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 an observational study. With the way the research question is phrased, it would not be possible to answer the question as is, because beauty is subjective. There are no set standards to measure beauty. Also since the study is observational, the variables can only show correlation, not causation.
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?
hist(evals$score)
The distribution is left skewed. This tells me that students typically only evaluate good teachers, which is expected.
Exercise 3 -
Excluding score, select two other variables and describe their relationship using an appropriate visualization (scatterplot, side-by-side boxplots, or mosaic plot).
boxplot(evals$cls_perc_eval~evals$gender)
I compared the gender of the professor to the percentage of the class that completed evaluations. BOth genders had wide spread of the class completing evaluations however, the female professors had a higher percentage overall.
plot(evals$score ~ evals$bty_avg)
Simple Linear Regression
Exercise 4 -
Replot the scatterplot, but this time use the function jitter() on the y- or the x-coordinate. (Use ?jitter to learn more.) What was misleading about the initial scatterplot?
?jitter
plot(evals$score~jitter(evals$bty_avg))
Using the jitter function allows us to see more noise surronding some of the data points which tells us that these points possibly hold more weight, which we couldn’t see in the initial plot.
Exercise 5 -
Let’s see if the apparent trend in the plot is something more than natural variation. Fit a linear model called m_bty to predict average professor score by average beauty rating and add the line to your plot using abline(m_bty). Write out the equation for the linear model and interpret the slope. Is average beauty score a statistically significant predictor? Does it appear to be a practically significant predictor?
m_bty <- lm(evals$score~evals$bty_avg)
plot(evals$bty_avg, evals$score)
abline(m_bty)
summary(m_bty)
##
## Call:
## lm(formula = evals$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
y = 3.88 + 0.067(bty_avg)
Statistically, average beauty score is a significant predictor since the p-value is practically 0. However, practically, average beauty score is not a signifcant predictor because the multiple R2 value is very small, 0.035. Plus, for every icrement the beauty average increases, the evaluation does not increase by much.
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.
plot(m_bty_gen$residuals~evals$bty_avg)
abline(h=0,lty=3)
There is no evident pattern so the distribution is linear and variable.
hist(m_bty_gen$residuals)
qqnorm(m_bty_gen$residuals)
qqline(m_bty_gen$residuals)
The QQ plot shows evidence of outliers and some skewing.
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?
Yes. After adding gender to the model, the p-value got even smaller.
multiLines(m_bty_gen)
Exercise 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?
scoreˆ=β̂0+β̂1×bty_avg+β̂2×(1)=β̂0+β̂1×bty_avg+β̂2
Males tend to have a higher evaluation despite receiving the same beauty rating.
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.
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 creates one less rank, leaving us with just ranktenure track and ranktenured.
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 think the number of professors variable would have the highest p-value because it shouldn’t affect how students would evaluate their professors.
Exercise 12
Check your suspicions from the previous exercise. Include the model output in your response.
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
My suspicions were correct as the the cls_profssingle variable had a p-value of 0.778.
Exercise 13
Interpret the coefficient associated with the ethnicity variable.
When all the other variables are constant, a professor who isn’t a minority would see an increase in their score by 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?
m_full.noprof <- 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.noprof)
##
## 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
Some of the coefficients and p-values for the remaining variables did change.
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.
m_full_best <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval + cls_credits + bty_avg + pic_color, data = evals)
summary(m_full_best)
##
## 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
y = 3.77 + 0.167(ethnicity) + 0.207(gender) - 0.206(language) - 0.006(age) + 0.004(percent) + 0.505(credits) + 0.051(beaty average) - 0.19(color)
Exercise 16
Verify that the conditions for this model are reasonable using diagnostic plots.
plot(m_full_best$residuals~evals$bty_avg)
abline(h=0,lty=3)
Distribution is constantly variable and linear.
hist(m_full_best$residuals)
qqnorm(m_full_best$residuals)
qqline(m_full_best$residuals)
Despite some outliers at the tails, the distribution still seems nearly normal.
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.
Non-minority males who graduated from an English-speaking university and teaches a one credit course, would be associated with a high evaluation score.
Exercise 19
Would you be comfortable generalizing your conclusions to apply to professors generally (at any university)? Why or why not?
No because the importance of these variables could vary based on region and/or student and professor demographics.
