Description

David Simbandumwe
Grading the professor
Here, you will analyze the data from this study in order to learn what goes into a positive professor evaluation.
Getting Started
## Rows: 463
## Columns: 23
## $ course_id <int> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 1~
## $ prof_id <int> 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5,~
## $ score <dbl> 4.7, 4.1, 3.9, 4.8, 4.6, 4.3, 2.8, 4.1, 3.4, 4.5, 3.8, 4~
## $ rank <fct> tenure track, tenure track, tenure track, tenure track, ~
Load packages
In this lab, you will explore and visualize the data using the tidyverse suite of packages. The data can be found in the companion package for OpenIntro resources, openintro.
Let’s load the packages.
library(tidyverse) library(openintro) library(GGally)
This is the first time we’re using the GGally package. You will be using the ggpairs function from this package later in the lab.
The data
glimpse(evals)
## $ ethnicity <fct> minority, minority, minority, minority, not minority, no~
## $ gender <fct> female, female, female, female, male, male, male, male, ~
## $ language <fct> english, english, english, english, english, english, en~
## $ age <int> 36, 36, 36, 36, 59, 59, 59, 51, 51, 40, 40, 40, 40, 40, ~
## $ cls_perc_eval <dbl> 55.81395, 68.80000, 60.80000, 62.60163, 85.00000, 87.500~
## $ cls_did_eval <int> 24, 86, 76, 77, 17, 35, 39, 55, 111, 40, 24, 24, 17, 14,~
## $ cls_students <int> 43, 125, 125, 123, 20, 40, 44, 55, 195, 46, 27, 25, 20, ~
## $ cls_level <fct> upper, upper, upper, upper, upper, upper, upper, upper, ~
## $ cls_profs <fct> single, single, single, single, multiple, multiple, mult~
## $ cls_credits <fct> multi credit, multi credit, multi credit, multi credit, ~
## $ bty_f1lower <int> 5, 5, 5, 5, 4, 4, 4, 5, 5, 2, 2, 2, 2, 2, 2, 2, 2, 7, 7,~
## $ bty_f1upper <int> 7, 7, 7, 7, 4, 4, 4, 2, 2, 5, 5, 5, 5, 5, 5, 5, 5, 9, 9,~
## $ bty_f2upper <int> 6, 6, 6, 6, 2, 2, 2, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 9, 9,~
## $ bty_m1lower <int> 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 7, 7,~
## $ bty_m1upper <int> 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 6, 6,~
## $ bty_m2upper <int> 6, 6, 6, 6, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 6, 6,~
## $ bty_avg <dbl> 5.000, 5.000, 5.000, 5.000, 3.000, 3.000, 3.000, 3.333, ~
## $ pic_outfit <fct> not formal, not formal, not formal, not formal, not form~
## $ pic_color <fct> color, color, color, color, color, color, color, color, ~
We have observations on 21 different variables, some categorical and some numerical. The meaning of each variable can be found by bringing up the help file:
?evals
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 an observational study – There is no control group defined for this study
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?
evals %>% ggplot(aes(x=score)) + geom_histogram(binwidth=0.1) + labs (title = “Rate my Teacher” )
Rate my Teacher

evals %>%
ggplot() + geom_boxplot(mapping = aes(x=score, color=gender)) + labs (title = “Rate my Teacher (by gender)” )
Rate my Teacher (by gender)

summary(evals$score)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 2.300 3.800 4.300 4.175 4.600 5.000
The score is normally distributed by skewed right with a mean of 4.18 and a median 4.3
3. Excluding score, select two other variables and describe their relationship with each other using an appropriate visualization.
evals %>% ggplot(aes(x=age, y=bty_avg, color=gender )) + geom_point() + geom_jitter() + geom_smooth(method = lm) + labs (title = “Beauty Average mapped Against Age (grouped by gender)” )
Beauty Average mapped Against Age (grouped by gender)

There is a negative relationship between age and beauty. As the age of the professor goes up there is a tendency to rate the average beauty lower. Interestingly the age penalty for beauty is higher for male vs female lectures
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:
ggplot(data = evals, aes(x = bty_avg, y = score)) +
geom_point()

