Assignment_9

library(tidyverse)

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## ✔ tibble  3.1.8     ✔ dplyr   1.0.9
## ✔ tidyr   1.2.0     ✔ stringr 1.4.1
## ✔ readr   2.1.2     ✔ forcats 0.5.2
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag()    masks stats::lag()

library(openintro)

## Loading required package: airports
## Loading required package: cherryblossom
## Loading required package: usdata

library(GGally)

## Registered S3 method overwritten by 'GGally':
##   method from   
##   +.gg   ggplot2

Set up

#Code snippet found @https://www.roelpeters.be/how-to-add-a-regression-equation-and-r-squared-in-ggplot2/
eq <- function(x,y) {
  m <- lm(y ~ x)
  as.character(
    as.expression(
      substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2,
                list(a = format(coef(m)[1], digits = 4),
                b = format(coef(m)[2], digits = 4),
                r2 = format(summary(m)$r.squared, digits = 3)))
    )
  )
}
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.

Looking at the design & methods, I believe this would be an observational study rather than an experiment, because this study observes responses and does not attempt to influence the response of the participants.

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?

mean <- mean(evals$score)
sd   <- sd(evals$score)
ggplot(data = evals, aes(x = score)) +
        geom_blank() +
        geom_histogram(aes(y = ..density..)) +
        stat_function(fun = dnorm, args = c(mean = mean, sd = sd), col = "tomato")

## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

Looking at the data without context, I would say that this data is left skewed. This is due to the fact that there are a lot of high score values. What this indicates is one of a few things: 1. All the professors are excellent 2. The students loved their courses, regardless of who taught it.

In terms of expectations, I would expect to see a skew to higher score values because people generally leave extremes of feedback (either really good feedback, or really bad feedback). Given that most people want to learn, I would always expect higher values (and thus more left skew) in most cases.

Exercise 3

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

metrics <- evals %>%
  group_by(gender,pic_outfit ) %>%
  summarise(count = n())

## `summarise()` has grouped output by 'gender'. You can override using the
## `.groups` argument.

metrics

box <- ggplot(metrics, aes(x=pic_outfit, y=count, fill=gender)) + 
  geom_bar(stat = "identity")
box

Looking at the prevalence of formal photos vs not formal photos and gender, it appeas as though higher rates of men prefer formalwear to their photo shoots. However, regardless of gender, people preferred not formal photos.

Exercise 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_point()

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

Looking at the two scatter plots, it appears as though there were a lot of similar scores, the jitters helped separate the points!

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?

ggplot(data = evals, aes(x = bty_avg, y = score)) +
  geom_jitter() +
  ylim(2, 6) +
  geom_smooth(method = "lm") +
   geom_text(x = 4, y = 5.5, label = eq(evals$bty_avg,evals$score), parse = TRUE)

## `geom_smooth()` using formula 'y ~ x'

b_score <- lm(evals$score ~ evals$bty_avg)
summary(b_score)

## 
## 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

Looking at the formula, the slope of the line is 0.06664. The Low p-value indicates that there is statistical significance of beauty affecting the score, but given the small coefficient, it is not a major impact.

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).

ggplot(data = b_score, aes(x = b_score$fitted, y = b_score$resid)) +
  geom_point() +
  geom_hline(yintercept = 0, linetype = "dashed") +
  ylab("Residuals")

ggplot(data = b_score, aes(x = .resid)) +
  geom_histogram(binwidth = .5) +
  xlab("Residuals")

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

Practically, we look for 4 key parts to analyze if a simple linear regression is reasonable.

1. Independence-This is a given spec of the data set
2. There is a linear relationship, which is demonstrated in the scatter plot above.
3. There is constant variance, which is shown by the uniform spread of residuals.
4. There is a normality of Residuals across a reasonable range, which is demonstrated by the chart above.

