load("more/evals.RData")
  1. Is this an observational study or an experiment? The original research question posed in the paper is whether beauty leads directly to the differences in course evaluations. Given the study design, is it possible to answer this question as it is phrased? If not, rephrase the question.

It is an observational study. This could also be considered as a classification problem and we could use logistical regression to answer the question.

  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?

This is unimodal distribution that is highly right skewed. Based on the plots below, high percentage of students rate courses favourably. This doesn’t surprise me because of most of the time student take the courses they like. Students often take classes that have favourable opinions or positive reviews(ratemyprofessor.com).

hist(evals$score, probability = TRUE)

qqnorm(evals$score)
qqline(evals$score)

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

I am selecting beauty and age. There is negative linear relationship between age and average beauty. As age increas, average beauty decreases.

plot(evals$bty_avg~evals$age)

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?

  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?
evals_jitter <- evals
evals_jitter$bty_avg <- jitter(evals_jitter$bty_avg)
plot(evals_jitter$score ~ evals_jitter$bty_avg)

  1. Let’s see if the apparent trend in the plot is something more than natural variation. Fit a linear model called m_bty to predict average professor score by average beauty rating and add the line to your plot using abline(m_bty). Write out the equation for the linear model and interpret the slope. Is average beauty score a statistically significant predictor? Does it appear to be a practically significant predictor?
m_bty<- lm(evals$score~evals$bty_avg)
summary(m_bty)
## 
## Call:
## lm(formula = evals$score ~ evals$bty_avg)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.9246 -0.3690  0.1420  0.3977  0.9309 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    3.88034    0.07614   50.96  < 2e-16 ***
## evals$bty_avg  0.06664    0.01629    4.09 5.08e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5348 on 461 degrees of freedom
## Multiple R-squared:  0.03502,    Adjusted R-squared:  0.03293 
## F-statistic: 16.73 on 1 and 461 DF,  p-value: 5.083e-05
plot(evals$score~evals$bty_avg)
abline(m_bty)

\[ \begin{aligned} score = 3.87857 + 0.06702 \times bty\_avg\end{aligned} \]

This doesnt appear to be a good predictor. Multiple R-squared: 0.03502, so this model only explains .03% of the variability.

  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).

Based on the plots below, the residuals distribution is not normal and aslo the variability does not seems to be constant. There are points below -1 but there are no points above 1. Normal qq plot also shows the distribution is not normal and right skewed.

hist(m_bty$residuals)

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

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

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)
## [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:

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.

Based on the plots below, the residuals distribution is not nearly normal and the variability does not seems to be constant. There are points below -1 but there are no points above 1. Normal qq plot also shows the distribution is right skewed.

hist(m_bty_gen$residuals)

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

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

  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?

bty_avg is not a significant predictor of score. Adding gender has made the model a slightly better predictor. Multiple R-squared: 0.05912. This model explains 2% more variability than previous model with only 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?

Based on the model, male professors received on averge 0.17239 more than female professors. \[ \begin{aligned} \widehat{score} &= \hat{\beta}_0 + \hat{\beta}_1 \times bty\_avg + \hat{\beta}_2 \times (1) \\ &= 3.74734 +0.074161 \times bty\_avg\ + 0.17239 \end{aligned} \]

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.)

  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.

Linear model will have one less Coefficient. In this example,tenure trackand tenuredare displayed in the summary data. bty_avg is really bty_avg + teaching.

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

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.

cls_levelupper. I simply dont see how the class level impacts how students rates the professor.

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

This has a pvalue of .29 and doesn’t improve Adjusted R-squared. Adjusted R-squared includingcls_levelupper : 0.1617 Adjusted R-squared excludingcls_levelupper : 0.1615

m_full <- lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval 
             + cls_students  + 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_profs + cls_credits + 
##     bty_avg + pic_outfit + pic_color, data = evals)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.76617 -0.31818  0.09325  0.35169  0.93485 
## 
## Coefficients:
##                         Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.0938164  0.2905587  14.089  < 2e-16 ***
## ranktenure track      -0.1419977  0.0819039  -1.734 0.083656 .  
## ranktenured           -0.0895639  0.0659238  -1.359 0.174957    
## ethnicitynot minority  0.1383380  0.0773580   1.788 0.074405 .  
## gendermale             0.2046338  0.0514798   3.975  8.2e-05 ***
## languagenon-english   -0.2109231  0.1099296  -1.919 0.055655 .  
## age                   -0.0087375  0.0031258  -2.795 0.005407 ** 
## cls_perc_eval          0.0053940  0.0015382   3.507 0.000499 ***
## cls_students           0.0003430  0.0003622   0.947 0.344081    
## cls_profssingle       -0.0150776  0.0519931  -0.290 0.771956    
## cls_creditsone credit  0.4692494  0.1116766   4.202  3.2e-05 ***
## bty_avg                0.0412068  0.0174728   2.358 0.018785 *  
## pic_outfitnot formal  -0.1216791  0.0733912  -1.658 0.098026 .  
## pic_colorcolor        -0.1955247  0.0684550  -2.856 0.004486 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.498 on 449 degrees of freedom
## Multiple R-squared:  0.1851, Adjusted R-squared:  0.1615 
## F-statistic: 7.845 on 13 and 449 DF,  p-value: 3.699e-14
  1. Interpret the coefficient associated with the ethnicity variable.

Non minority professors were rated higher on average by 0.1383380points.

  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?

cls_profs has the highest p-value 0.771956 Adjusted R-squared includingcls_profs : 0.1617 Adjusted R-squared excludingcls_profs : 0.1634 Coefficients of other variabels changed after dropping cls_profs from the model. This tells us that there is correlation among predictor variables.

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

All Adjusted R-squared: 0.1617 excluding cls_profs : 0.1634 Remoing cls_profs improved R-squared value. Removing other variables decreased the value of Adjusted R-squared.

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

Near Normal Residuals: Based on these plots, the distribution is not normal. Residuals distribution is skewed.

Constant variability : No. There are alot of points below -1.5 and no points above 1.

hist(m_full$residuals)

qqnorm(m_full$residuals)
qqline(m_full$residuals)

plot(m_full$residuals)
abline(h = 0, lty = 3)

  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?

Each observations should be independent. If a student took more than one course by a profesosr, then the observations by that student wouldn’t be independent.

  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.

Non minority male whose primary language is english and teaches one credit/ low credit course.

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

It can be generalized to these university’s population. It could also be generalized to schools with similar populations. However, to generalize this to entire US would require more work.

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.