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.

download.file("http://www.openintro.org/stat/data/evals.RData", destfile = "evals.RData")
load("evals.RData")
head(evals)
##   score         rank    ethnicity gender language age cls_perc_eval
## 1   4.7 tenure track     minority female  english  36      55.81395
## 2   4.1 tenure track     minority female  english  36      68.80000
## 3   3.9 tenure track     minority female  english  36      60.80000
## 4   4.8 tenure track     minority female  english  36      62.60163
## 5   4.6      tenured not minority   male  english  59      85.00000
## 6   4.3      tenured not minority   male  english  59      87.50000
##   cls_did_eval cls_students cls_level cls_profs  cls_credits bty_f1lower
## 1           24           43     upper    single multi credit           5
## 2           86          125     upper    single multi credit           5
## 3           76          125     upper    single multi credit           5
## 4           77          123     upper    single multi credit           5
## 5           17           20     upper  multiple multi credit           4
## 6           35           40     upper  multiple multi credit           4
##   bty_f1upper bty_f2upper bty_m1lower bty_m1upper bty_m2upper bty_avg
## 1           7           6           2           4           6       5
## 2           7           6           2           4           6       5
## 3           7           6           2           4           6       5
## 4           7           6           2           4           6       5
## 5           4           2           2           3           3       3
## 6           4           2           2           3           3       3
##   pic_outfit pic_color
## 1 not formal     color
## 2 not formal     color
## 3 not formal     color
## 4 not formal     color
## 5 not formal     color
## 6 not formal     color
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.

    This is observational study since we have no control over the variables. An observational study can’t establish causation between the explanatory and response variables. Instead we can calculate a correlation. I would change the original question to whether beauty has positive or negative effect on course evaluation scores.

  2. 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?

hist(evals$score)

The evaluation scores are skewed to the left. It means that there far more positive evaluations than negative evaluations. I expected that evaluation scores are normally distributed (most teachers receive average scores and a few teachers receive very low or very high scores).

  1. Excluding score, select two other variables and describe their relationship using an appropriate visualization (scatter plot, side-by-side box plots, or mosaic plot).
boxplot(evals$bty_avg ~ evals$language)
regression_line <- lm(evals$bty_avg ~ evals$language)
abline(regression_line)

I selected the variable average beauty score and the variable language, The median of average beauty score is higher for professors that graduated from non-english speaking schools than for english speaking schools. However, 50% of professors that graduated from english speaking schools (the range from 1st to 3rd quartiles) have higher average beauty scores than 50% of professors that graduated from non-english speaking schools. Furthermore, the maximum average beauty score for professors that graduated from english speaking schools is higher than the maximum average beauty score for professors that graduated from non-english speaking schools. While the minimum average beauty score of professors that graduated from english speaking schools is higher than the minimum average beauty score of professors that graduated from non-english speaking schools. Moreover, average beauty scores for professors that graduated from non-english speaking schools contain an outlier while average beauty scores for professors that graduated from english speaking schools don’t contain an outlier.

By looking at the regression line we can conclude that there is a slight positive linear trend.

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?

dim(evals)
## [1] 463  21

There are 463 observations in the data set. However, the number of points on the scattlebutt is much less than 463. I guess that we can’t see all points because some points overlaps with each other.

  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?
plot(jitter(evals$score,factor=100) ~ jitter(evals$bty_avg,factor=100))

Now we can see points that overlapped with other points in the previous scatterplot.

  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.
m_bty <- lm(evals$score ~ evals$bty_avg)
plot(jitter(evals$score,factor=1000) ~ jitter(evals$bty_avg,factor=1000))
abline(m_bty)

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
The equation of the linear module is:

score = 3.88034 + 0.06664*bty_avg

The slope of 0.06664 means that a single increase in beauty average score increases evulation score by 0.06664.

Is average beauty score a statistically significant
predictor? Does it appear to be a practically significant predictor?

As p-value is close to 0 average beauty score is a statistically significant predictor of evaluation score. However, in practice, it might not be a significant predictor since for a single increase in average beauty score increases evaluation score by a very small number of 0.06664 (that doesn't change evaluation score sagnificatly).

  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).