2 4 6 8 bty_avg
dim(evals)
## [1] 463 23
glimpse(evals$bty_avg)
## num [1:463] 5 5 5 5 3 …
glimpse(evals$score)
## num [1:463] 4.7 4.1 3.9 4.8 4.6 4.3 2.8 4.1 3.4 4.5 …
Before you 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?
There are far fewer visual data points than the 463 that should be graphed
4. Replot the scatterplot, but this time use geom_jitter as your layer. What was misleading about the initial scatterplot?
ggplot(data = evals, aes(x = bty_avg, y = score)) +
geom_jitter()
given the limited real estate for plotting it was difficult to get sense of density because points were being plotted directly over each other
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. 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?
Add the line of the best fit model to your plot using the following:
ggplot(data = evals, aes(x = bty_avg, y = score)) +
geom_jitter() + geom_smooth(method = “lm”) y = b1x + b0
m_bty <- lm(data=evals, score ~ bty_avg) 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
y = 0.067x + 3.880

intercept: at an average beauty score of zero the teachers score will be 3.880 slope: for each additional 1 unit increase in beauty score the teachers score will increase by 0.0.67 the p-value is less then 0.05 so we would reject the H0 in favor of the alternative hypothesis at the 0.05 confidence
The blue line is the model. The shaded gray area around the line tells you about the variability you might expect in your predictions. To turn that off, use se = FALSE.
ggplot(data = evals, aes(x = bty_avg, y = score)) +
geom_jitter() +
geom_smooth(method = “lm”, se = TRUE)

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).
par(mfrow = c(1,1)) plot(m_bty)

Fitted values lm(score ~ bty_avg)

Theoretical Quantiles lm(score ~ bty_avg)

Fitted values lm(score ~ bty_avg)

Leverage lm(score ~ bty_avg)
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.
ggplot(data = evals, aes(x = bty_f1lower, y = bty_avg)) +
geom_point()

2 4 6 8 bty_f1lower
evals %>% summarise(cor(bty_avg, bty_f1lower))
## # A tibble: 1 x 1
## ‘cor(bty_avg, bty_f1lower)‘
## <dbl>
## 1 0.844
As expected, the relationship is quite strong—after all, the average score is calculated using the individual scores. You can actually look at the relationships between all beauty variables (columns 13 through 19) using the following command:
evals %>% select(contains(“bty”)) %>% ggpairs()

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 you’ve accounted for the professor’s gender, you 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
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.
par(mfrow = c(1,1)) plot(m_bty_gen)

Fitted values lm(score ~ bty_avg + gender)

Theoretical Quantiles lm(score ~ bty_avg + gender)

Fitted values lm(score ~ bty_avg + gender)

Leverage lm(score ~ bty_avg + gender)
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?
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
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
yes – the addition of gender slightly improved 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 male and female to being an indicator variable called gendermale that takes a value of 0 for female professors and a value of 1 for male professors. (Such variables are often referred to as “dummy” variables.)
As a result, for female professors, the parameter estimate is multiplied by zero, leaving the intercept and slope form familiar from simple regression.
score[ = βˆ0 + βˆ1× bty_avg + βˆ2× (0) = βˆ0 + βˆ1× bty_avg
ggplot(data = evals, aes(x = bty_avg, y = score, color = pic_color)) +
geom_smooth(method = “lm”, formula = y ~ x, se = FALSE)

