Grading the professor

Many college courses conclude by giving students the opportunity to evaluate the course and the instructor anonymously. However, the use of these student evaluations as an indicator of course quality and teaching effectiveness is often criticized because these measures may reflect the influence of non-teaching related characteristics, such as the physical appearance of the instructor. The article titled, “Beauty in the classroom: instructors’ pulchritude and putative pedagogical productivity” (Hamermesh and Parker, 2005) found that instructors who are viewed to be better looking receive higher instructional ratings. (Daniel S. Hamermesh, Amy Parker, Beauty in the classroom: instructors pulchritude and putative pedagogical productivity, Economics of Education Review, Volume 24, Issue 4, August 2005, Pages 369-376, ISSN 0272-7757, 10.1016/j.econedurev.2004.07.013. http://www.sciencedirect.com/science/article/pii/S0272775704001165.)

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

The data

The data were gathered from end of semester student evaluations for a large sample of professors from the University of Texas at Austin. In addition, six students rated the professors’ physical appearance. (This is aslightly modified version of the original data set that was released as part of the replication data for Data Analysis Using Regression and Multilevel/Hierarchical Models (Gelman and Hill, 2007).) The result is a data frame where each row contains a different course and columns represent variables about the courses and professors.

rm(list = ls())
load("more/evals.RData")
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.

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, as this is already a process which is followed, and t was not specially done for this analysis.

  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)

As we see from the histogram, the score is strongly skewed on the left. That means students prefer not to rate the professors below a certain level. Also we see that most of the professors are rated towards the highest rating.

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

Answer: We will use 2 variables: cls_perc_eval and bty_avg. Both of these are continuous numeric variables. And we can try to see if the average beauty rating of a professor and the percentage of students in the class who voted are related to each other. Creating a scatter plot to see how the relationship looks like:

library(ggplot2)

ggplot(evals, aes(x=bty_avg, y=cls_perc_eval)) +
  geom_point()

bty_avg.cls_perc_eval.model <- lm(formula = cls_perc_eval ~ bty_avg, data = evals)
summary(bty_avg.cls_perc_eval.model)
## 
## Call:
## lm(formula = cls_perc_eval ~ bty_avg, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -61.634 -11.804   3.067  12.391  29.068 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   67.021      2.360   28.40  < 2e-16 ***
## bty_avg        1.677      0.505    3.32  0.00097 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 16.58 on 461 degrees of freedom
## Multiple R-squared:  0.02336,    Adjusted R-squared:  0.02124 
## F-statistic: 11.02 on 1 and 461 DF,  p-value: 0.0009704

From above, it is clear that there is statistical relationship between the 2 variables we chose above, but the R value is quite less.

Simple linear regression

The fundamental phenomenon suggested by the study is that better looking teachers are evaluated more favorably. Let’s create a scatterplot to see if this appears to be the case:

plot(evals$score ~ evals$bty_avg)

Before we draw conclusions about the trend, compare the number of observations in the data frame with the approximate number of points on the scatterplot. Is anything awry? Answer: Apparently, as the average beauty range increases, the score is also raising. The number of points on the plot does not look to be equal to the number of observations which is 463. Some points seem to overlap hence looking like a lesser number.

  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? Answer:
plot(evals$score ~ jitter(evals$bty_avg, 1))

This gives jitter (noise) on the x-axis, meaning the points are made apart so that more number of points could be plotted and are visible.

  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?

Answer:

m_bty <- lm(formula = 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

Slope is a very less value : 0.06664. That means for increase of 1 score in average beauty, the score increases by 0.06664

ggplot(evals, aes(x=bty_avg, y=score)) +
  geom_point(color = "red") +
  geom_line(aes(x=bty_avg, y=predict(m_bty, newdata = evals)), color = "blue")

  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).

