library(tidyverse)
## -- Attaching packages --------------------------------------- tidyverse 1.3.1 --
## v ggplot2 3.3.5     v purrr   0.3.4
## v tibble  3.1.6     v dplyr   1.0.7
## v tidyr   1.2.0     v stringr 1.4.0
## v readr   2.1.1     v forcats 0.5.1
## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()
library(openintro)
## Loading required package: airports
## Loading required package: cherryblossom
## Loading required package: usdata
library(GGally)
## Warning: package 'GGally' was built under R version 4.1.3
## Registered S3 method overwritten by 'GGally':
##   method from   
##   +.gg   ggplot2
glimpse(evals)
## 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, ~
## $ 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, ~

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 because the data is gathered from a survey on student evaluation. The current question cannot be answered as the way it is phrased. Maybe a better phrased question can be Does differences in course evaluation be explained by the professor’s appearances?

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 looks left tail skewed this tells me that the students rate the courses pretty high on average.

ggplot(evals,aes(x=score)) +
  geom_histogram()+
  labs(
    title = "Student score distribution"
  )
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

Exercise 3

Excluding score, select two other variables and describe their relationship with each other using an appropriate visualization.

Seems like on average male students are older than female students,maybe each specific gender influenced the evaluations?

ggplot(evals,aes(x=gender,y=age))+
  geom_boxplot() + 
  coord_flip()

Exercise 4

Replot the scatterplot, but this time use geom_jitter as your layer. What was misleading about the initial scatterplot?

The initial scatter plot doesn’t seem to show any correlation but the jitter helps show insight into the complex relationship between the beauty avg and the score.

ggplot(data = evals, aes(x = bty_avg, y = score)) +
  geom_jitter() +
  geom_point()

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. 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 of the linear model is 3.88 + 0.066*x. This is telling us that the score is influenced by per beauty score. The p value is 0.0000508 which is less than the significance level which tells us that this is a statistically 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
ggplot(data = evals, aes(x = bty_avg, y = score)) +
  geom_jitter() +
  geom_smooth(method = "lm")
## `geom_smooth()` using formula 'y ~ x'

### 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)

## For linearity it seems like the numbers are distributed all over the place but can be considered a linear relationship 

ggplot(data=m_bty,aes(x=.fitted,y=.resid)) + 
  geom_point() + 
  geom_hline(yintercept=0,linetype="dashed") +
  xlab("Fitted Values") + 
  ylab("Residuals")

## histogram
##The distribution looks left skewed 

hist(m_bty$residuals)

## Constant v arability
## The graph seems to follow a mostly diagonal line so we can assume the data seems to have constant variability

ggplot(data = m_bty,aes(sample=.resid)) + 
  stat_qq()

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.

From the following plots we can verify that the data is slightly linear and it is positive linearly. From the histogram we can see that the distribution of the residuals are left skewed and the normal probability plot seems to follow the lines for the upper quartiles and the residuals plot show that the data to have constant varability. For independence we can assume that the sample was randomized and each observation was randomly collected.

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
## check linearity

## Data looks to be distributed all over the place but can be considered to have a linear relationship

ggplot(m_bty_gen,aes(x=.fitted,y=.resid)) +
  geom_point() + 
  geom_hline(yintercept=0,linetype="dashed") +
  xlab("Fitted Values") + 
  ylab("Residuals")

## Data for the distribution of the residuals looks to be left-skewed
hist(m_bty_gen$residuals)

## The probability plot suggests that the lines doesn't follow the line for upper quadrilles. Since we don't know the manner in which the sample was gathered we can assume independence. 

plot(m_bty_gen) 

## Seems like women on have a slightly higher beauty average compared to their male counterparts. 
ggplot(evals,aes(x=gender,y=bty_avg)) + 
  geom_boxplot() +
  coord_flip()

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? If we look at the models for m_bty_gen and m_bty we can see that the addition of gender is a significant predictor of score.Comparing the score we can see that the p-value computed in m_bty_gen is 6.48e-06 which is less than 5.08e-05 making the model with the gender category more statistically significant.

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
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 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?

I’ve created a new model and this model calculates the score based on average beauty plus the picture color and the equation of the line I believe is: 4.06 + 0.055 * bty_avg + (-0.160) * pic_color where pic_color takes a value of 1(white), otherwise if pic_color is 0(black) then the equation is 4.06 + 0.055 * bty_avg_. Comparing the two equations we can see that the equation where the pic_color has a value of 1 will have a score of -0.160 less compared to the equation where the pic_color has a value of 0.

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

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.