    Let’s draw a scatterplot and a regression line.

regression_line <- lm(evals$score ~ evals$bty_avg)
plot(jitter(evals$score,factor=100) ~ jitter(evals$bty_avg,factor=100),xlab = "bty_avg",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'bty_avg'")
abline(regression_line)

summary(regression_line)
## 
## 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$bty_avg  0.06664    0.01629    4.09 5.08e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5348 on 461 degrees of freedom
## Multiple R-squared:  0.03502,    Adjusted R-squared:  0.03293 
## F-statistic: 16.73 on 1 and 461 DF,  p-value: 5.083e-05
cor(evals$score,evals$bty_avg)
## [1] 0.1871424

In order to apply the least squares method the linear regression should meet four conditions.

The first condition is linearity.

The relationships between evaluation score and average beauty score are week since the correlation coefficient is mush less than 0.5. Looking at the correlation of 0.1871424 between two variables we can say that there is very week positive linear association between evaluation score and average beauty score. The plot travels upwards from left to right. It shows a tiny rate of increase. The condition is met as we can see slight linear trend.

The second condition is nearly normal residuals.

hist(regression_line$residuals)

qqnorm(regression_line$residuals)
qqline(regression_line$residuals)  # adds diagonal line to the normal prob plot

By looking at the histogram we can state that the residuals distribution is unimodal (one clear peak) and bell-shaped but slightly left skewed. Moreover, by looking at normal probability plot we can see that most of the points either fall on the regression line or slightly deviate from the line. So that, we can conclude that distribution of residual is close to normal distribution. It means that the nearly normal residuals condition appear to be met.

The third condition is constant variability.

plot(jitter(regression_line$residuals,10000) ~ jitter(evals$bty_avg,10000))
abline(h = 0, lty = 3)  # adds a horizontal dashed line at y = 0

I believe that the variability condition appear to be met as residuals values are pretty equally and randomly spaced around the horizontal axis.

The forth condition is independent observations. By looking the scatter plot we can conclude that there is no pattern that suggests that observations depend on each other. So, most likely the observations are independent.

We’ve verified that all conditions of least squared regression are reasonable because all four conditions are met.

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.

Let’s verify that all four condition for this model are reasonable.

The first condition is linearity. Let’s check whether each variable is linearly related to the outcome.

Let’s start with avg_bty variable draw a scatterplot and a regression line.

regression_line <- lm(evals$score ~ evals$bty_avg)
plot(jitter(evals$score,factor=100) ~ jitter(evals$bty_avg,factor=100),xlab = "bty_avg",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'bty_avg'")
abline(regression_line)

summary(regression_line)
## 
## 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$bty_avg  0.06664    0.01629    4.09 5.08e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5348 on 461 degrees of freedom
## Multiple R-squared:  0.03502,    Adjusted R-squared:  0.03293 
## F-statistic: 16.73 on 1 and 461 DF,  p-value: 5.083e-05
cor(evals$score,evals$bty_avg)
## [1] 0.1871424

The relationships between evaluation score and average beauty score are week since the correlation coefficient is mush less than 0.5. Looking at the correlation of 0.1871424 between two variables we can say that there is very week positive linear association between evaluation score and average beauty score. The plot travels upwards from left to right. It shows a tiny rate of increase. We can observe a slight linear trend.

Now let’s take a look at the variable ‘gender’.

plot(jitter(evals$score,factor=100) ~ evals$gender,xlab = "bty_avg",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'bty_avg'")
m_bty_gen2 <- lm(score ~ gender, data = evals)
abline(m_bty_gen2)

By looking at the boxplots and the regression line we can see a slight positive linear trend.

It means that the linearity condition appear to be met.

The second condition is nearly normal residuals.

hist(m_bty_gen$residuals)

qqnorm(m_bty_gen$residuals)
qqline(m_bty_gen$residuals)  # adds diagonal line to the normal prob plot

By looking at the histogram we can state that the residuals distribution is unimodal (one clear peak) and bell-shaped but slightly left skewed. Moreover, by looking at normal probability plot we can see that most of the points fall to the regression line or slightly deviate from the regression line. So that, we can conclude that distribution of residual is close to normal distribution. It means that the nearly normal residuals condition appear to be met.

The third condition is constant variability.

plot(jitter(m_bty_gen$residuals,10000) ~ jitter(evals$bty_avg,10000))
abline(h = 0, lty = 3)  # adds a horizontal dashed line at y = 0

I believe that the variability condition appear to be met as residuals values are pretty equally and randomly spaced around the horizontal axis.