9. What is the equation of the line corresponding to those with color pictures? (Hint: For those with color pictures, the parameter estimate is multiplied by 1.) For two professors who received the same beauty rating, which color picture tends to have the higher course evaluation score?
m_bty_pic <- lm(score ~ bty_avg + pic_color, data = evals) summary(m_bty_pic)
##
## Call:
## lm(formula = score ~ bty_avg + pic_color, data = evals)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.8892 -0.3690 0.1293 0.4023 0.9125
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.06318 0.10908 37.249 < 2e-16 ***
## bty_avg 0.05548 0.01691 3.282 0.00111 **
## pic_colorcolor -0.16059 0.06892 -2.330 0.02022 *
## —
## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
##
## Residual standard error: 0.5323 on 460 degrees of freedom
## Multiple R-squared: 0.04628, Adjusted R-squared: 0.04213
## F-statistic: 11.16 on 2 and 460 DF, p-value: 1.848e-05
× bty_avg + b2× pic_colorcolor
× bty_avg − 0.161 × pic_colorcolor
× bty_avg − 0.161 × 0
× bty_avg
The decision to call the indicator variable gendermale instead of genderfemale 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.)
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
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
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.
cls_credits – I am not sure if the credits would impact the teachers evaluation score Let’s run the model…
m_full <- lm(score ~ rank + gender + ethnicity + 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 + ethnicity + 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
## gendermale 0.2109481 0.0518230 4.071 5.54e-05 ***
## ethnicitynot minority 0.1234929 0.0786273 1.571 0.11698
## 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
12. Check your suspicions from the previous exercise. Include the model output in your response.
i was wrong it looks like the number of professors teaching a course in the sample has the highest p-value of 0.778
13. Interpret the coefficient associated with the ethnicity variable.
all other variables being equal if the professor is a minority the associated score will increase by 0.123 units
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_partial <- lm(score ~ rank + gender + ethnicity + language + age + cls_perc_eval
+ cls_students + cls_level + cls_credits + bty_avg
+ pic_outfit + pic_color, data = evals)
summary(m_partial)
##
## Call:
## lm(formula = score ~ rank + gender + ethnicity + 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
## gendermale 0.2101231 0.0516873 4.065 5.66e-05 ***
## ethnicitynot minority 0.1274458 0.0772887 1.649 0.099856 .
## 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
yes – the coefficients and teh singnficance measures changed the dropped variable are not collinear because they are correlated and cannot be used to independently predict the dependent variable
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.
m1 <- lm(score ~ gender + + ethnicity + language + age + cls_perc_eval
+ cls_credits + bty_avg
+ pic_color, data = evals)
summary(m1)
##
## Call:
## lm(formula = score ~ gender + +ethnicity + 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 ***
## gendermale 0.207112 0.050135 4.131 4.30e-05 ***
## ethnicitynot minority 0.167872 0.075275 2.230 0.02623 *
## 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
m1 <- lm(score ~ gender + + ethnicity + age + cls_perc_eval
+ cls_credits + bty_avg
+ pic_color, data = evals)
summary(m1)
##
## Call:
## lm(formula = score ~ gender + +ethnicity + age + cls_perc_eval +
## cls_credits + bty_avg + pic_color, data = evals)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.85434 -0.33568 0.09247 0.38288 0.93903
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 3.690771 0.229181 16.104 < 2e-16 ***
## gendermale 0.201574 0.050220 4.014 6.99e-05 ***
## ethnicitynot minority 0.216955 0.071348 3.041 0.00250 **
## age -0.006034 0.002621 -2.302 0.02176 *
## cls_perc_eval 0.004719 0.001439 3.278 0.00113 **
## cls_creditsone credit 0.527806 0.103839 5.083 5.44e-07 ***
## bty_avg 0.052431 0.016975 3.089 0.00213 **
## pic_colorcolor -0.170149 0.066780 -2.548 0.01116 *
## —
## Signif. codes: 0 ’***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
##
## Residual standard error: 0.5008 on 455 degrees of freedom
## Multiple R-squared: 0.1649, Adjusted R-squared: 0.1521
## F-statistic: 12.84 on 7 and 455 DF, p-value: 4.344e-15
m_final <- lm(score ~ gender + + ethnicity + language + age + cls_perc_eval
+ cls_credits + bty_avg
+ pic_color, data = evals)
summary(m_final)
##
## Call:
## lm(formula = score ~ gender + +ethnicity + 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 ***
## gendermale 0.207112 0.050135 4.131 4.30e-05 ***
## ethnicitynot minority 0.167872 0.075275 2.230 0.02623 *
## 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
×gendermale+0.168×ethnicitynot_minority−0.206×languagenon_english−0.006×age+0.005×cls_pe
16. Verify that the conditions for this model are reasonable using diagnostic plots.
par(c(1,1))
## NULL
plot(m_final)

Fitted values
lm(score ~ gender + +ethnicity + language + age + cls_perc_eval + cls_credi …

Theoretical Quantiles lm(score ~ gender + +ethnicity + language + age + cls_perc_eval + cls_credi …

Fitted values lm(score ~ gender + +ethnicity + language + age + cls_perc_eval + cls_credi …

Leverage
lm(score ~ gender + +ethnicity + language + age + cls_perc_eval + cls_credi …
no there are outlines in the residual plot that do not conform to normal distribution or near constant residuals
yes the variables would not be independent because the same teacher would be graded on multiple observations.
** – male – ethnic minority – non English speaker – younger – with a higher percentage of students fill out the evaluation – teacher of multi credit courses – attractive – with a black and white picture **
no – with an r squared of only 0.158 this model has limited predicting power