R appears to handle the two levels as they are two different variables we can see ranktenure track and ranktenured

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

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 predict the variable that will have the highest p-value in this model is the class size Cls_students in this case since I would believe that class size would have no association with the professor score.

Exercise 12

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

I was way off the mark the p value related cls_students has a p-value of 0.22 whereas the variable with the least association is cls_profssingle where the p value for a professor teaching 1 section has a p-value of 0.77

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

Exercise 13

Interpret the coefficient associated with the ethnicity variable.

The coefficient associated with the ethnicity variable for not minority is 0.12349 so in other words with everything being equal. The professor score regarding ethnicity is 0.12349 higher.

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?

Some of the coefficients and the significance changed and some of them didn’t change. The only noticeable change I can see in the p-value in the revised model is the ethnicity variable where it changed from 0.12349 to 0.99856.For the other variable the dropped variable was collinear since the cls_profs didn’t seem to have much of an effect on the explanatory variables.

## I've removed the cls_profs since that had the highest p-value

m_revised <- 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_revised)
## 
## 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

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

Equation score = B0 + B1 * gendermale + B2 * ethnicitynot_minority + B3 * languagenon_english + B4 * age + B5 * cls_perc_eval + B6 * cls_creditsone_credit + B7 * bty_avg + B8 * pic_outfitnot_formal + B9 * pic_colorcolor

## Here I removed the variable in the following order I removed cls_level,then cls_students and then rank..
m_final <- lm(score ~ gender + ethnicity + language + age + cls_perc_eval 
             + cls_credits + bty_avg 
             + pic_outfit + pic_color, data = evals)
summary(m_final)
## 
## Call:
## lm(formula = score ~ gender + ethnicity + language + age + cls_perc_eval + 
##     cls_credits + bty_avg + pic_outfit + pic_color, data = evals)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.8455 -0.3221  0.1013  0.3745  0.9051 
## 
## Coefficients:
##                        Estimate Std. Error t value Pr(>|t|)    
## (Intercept)            3.907030   0.244889  15.954  < 2e-16 ***
## gendermale             0.202597   0.050102   4.044 6.18e-05 ***
## ethnicitynot minority  0.163818   0.075158   2.180 0.029798 *  
## languagenon-english   -0.246683   0.106146  -2.324 0.020567 *  
## age                   -0.006925   0.002658  -2.606 0.009475 ** 
## cls_perc_eval          0.004942   0.001442   3.427 0.000666 ***
## cls_creditsone credit  0.517205   0.104141   4.966 9.68e-07 ***
## bty_avg                0.046732   0.017091   2.734 0.006497 ** 
## pic_outfitnot formal  -0.113939   0.067168  -1.696 0.090510 .  
## pic_colorcolor        -0.180870   0.067456  -2.681 0.007601 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4982 on 453 degrees of freedom
## Multiple R-squared:  0.1774, Adjusted R-squared:  0.161 
## F-statistic: 10.85 on 9 and 453 DF,  p-value: 2.441e-15

Exercise 16

Verify that the conditions for this model are reasonable using diagnostic plots. I’ve graphed some of the plots for the models and for linearity we don’t see that much of a linear trend and for normality the qq plot seems to follow the lines except it breaks off around the upper quartiles which shows the residuals are normally distributed. The residuals seems to have varying levels of variance but it doesn’t seem to have the tube shaped. Finally,assuming that the sample is randomized we can assume each observation is independent.

plot(m_final)

hist(m_final$residuals)

## checking Linearity with each explantory variables with the response variable I've graphed a bunch of scatterplot to check. 

ggplot(evals,aes(x=gender,y=score)) +
  geom_boxplot() + 
  coord_flip()

ggplot(evals,aes(x=ethnicity,y=score)) +
  geom_boxplot() +
  coord_flip()

ggplot(evals,aes(x=language,y=score)) +
  geom_boxplot() + 
  coord_flip()

ggplot(evals,aes(x=age,y=score)) +
  geom_point() + 
  geom_jitter()

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?

Since each row represents a course a professor taught, this should not have any impact on the conditions of linear regression since each course taught is independent of other courses and shouldn’t have any impact on the score evaluation.

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

According to my model, the professor would have to teach a one credit class who is a reasonably old male who is not a minority and can speak English, the teacher is not wearing a formal outfit and the color of the picture is white in order to get a high evaluation score. This model is pretty Cool!

Exercise 19

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

My model would be useless to professors outside the university of Texas since I believe survey questions vary from university to university and thus my model would incorrectly assess the professors evaluation score wrong.