethnicity, data=evals, main='Boxplot depicting the distribution of average beauty scores by Ethnicity', xlab='ethnicity', ylab='average beauty score')

Taking the above boxplot into consideration, professors belonging to non-minority classes, seem to receive on an average, higher beauty scores.

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()

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?

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

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?

Answer: If we compare the two scatter plots (the one drawn using the geom_point function and the one drawn using the geom_jitter function), we can clearly see that the first scatterplot has not plotted all the overlapping observations.

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.

m1 <- lm(evals$score~evals$bty_avg)
summary(m1)

## 
## 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

Answer: The regression equation as per the output from the linear model is as follows:

\[ \hat{y} = 3.88034 + 0.06664\times bty\_avg \]

Interpretation: If the average beauty score of a professor increases by 1 point, then on an average, we can expect a professor’s average evaluation score to increase by 0.067 points.

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

Answer: Judging by the p-value and the magnitude of the calculated value of the t-score (whose absolute value is more than two standard deviations of the mean), we can reject the null hypothesis that the slope coefficient estimate for the explanatory variable bty_avg is 0 and conclude that the estimate of the slope coefficient is significantly different from 0. Hence, the average beauty score of a professor seems to be a significant predictor of the average evaluation score of a professor.

5. 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.

#Normality
hist(m1$residuals)

qqnorm(m1$residuals)
qqline(m1$residuals)

#Linear Association - plot of residuals vs. the explanatory variable
plt <- plot(m1$residuals~evals$bty_avg)
abline(h=0, lty=3)

#Constant spread of residuals - the residuals do not increase / decrease with an increase in the values of the dependent variable (no fan shape)
plot(m1$residuals~m1$fitted.values)
abline(h=0, lty=3)

From the plots, the normality condition seems to be violated - the residuals seem to have a long left tail and are not centered at 0. The conditions pertaining to linearity between the dependent and the explanatory variable and the constant spread of residuals, appears to have been met to a reasonable degree.

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                       0.844

As expected the relationship is quite strong - after all, the average score is calculated using the individual scores. We can actually take a look at the relationships between all beauty variables (columns 13 through 19) using the following command:

ggpairs(evals, columns = 13:19)

These variables are collinear (correlated), and adding more than one of these variables to the model would not add much value to the model. In this application and with these highly-correlated predictors, it is reasonable to use the average beauty score as the single representative of these variables.

In order to see if beauty is still a significant predictor of professor score after we’ve accounted for the gender of the professor, we can add the gender term into the model.

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

6. 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

#Normality
hist(m_bty_gen$residuals)

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

#Linear Association - plot of residuals vs. the explanatory variable
plt <- plot(m_bty_gen$residuals~evals$bty_avg)
#Use the abline function to draw a line at y=0, so as to examine whether the residuals have a random scatter about 0
#The argument lty is line type
abline(h=0, lty=3)

plt <- plot(m_bty_gen$residuals~evals$gender)
abline(h=0, lty=3)

#Constant spread of residuals - the residuals do not increase / decrease with an increase in the values of the dependent variable (no fan shape)
plot(m_bty_gen$residuals~m_bty_gen$fitted.values)
abline(h=0, lty=3)

Answer: The normality conditions for the residuals has been violated. The residuals are not centered at 0. Looking at the histogram and the normal probability plot, the distribution of the residuals is left skewed (having a longer left tail). As far as the condition for linearity is concerned, the residuals do appear to be randomly scattered about 0 in the case of both explanatory variables in both residual plots. Also, it appears that the condition of homoskedasticity or a constant spread of residuals (the residuals not increasing or decreasing as the predicted value of the response variable increases) has been reasonably met.

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?

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

summary(m1)

## 
## 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

Answer: It appears that in both the models, that the average beauty score of a professor is a statistically significant predictor of the average evaluation score. In fact, with the addition of the categorical variable gender, the statistical significance of the explanatory variable ‘bty_avg’ appears to have increased, judging by:

1. A further decline in the p-value from 0.00005083 (when ‘bty_avg’ was the sole explantory variable in the model) to 0.00000648 (when gender was included as a second predictor in the model)

2. The absolute value of the t-statistic associated with the explanatory variable ‘bty_avg’ has increased marginally from 4.09 to 4.563

The estimated slope coefficient associated with the explanatory variable ‘bty_avg’, has also increased from 0.06664 to 0.07416, with the inclusion of the gender 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 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.

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

7. 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: This is true.

Consider two separate linear regression equations for one male professor and one female professor. Both have received an average beauty rating of 3

The original linear regression equation is:

\[ \hat{y} = 3.74734 + 0.07416 \times bty\_avg + 0.17239 \times gender \]

The regression equation for the male professor is:

\[ \hat{y} = 3.74734 + 0.07416(3) + 0.17239(1) = 4.13827 \]

The regression equation for the female professor is:

\[ \hat{y} = 3.74734 + 0.07416(3) + 0.17239(0) = 3.96588 \]

Clearly, the expected average evaluation score is lower for the female professor.

The decision to call the indicator variable gendermale instead ofgenderfemalehas 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.

To answer this question, first examine the structure of the rank column in the evals data frame:

str(evals$rank)

##  Factor w/ 3 levels "teaching","tenure track",..: 2 2 2 2 3 3 3 3 3 3 ...

There are three levels - teaching (coded as 2), tenure track (coded as 3) and tenured (coded as 4).

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

It appears that the lm function has estimated two separate slope coefficients for the two higher ordered levels - tenure track and tenure. Both tenure track and tenure are the non-reference levels, while teaching is the reference level.

8. 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: As per the separately estimated slope coefficients for the two higher ordered levels of the rank variable, the expected value of the average evaluation scores seem lowest for tenure track and highest for 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.