The forth condition is independent observations. By looking at the order of data collection plot we can conclude that there is no pattern that suggests that observations depend on each other. So, most likely the observations are independent.

We verified that all four condition for this model are reasonable.

  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?

Average beauty score is still a significant variable since p of bty_avg is even smaller than in a previous model.

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.)

    \[ \begin{aligned} \widehat{score} &= \hat{\beta}_0 + \hat{\beta}_1 \times bty\_avg + \hat{\beta}_2 \times (1) \\ &= \hat{\beta}_0 + \hat{\beta}_1 \times bty\_avg\end{aligned}+\hat{\beta}_2 \]

For two professors who received the same beauty rating, which gender tends to have the higher course evaluation score?

score =3.74734 + 0.07416bty_avg + 0.17239 (1)

Male professors tend to have an evaluation score higher by 0.17239.

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.
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)

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.

    I think that the variable that logically is not related to professors evaluation score will have the highest p-value. For example, it can be number of professors teaching sections in course in sample.

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.

The variable ‘cls_profs’ (number of professors teaching sections in course in sample) has higher p (0.77806) value in the model. Moreover, it’s greater than significant level of 0.05. It means that the variable ‘cls_profs’ has no assassination with professor score.

  1. Interpret the coefficient associated with the ethnicity variable.

The variable ‘ethnicity’ is not statistically significant since it has p-value (0.11698 ) that is higher than significant level of 0.05. If the variable ‘ethnicity’ were statistically significant professor who don’t belong to minority tend to score 0.1234929 higher while holding all other variables constant.

  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?
m_full <- lm(score ~ gender + language + age + cls_perc_eval 
             + cls_credits + bty_avg + pic_color, data = evals)
summary(m_full)
## 
## Call:
## lm(formula = score ~ gender + language + age + cls_perc_eval + 
##     cls_credits + bty_avg + pic_color, data = evals)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.81919 -0.32035  0.09272  0.38526  0.88213 
## 
## Coefficients:
##                        Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            3.967255   0.215824  18.382  < 2e-16 ***
## gendermale             0.221457   0.049937   4.435 1.16e-05 ***
## languagenon-english   -0.281933   0.098341  -2.867  0.00434 ** 
## age                   -0.005877   0.002622  -2.241  0.02551 *  
## cls_perc_eval          0.004295   0.001432   2.999  0.00286 ** 
## cls_creditsone credit  0.444392   0.100910   4.404 1.33e-05 ***
## bty_avg                0.048679   0.016974   2.868  0.00432 ** 
## pic_colorcolor        -0.216556   0.066625  -3.250  0.00124 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5014 on 455 degrees of freedom
## Multiple R-squared:  0.1631, Adjusted R-squared:  0.1502 
## F-statistic: 12.67 on 7 and 455 DF,  p-value: 6.996e-15

After statistically insignificant variables (which p-values is greater than significant level of 0.05) were removed there was a slight change in the coefficients and significance of the remaining explanatory variables. All remaining variables became even more significant since their p-values decreased.

  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.
m_full_back <- lm(score ~ gender + language + age + cls_perc_eval 
             + cls_credits + bty_avg + pic_color, data = evals)
summary(m_full_back)
## 
## Call:
## lm(formula = score ~ gender + language + age + cls_perc_eval + 
##     cls_credits + bty_avg + pic_color, data = evals)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.81919 -0.32035  0.09272  0.38526  0.88213 
## 
## Coefficients:
##                        Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            3.967255   0.215824  18.382  < 2e-16 ***
## gendermale             0.221457   0.049937   4.435 1.16e-05 ***
## languagenon-english   -0.281933   0.098341  -2.867  0.00434 ** 
## age                   -0.005877   0.002622  -2.241  0.02551 *  
## cls_perc_eval          0.004295   0.001432   2.999  0.00286 ** 
## cls_creditsone credit  0.444392   0.100910   4.404 1.33e-05 ***
## bty_avg                0.048679   0.016974   2.868  0.00432 ** 
## pic_colorcolor        -0.216556   0.066625  -3.250  0.00124 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5014 on 455 degrees of freedom
## Multiple R-squared:  0.1631, Adjusted R-squared:  0.1502 
## F-statistic: 12.67 on 7 and 455 DF,  p-value: 6.996e-15

The equation of the linear module is:

score= 3.967255 + 0.221457gendermale - 0.281933languagenon-english - 0.005877age + 0.004295cls_perc_eval + 0.444392cls_creditsone credit + 0.048679bty_avg - 0.216556*pic_colorcolor

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

Let’s verify that all four condition for this model are reasonable.