Plotting the residual plot:

plot(m_bty$residuals ~ evals$bty_avg)
abline(h = 0, lty = 3)

Creating the residuals histogram

hist(m_bty$residuals)

Q-Q Plot for residuals:

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

Let us evaluate the 4 conditions for fitting the least squared line for this scenario: 1) Linearity: From the linear model and the plot made from the lm model, it is clear that the data trend is linear for this one. Hence this condition is satisfied. 2) Nearly normal residuals: As seen from the histogram, the residuals are not nearly normal for this. Hence this condition is not satisfied too. 3) Constant Variability: From the residuals plot drawn above, the variability s not constant as the positive residuals vary from 0 to 1 which the negative ones go as far as -2. Hence the constant variability condition is also not satisfied. 4) Independent observations: The observations can be said to be independent as this is not a time based observation.

As the conditions for normal residuals and constant variability are not satisfied, we can say that the least squared line cannot be fitted for this scenario.

Multiple linear regression

The data set contains several variables on the beauty score of the professor: individual ratings from each of the six students who were asked to score the physical appearance of the professors and the average of these six scores. Let’s take a look at the relationship between one of these scores and the average beauty score.

plot(evals$bty_avg ~ evals$bty_f1lower)
cor(evals$bty_avg, evals$bty_f1lower)

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

plot(evals[,13:19])

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

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

m_bty_gen <- lm(score ~ bty_avg + gender, data = evals)
summary(m_bty_gen)
## 
## Call:
## lm(formula = score ~ bty_avg + gender, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.8305 -0.3625  0.1055  0.4213  0.9314 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  3.74734    0.08466  44.266  < 2e-16 ***
## bty_avg      0.07416    0.01625   4.563 6.48e-06 ***
## gendermale   0.17239    0.05022   3.433 0.000652 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5287 on 460 degrees of freedom
## Multiple R-squared:  0.05912,    Adjusted R-squared:  0.05503 
## F-statistic: 14.45 on 2 and 460 DF,  p-value: 8.177e-07
  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:

Building the residual plot:

plot(m_bty_gen$residuals ~ evals$bty_avg)
abline(h = 0, lty = 3)

Histogram of residuals:

hist(m_bty_gen$residuals)

Q-Q Plot:

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

Checking for the regression: 1) Linearity: The data looks to be linear. 2)Nearly normal Residuals: From the residuals histogram, it is clear the histogram is skewed on the left, and is not nearly normal. 3) Constant Variability: From the residual plot, similar to the previous one with the model with only bty_avg, the variability is not constant. 4) Indenpendent Observations : The observations are all independent.

However 1 and 4 conditions are met, 2 and 4 are not satisfied. Hence all the conditions are not met.

  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?

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.

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

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?

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.)

Answer:

score = 3.75 + 0.074 X bty_avg + 0.173 X gender where gender = 1 for male and gender = 0 for female

For males: score = 3.75 + 0.074 X bty_avg + 0.173

  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
multiLines(m_bty_rank)

R adds (n - 1) variables for a categorical value with n categories.

The interpretation of the coefficients in multiple regression is slightly different from that of simple regression. The estimate for bty_avg reflects how much higher a group of professors is expected to score if they have a beauty rating that is one point higher while holding all other variables constant. In this case, that translates into considering only professors of the same rank with bty_avg scores that are one point apart.

The search for the best model

We will start with a full model that predicts professor score based on rank, ethnicity, gender, language of the university where they got their degree, age, proportion of students that filled out evaluations, class size, course level, number of professors, number of credits, average beauty rating, outfit, and picture color.

  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 variable pic_outfit is expected to be not a predictor for the 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)
  1. Check your suspicions from the previous exercise. Include the model output in your response.

Answer: I was expecting the variable cls_profs (number of professors teaching sections in course in sample: single, multiple.) to be at least at some level a predictor of the score. But as we see the p-value of this variable is 0.77 which is very high. That means this variable cannot be a predictor of the score.

  1. Interpret the coefficient associated with the ethnicity variable. Answer: The coefficient for ethnicity of non-minority is 0.12 while 0 in case of a minority. That means for professors from non-minority have a score of 0.12 greater than the score for a minor ethiicity background professor.