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.

m1 <- lm(score ~ bty_avg + gender, data = evals)

ggplot(data = m1, aes(x = .fitted, y = .resid)) +
  geom_point() +
  geom_hline(yintercept = 0, linetype = "dashed") +
  xlab("Predicted Value") +
  ylab("Residuals") + 
  geom_jitter()

ggplot(data = m1, aes(x = .resid)) +
  geom_histogram(binwidth = 0.5) +
  xlab("Residuals")

ggplot(data = m1, aes(sample = .resid)) +
  stat_qq()

Practically, we look for 4 key parts to analyze if a regression is reasonable.

1. Independence-This is a given spec of the data set
2. There is a linear relationship, which is demonstrated in the scatter plot above.
3. There is constant variance, which is shown by the uniform spread of residuals.
4. There is a normality of Residuals across a reasonable range, which is demonstrated by the chart above.

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?

summary(lm(score ~ bty_avg + gender, data = evals))

## 
## 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

For this multi-variable model, bty_avg still heavily weights on score (0.07416), which is an increase from the model of beauty alone (0.06664). However, the fit is still low, so there are probably other parameters that have more weight.

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?

summary(lm(score ~ bty_avg + pic_color, data = evals))

## 
## 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

Score = C1 + C2 * bty_avg + -C3 * color

In color photos, it would simply be score = C1 + C2 * bty_avg For non color photos it would be score = c1 + C2 * bty_avg - C3

Which would indicate that any two professors with the same bty_avg, the one with a color photo would have a higher 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.

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

Honestly its pretty much just 1 hot encoding.

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.

Going off personal experience, I believe age would have the highest p-value, as it seems the most irrelevant to me from my personal experiences.

Exercise 12

Let’s check exercise 11

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

Unfortunately I was incorrect, cls_profs has the highest p-value. Although, in hindsight, this makes sense to me.

Exercise 13

Interpret the coefficient associated with the ethnicity variable.

In plain English, professors who are a minority would receive a score decrease of 0.1234929, all other parameters held constant.

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?

summary( lm(score ~ rank + ethnicity + gender + language + age + cls_perc_eval 
             + cls_students + cls_level + cls_credits + bty_avg 
             + pic_outfit + pic_color, data = evals))

## 
## 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

There is no significant change to the coefficients. This indicates that there are variables that correlate with the cls_profs.

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.

summary(lm(score ~ ethnicity + gender + language + age + cls_perc_eval 
             + cls_credits + bty_avg 
             + pic_color, data = evals))

## 
## 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

The linear model is score = 3.771922 + 0.167872 * ethnicity + 0.207112 * gender - 0.206178 * language - 0.006046 * age + 0.004656 * cls_perc_eval + 0.505306 * cls_credits + 0.051069 * bty_avg - 0.190579 * pic

Exercise 16

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

m2 <- lm(score ~ ethnicity + gender + language + age + cls_perc_eval 
             + cls_credits + bty_avg 
             + pic_color, data = evals)

ggplot(data = m2, aes(x = .fitted, y = .resid)) +
  geom_point() +
  geom_hline(yintercept = 0, linetype = "dashed") +
  xlab("Avg") +
  ylab("Residuals") + 
  geom_jitter()

ggplot(data = m2, aes(x = .resid)) +
  geom_histogram(binwidth = 0.5) +
  xlab("Residuals")

ggplot(data = m2, aes(sample = .resid)) +
  stat_qq()

Practically, we look for 4 key parts to analyze if a regression is reasonable.

1. Independence-This is a given spec of the data set
2. There is a linear relationship, which is demonstrated in the scatter plot above.
3. There is constant variance, which is shown by the uniform spread of residuals.
4. There is a normality of Residuals across a reasonable range, which is demonstrated by the chart above.

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, there is an impact on the conditions of linear regression since independence is removed as a professor may be in this data multiple times.

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.

A young non-minority male, who speaks English, that teaches the most credit per class, and is handsome with a lot of responses, would have a high eval score.

Exercise 19

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

No I would not, if you just take the concept and stretch it to universities abroad, the weight of English speaking would be much too high. In addition, this is simply a localized sample of people, that I would not expect to apply elsewhere. I would also expect that as time goes on, preferences change, and, given new data, that this conclusion will change.