The first condition is linearity. Let’s check whether each variable is linearly related to the outcome.

Let’s start with avg_bty variable draw a scatterplot and a regression line.

##linear condition for the variable "gender"
regression_line_gender <- lm(evals$score ~ evals$gender)
plot(evals$score ~ evals$gender,xlab = "gender",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'gender'")
abline(regression_line_gender)

##linear condition for the variable "language"
regression_line_language <- lm(evals$score ~ evals$language)
plot(evals$score ~ evals$language, xlab = "language",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'language'")
abline(regression_line_language)

##linear condition for the variable "age"
regression_line_age <- lm(evals$score ~ evals$age)
plot(jitter(evals$score,factor=100) ~ jitter(evals$age,factor=100),xlab = "age",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'age'")
abline(regression_line_age)

cor(evals$score,evals$age)
## [1] -0.107032
##linear condition for the variable "cls_perc_eval"
regression_line_cls_perc_eval <- lm(evals$score ~ evals$cls_perc_eval)
plot(evals$score ~ evals$cls_perc_eval, xlab = "cls_perc_eval",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'cls_perc_eval'")
abline(regression_line_cls_perc_eval)

##linear condition for the variable "cls_credits"
regression_line_cls_credits <- lm(evals$score ~ evals$cls_credits)
plot(evals$score ~ evals$cls_credits, xlab = "cls_credits",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'cls_credits'")
abline(regression_line_cls_credits)

##linear condition for the variable "bty_avg"
regression_line_bty_avg <- lm(evals$score ~ evals$bty_avg)
plot(jitter(evals$score,factor=100) ~ jitter(evals$bty_avg,factor=100),xlab = "bty_avg",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'bty_avg'")
abline(regression_line_bty_avg)

cor(evals$score,evals$bty_avg)
## [1] 0.1871424
##linear condition for the variable "pic_color"
regression_line_pic_color <- lm(evals$score ~ evals$pic_color)
plot(evals$score ~ evals$bty_avg, xlab = "pic_color",             
     ylab = "score",
     main = "Relationships between the variables 'score' and 'pic_color'")
abline(regression_line_pic_color)

By looking at the plots and the regression line we can conclude that there are linear trend in all plots. Some of the linear relationships are stronger wile other are weeker. Also, some of the linear relationships are positive whereas other linear relationships are negative.

The second condition is nearly normal residuals.

hist(m_full_back$residuals)

qqnorm(m_full_back$residuals)
qqline(m_full_back$residuals)  # adds diagonal line to the normal prob plot

By looking at the histogram we can state that the residuals distribution is unimodal (one clear peak) and bell-shaped but slightly left skewed. Moreover, by looking at normal probability plot we can see that most of the points fall to the regression line or slightly deviate from the regression line. So that, we can conclude that distribution of residual is close to normal distribution. It means that the nearly normal residuals condition appear to be met.

The third condition is constant variability.

plot(jitter(m_full_back$residuals,100000) ~ jitter(evals$bty_avg,100000))
abline(h = 0, lty = 3)  # adds a horizontal dashed line at y = 0

I believe that the variability condition appear to be met as residuals values are pretty equally and randomly spaced around the horizontal axis.

The forth condition is independent observations. By looking at the order of data collection plot we can conclude that there is no pattern that suggests that observations depend on each other. So, most likely the observations are independent.

We’ve verified that all four condition for this model are reasonable.

  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?

    This new information won’t impact any of the conditions of linear regression because courses are independent of each other. Hence, evaluation scores from one course is indpendent of the other even if the course is being taught by the same professor.

  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.

    According to the final linear model, the highest evaluation scores have relatively young male professors who graduated from English speaking schools and teaches a one credit course. Also, they have a high beauty average score from the students and their class photo is in black and white. He must also be relatively young. Moreover, most of their tuents must have completed the evaluation.

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

I would say no because the sample size of six is not big enough and the sense of beauty can vary from culture to culture.

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.