download.file("http://www.openintro.org/stat/data/evals.RData", destfile = "evals.RData")
load("evals.RData")

Exercise 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 a observational study but because there isn’t a control group or a experimental group we cannot answer the question as it is right now.

Exercise 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?

The distribution is left skewed. I was expecting for the data to show me a normal skew distribution. This would mean that the students rate the courses high.

hist(evals$score)

Exercise 3

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

Woman average higher in beauty than males do.

boxplot(evals$bty_avg ~ evals$gender)

Simple linear regression

plot(evals$score ~ evals$bty_avg)

Exercise 4

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?

In the new scatterplot using the jitter fuction. It shows more points than the origical scatter plot. Also the points in the new scatter plot show some point overlapping each other.

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

Exercise 5

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?

The equation for the linear model is y = 0.0666 + 3.880. The p-value for this scatter plot is 0.03 which would show a strong evidence of being statistically significant but because most of the points doing fall on to the line it makes it not a practical significant predictor.

m_bty <-lm(evals$score ~ evals$bty_avg)
plot(jitter(evals$score) ~ evals$bty_avg)
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

Exercise 6

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).

The residual seems to look reasonable based on the plots. The only thing I point out is that the distribution is left skewed showing both in the Q-Q plot and the histogram.

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

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

hist(m_bty$residuals)

boxplot(m_bty$residuals)

Multiple linear regression

plot(evals$bty_avg ~ evals$bty_f1lower)

cor(evals$bty_avg, evals$bty_f1lower)
## [1] 0.8439112
plot(evals[,13:19])

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

Exercise 7

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.

The condition seem to be reasonable based on the plot. It also shows the distribution to be left skewed on the Q-Q plot.

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

Exercise 8

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 still a significant predictor of score. Based on the boxplot the males seem to have a higher score than the females.

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
boxplot(evals$score ~ evals$gender)
abline(h = 0)

multiLines(m_bty_gen)

Exerise 9

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 equation for the males is y = 0.172 + 3.747. Males have a higher course evaluation score.

Exercise 10

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.

It appear that R used rank tenure track and rank tenured. What i think R does is instead of using 3 ranks it splits it and uses 2 rank variables.

m_bty_rank <- lm(evals$score ~ evals$bty_avg + evals$rank)
summary(m_bty_rank)
## 
## Call:
## lm(formula = evals$score ~ evals$bty_avg + evals$rank)
## 
## 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 ***
## evals$bty_avg           0.06783    0.01655   4.098 4.92e-05 ***
## evals$ranktenure track -0.16070    0.07395  -2.173   0.0303 *  
## evals$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 search for the best model

Exercise 11

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 would think that possibly the credits would have the highest p-value. I dont think pic_outfit or pic_color would have a high value but lets see and run the test!

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

Exercise 12

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

Cls_profssingle has the highest p-value which I honestly didn’t think would have a high p-value. The cls_credits has the second highest p-value.

Exercise 13

Interpret the coefficient associated with the ethnicity variable.

0.123 is the p-value for the ethnicity not minority variable. Which means If the professor isn’t a minority, the professor would have a score of 0.123.

Exercise 14

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?

Defiantly dropping the highest p-value did change the coefficient and significance of the other explanatory variable.It does matter about the professor status of being a minority or not. The ethnicity not minority has a big role in the evaluation score of the professors.

m_full <- lm(score ~ rank + 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 + 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 
## -1.73681 -0.32734  0.08283  0.35834  0.98639 
## 
## Coefficients:
##                         Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            4.2676351  0.2694274  15.840  < 2e-16 ***
## ranktenure track      -0.1660677  0.0813523  -2.041 0.041801 *  
## ranktenured           -0.1127978  0.0657022  -1.717 0.086705 .  
## gendermale             0.2241744  0.0512176   4.377 1.50e-05 ***
## languagenon-english   -0.2862448  0.1055924  -2.711 0.006968 ** 
## age                   -0.0092040  0.0031385  -2.933 0.003534 ** 
## cls_perc_eval          0.0051119  0.0015357   3.329 0.000944 ***
## cls_students           0.0004785  0.0003777   1.267 0.205899    
## cls_levelupper         0.0767503  0.0567182   1.353 0.176677    
## cls_profssingle       -0.0292174  0.0512393  -0.570 0.568817    
## cls_creditsone credit  0.4589918  0.1128358   4.068 5.61e-05 ***
## bty_avg                0.0375980  0.0174661   2.153 0.031880 *  
## pic_outfitnot formal  -0.1208610  0.0738165  -1.637 0.102265    
## pic_colorcolor        -0.2400696  0.0701264  -3.423 0.000675 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4988 on 449 degrees of freedom
## Multiple R-squared:  0.1826, Adjusted R-squared:  0.159 
## F-statistic: 7.717 on 13 and 449 DF,  p-value: 6.792e-14

Exercise 15

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.

score = 3.887 + 0.218 + (-0.290) + (-0.006)+ 0.005 + 0.001 + 0.005 + 0.001 + 0.080 + (-0.019) + 0.507 + 0.044 + (-0.244)

best_model <- lm(score ~ gender + language + age + cls_perc_eval 
             + cls_students + cls_level + cls_profs + cls_credits + bty_avg 
             + pic_color, data = evals)
summary(best_model)
## 
## Call:
## lm(formula = score ~ gender + language + age + cls_perc_eval + 
##     cls_students + cls_level + cls_profs + cls_credits + bty_avg + 
##     pic_color, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.8074 -0.3220  0.0918  0.3767  0.9527 
## 
## Coefficients:
##                         Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            3.8871312  0.2203699  17.639  < 2e-16 ***
## gendermale             0.2185742  0.0504303   4.334 1.80e-05 ***
## languagenon-english   -0.2896975  0.0999268  -2.899 0.003925 ** 
## age                   -0.0061869  0.0026414  -2.342 0.019598 *  
## cls_perc_eval          0.0050854  0.0015391   3.304 0.001028 ** 
## cls_students           0.0005987  0.0003533   1.695 0.090852 .  
## cls_levelupper         0.0797485  0.0560934   1.422 0.155800    
## cls_profssingle       -0.0188742  0.0502694  -0.375 0.707494    
## cls_creditsone credit  0.5070844  0.1090804   4.649 4.39e-06 ***
## bty_avg                0.0437243  0.0172619   2.533 0.011646 *  
## pic_colorcolor        -0.2438181  0.0689580  -3.536 0.000449 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.5007 on 452 degrees of freedom
## Multiple R-squared:  0.1708, Adjusted R-squared:  0.1525 
## F-statistic: 9.312 on 10 and 452 DF,  p-value: 4.347e-14

Exercise 16

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

Based on the the plots it shows the distribution is left skewed but the condition seem to be reasonable. We can possibly assume it’s a normal distribution.

qqnorm(best_model$residuals)
qqline(best_model$residuals)

boxplot(best_model$residuals)

plot(best_model$residuals)
abline(h = 0)

Exercise 17

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?

Yes this would impact the conditions of the the linear regression because these observations are based on the course that each professor teaches. It is also based on the score.

Exercise 18

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.

gender male, language-english, age, high percent eval score, number of students in class, class level upper, professor has to be single, depends on course credit, beauty average, presentation in picture color.

Exercise 19

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

I wouldn’t feel comfortable because each university is different, which means different professors. The features of the professor will be different in each university including who they are and the level of the course they teach. Could be a high or lower credit class. That’s why I wouldn’t feel comfortable generalizing my application to any unveristy.