  2. 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:Highest p-value from the full model is for cls_profssingle. Hence removing the variable cls_profs, and rebuilding the model.

m_full_minus_cls_profs <- lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval 
                             + cls_students + cls_level + cls_credits + bty_avg 
                             + pic_outfit + pic_color, data = evals)
summary(m_full_minus_cls_profs)
## 
## Call:
## lm(formula = score ~ rank + ethnicity + gender + language + age + 
##     cls_perc_eval + cls_students + cls_level + cls_credits + 
##     bty_avg + pic_outfit + pic_color, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.7836 -0.3257  0.0859  0.3513  0.9551 
## 
## Coefficients:
##                         Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.0872523  0.2888562  14.150  < 2e-16 ***
## ranktenure track      -0.1476746  0.0819824  -1.801 0.072327 .  
## ranktenured           -0.0973829  0.0662614  -1.470 0.142349    
## ethnicitynot minority  0.1274458  0.0772887   1.649 0.099856 .  
## gendermale             0.2101231  0.0516873   4.065 5.66e-05 ***
## languagenon-english   -0.2282894  0.1111305  -2.054 0.040530 *  
## age                   -0.0089992  0.0031326  -2.873 0.004262 ** 
## cls_perc_eval          0.0052888  0.0015317   3.453 0.000607 ***
## cls_students           0.0004687  0.0003737   1.254 0.210384    
## cls_levelupper         0.0606374  0.0575010   1.055 0.292200    
## cls_creditsone credit  0.5061196  0.1149163   4.404 1.33e-05 ***
## bty_avg                0.0398629  0.0174780   2.281 0.023032 *  
## pic_outfitnot formal  -0.1083227  0.0721711  -1.501 0.134080    
## pic_colorcolor        -0.2190527  0.0711469  -3.079 0.002205 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4974 on 449 degrees of freedom
## Multiple R-squared:  0.187,  Adjusted R-squared:  0.1634 
## F-statistic: 7.943 on 13 and 449 DF,  p-value: 2.336e-14
  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: Moving forward with the m_full_minus_cls_profs. The highest p-value is for the variable: cls_levelupper which is 0.29. Hence removing this variable cls_level

model2 <- lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval 
             + cls_students + cls_credits + bty_avg 
             + pic_outfit + pic_color, data = evals)
summary(model2)
## 
## Call:
## lm(formula = score ~ rank + ethnicity + gender + language + age + 
##     cls_perc_eval + cls_students + cls_credits + bty_avg + pic_outfit + 
##     pic_color, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.7761 -0.3187  0.0875  0.3547  0.9367 
## 
## Coefficients:
##                         Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.0856255  0.2888881  14.143  < 2e-16 ***
## ranktenure track      -0.1420696  0.0818201  -1.736 0.083184 .  
## ranktenured           -0.0895940  0.0658566  -1.360 0.174372    
## ethnicitynot minority  0.1424342  0.0759800   1.875 0.061491 .  
## gendermale             0.2037722  0.0513416   3.969 8.40e-05 ***
## languagenon-english   -0.2093185  0.1096785  -1.908 0.056966 .  
## age                   -0.0087287  0.0031224  -2.795 0.005404 ** 
## cls_perc_eval          0.0053545  0.0015306   3.498 0.000515 ***
## cls_students           0.0003573  0.0003585   0.997 0.319451    
## cls_creditsone credit  0.4733728  0.1106549   4.278 2.31e-05 ***
## bty_avg                0.0410340  0.0174449   2.352 0.019092 *  
## pic_outfitnot formal  -0.1172152  0.0716857  -1.635 0.102722    
## pic_colorcolor        -0.1973196  0.0681052  -2.897 0.003948 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4975 on 450 degrees of freedom
## Multiple R-squared:  0.185,  Adjusted R-squared:  0.1632 
## F-statistic:  8.51 on 12 and 450 DF,  p-value: 1.275e-14

Now we see that the highest p-value is for cls_students (0.31) which is greater than SL of 0.05. Hence moving forward and removing this field from the regression model:

model3 <- lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval 
             + cls_credits + bty_avg 
             + pic_outfit + pic_color, data = evals)
summary(model3)
## 
## Call:
## lm(formula = score ~ rank + ethnicity + gender + language + age + 
##     cls_perc_eval + cls_credits + bty_avg + pic_outfit + pic_color, 
##     data = evals)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.78424 -0.31397  0.09261  0.35904  0.92154 
## 
## Coefficients:
##                        Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.152893   0.280892  14.785  < 2e-16 ***
## ranktenure track      -0.142239   0.081819  -1.738 0.082814 .  
## ranktenured           -0.083092   0.065532  -1.268 0.205469    
## ethnicitynot minority  0.143509   0.075972   1.889 0.059535 .  
## gendermale             0.208080   0.051159   4.067 5.61e-05 ***
## languagenon-english   -0.222515   0.108876  -2.044 0.041558 *  
## age                   -0.009074   0.003103  -2.924 0.003629 ** 
## cls_perc_eval          0.004841   0.001441   3.359 0.000849 ***
## cls_creditsone credit  0.472669   0.110652   4.272 2.37e-05 ***
## bty_avg                0.043578   0.017257   2.525 0.011903 *  
## pic_outfitnot formal  -0.136594   0.068998  -1.980 0.048347 *  
## pic_colorcolor        -0.189905   0.067697  -2.805 0.005246 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4975 on 451 degrees of freedom
## Multiple R-squared:  0.1832, Adjusted R-squared:  0.1632 
## F-statistic: 9.193 on 11 and 451 DF,  p-value: 6.364e-15

Now we see that the variable ranktenuredhas the highest p-value of 0.2 which is greater than 0.05. Hence removing this column:

model4 <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval 
             + cls_credits + bty_avg 
             + pic_outfit + pic_color, data = evals)
summary(model4)
## 
## Call:
## lm(formula = score ~ ethnicity + gender + language + age + cls_perc_eval + 
##     cls_credits + bty_avg + pic_outfit + pic_color, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.8455 -0.3221  0.1013  0.3745  0.9051 
## 
## Coefficients:
##                        Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            3.907030   0.244889  15.954  < 2e-16 ***
## ethnicitynot minority  0.163818   0.075158   2.180 0.029798 *  
## gendermale             0.202597   0.050102   4.044 6.18e-05 ***
## languagenon-english   -0.246683   0.106146  -2.324 0.020567 *  
## age                   -0.006925   0.002658  -2.606 0.009475 ** 
## cls_perc_eval          0.004942   0.001442   3.427 0.000666 ***
## cls_creditsone credit  0.517205   0.104141   4.966 9.68e-07 ***
## bty_avg                0.046732   0.017091   2.734 0.006497 ** 
## pic_outfitnot formal  -0.113939   0.067168  -1.696 0.090510 .  
## pic_colorcolor        -0.180870   0.067456  -2.681 0.007601 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4982 on 453 degrees of freedom
## Multiple R-squared:  0.1774, Adjusted R-squared:  0.161 
## F-statistic: 10.85 on 9 and 453 DF,  p-value: 2.441e-15

Now wee that the column pic_outfitnot formal has the highest p-value of 0.09 which is greater than 0.05. Hence removing this field:

model5 <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval 
             + cls_credits + bty_avg 
             + pic_color, data = evals)
summary(model5)
## 
## 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

Now the biggest p-value is for languagenon-english which is 0.047. As it is less than 0.05, it can be ignored. And this is the final model.

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

Answer: Building the residual plot:

plot(model5$residuals ~ model5$fitted.values)
abline(h = 0, lty = 3)

Histogram of residuals:

hist(model5$residuals)

Q-Q Plot:

qqnorm(model5$residuals)
qqline(model5$residuals)

Checking for the regression: 1) Linearity: The data looks to be linear. 2)Nearly normal Residuals: From the residuals histogram, it is clear the histogram is skewed on the left, and is not nearly normal. 3) Constant Variability: From the residual plot, similar to the previous one with the model with only bty_avg, the variability is not constant. 4) Indenpendent Observations : The observations are all independent.

  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: It might be impacted. There might be multiple courses taken by the same professor, so in that case the sameprofessor might have got 2 different scores for the 2 different courses he or she takes.

  2. 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: High evaluation score means the categorical variables with the positive values have the category as represented by the model, and the categorical variables with a negative coefficient has the other category, so that the score becomes maximum. Hence, ethnicity of non-minority, gender male, language english, cls_credit of one level, pic_color value as black & white - will be the ones with the highest scores.

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

Answer: No, the conclusions cannot be generalized to other universities, as there can be many other factors impacting the scores.

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.