Are Emily and Greg More Employable than Lakisha and Jamal? A Field Experiment on Labor Market Discrimination.

• Researchers respond to help-wanted ads in Boston and Chicago newspapers with fictitious resumes.

• They randomly assign White sounding names to half the resumes and African American sounding names to the other half.

• They create high quality resumes (more experience, likely to have an email address etc.) and low quality resumes.

• For each job ad they send four resumes (two high quality and two low quality.)

Data

resume <- resume %>% 
  select(received_callback, race, years_experience, 
         job_city)
glimpse(resume)

## Rows: 4,870
## Columns: 4
## $ received_callback <lgl> FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, FAL…
## $ race              <chr> "white", "white", "black", "black", "white", "white"…
## $ years_experience  <int> 6, 6, 6, 6, 22, 6, 5, 21, 3, 6, 8, 8, 4, 4, 5, 4, 5,…
## $ job_city          <chr> "Chicago", "Chicago", "Chicago", "Chicago", "Chicago…

Response variable: received_callback

count(resume, received_callback) %>% 
  mutate(prop = n / sum(n))

## # A tibble: 2 x 3
##   received_callback     n   prop
##   <lgl>             <int>  <dbl>
## 1 FALSE              4478 0.920 
## 2 TRUE                392 0.0805

Notation

$y_i$ = whether a (fictitious) job candidate receives a call back.

${\pi }_i$ = probability that the $i$th job candidate will receive a call back.

$1-{\pi }_i$ = probability that the $i$th job candidate will not receive a call back.

Where is the line?

ggplot(resume, aes(x = race, y = received_callback)) +
  geom_point()

The Linear Model

We can model the probability of receiving a callback with a linear model.

$\text{transformation}\left({\pi }_i\right)={\beta }_0+{\beta }_1{x}_{1i}+{\beta }_2{x}_{2i}+....{\beta }_k{x}_{ki}$

$logit\left({\pi }_i\right)={\beta }_0+{\beta }_1{x}_{1i}+{\beta }_2{x}_{2i}+....{\beta }_k{x}_{ki}$

$logit\left({\pi }_i\right)=log\left(\frac{{\pi }_i}{1-{\pi }_i}\right)$

Note that log is natural log and not base 10. This is also the case for the log() function in R.

When race is black (0)

resume %>% 
  filter(race == "black") %>% 
  count(received_callback) %>% 
  mutate(prop = n / sum(n))

## # A tibble: 2 x 3
##   received_callback     n   prop
##   <lgl>             <int>  <dbl>
## 1 FALSE              2278 0.936 
## 2 TRUE                157 0.0645

Note that R assigns 0 an 1 to levels of categorical variables in alphabetical order. In this case black (0) and white(1)

When race is black (0)

p_b <- resume %>% 
  filter(race == "black") %>% 
  count(received_callback) %>% 
  mutate(prop = n / sum(n)) %>% 
  filter(received_callback == TRUE) %>% 
  select(prop) %>% 
  pull()

Probability of receiving a callback when the candidate has a Black sounding name is 0.0644764.

When race is white (1)

p_w <- resume %>% 
  filter(race == "white") %>% 
  count(received_callback) %>% 
  mutate(prop = n / sum(n)) %>% 
  filter(received_callback == TRUE) %>% 
  select(prop) %>% 
  pull()

Probability of receiving a callback when the candidate has a white sounding name is 0.0965092.

model_r <- glm(received_callback ~ race,
               data = resume,
               family = binomial)

tidy(model_r)

## # A tibble: 2 x 5
##   term        estimate std.error statistic   p.value
##   <chr>          <dbl>     <dbl>     <dbl>     <dbl>
## 1 (Intercept)   -2.67     0.0825    -32.4  1.59e-230
## 2 racewhite      0.438    0.107       4.08 4.45e-  5

$log\left(\frac{{\hat{\pi }}_i}{1-{\hat{\pi }}_i}\right)=-2.67+0.438\times racewhit{e}_i$

We will consider years of experience as an explanatory variable. Normally, we would also include race in the model and have multiple explanatory variables, however, for learning purposes, we will keep the model simple.

model_y <- glm(received_callback ~ years_experience,
               data = resume,
               family = binomial)

tidy(model_y)

## # A tibble: 2 x 5
##   term             estimate std.error statistic   p.value
##   <chr>               <dbl>     <dbl>     <dbl>     <dbl>
## 1 (Intercept)       -2.76     0.0962     -28.7  5.58e-181
## 2 years_experience   0.0391   0.00918      4.26 2.07e-  5

model_y_summary <- tidy(model_y)
intercept <- model_y_summary %>% 
  filter(term == "(Intercept)") %>% 
  select(estimate) %>% 
  pull()
slope <- model_y_summary %>% 
  filter(term == "years_experience") %>% 
  select(estimate) %>% 
  pull()

From logit to odds

Logit for a Candidate with 1 year of experience (rounded equation)

$-2.76+0.0391\times 1$

Odds for a Candidate with 1 year of experience

$odds=e^{logit}$

$\frac{{\pi }_i}{1-{\pi }_i}=e^{log\left(\frac{{\pi }_i}{1-{\pi }_i}\right)}$

$\frac{{\hat{\pi }}_i}{1-{\hat{\pi }}_i}=e^{-2.76+0.0391\times 1}$

From odds to probability

${\pi }_i=\frac{odds}{1+odds}$

${\pi }_i=\frac{\frac{{\pi }_i}{1-{\pi }_i}}{1+\frac{{\pi }_i}{1-{\pi }_i}}$

${\hat{\pi }}_i=\frac{e^{-2.76+0.0391\times 1}}{1+e^{-2.76+0.0391\times 1}}=0.0618$

Note you can use exp() function in R for exponentiating number e.

exp(1)

## [1] 2.718282

Logistic Regression model

Logit form:

$log\left(\frac{{\pi }_i}{1-{\pi }_i}\right)={\beta }_0+{\beta }_1{x}_{1i}+{\beta }_2{x}_{2i}+....{\beta }_k{x}_{ki}$

Probability form:

${\pi }_i=\frac{e^{{\beta }_0+{\beta }_1{x}_{1i}+{\beta }_2{x}_{2i}+....{\beta }_k{x}_{ki}}}{1+e^{{\beta }_0+{\beta }_1{x}_{1i}+{\beta }_2{x}_{2i}+....{\beta }_k{x}_{ki}}}$

model_ryc <- glm(received_callback ~ race + 
                  years_experience + job_city,
               data = resume,
               family = binomial)

tidy(model_ryc)

## # A tibble: 4 x 5
##   term             estimate std.error statistic  p.value
##   <chr>               <dbl>     <dbl>     <dbl>    <dbl>
## 1 (Intercept)       -2.78     0.134      -20.8  6.18e-96
## 2 racewhite          0.440    0.108        4.09 4.39e- 5
## 3 years_experience   0.0332   0.00940      3.53 4.11e- 4
## 4 job_cityChicago   -0.329    0.108       -3.04 2.33e- 3

The estimated probability that a Black candidate with 10 years of experience, residing in Boston, would receive a callback.

${\hat{\pi }}_i=\frac{e^{-2.78+(0.0440\times 0)+(0.0332\times 10)+(-0.0329\times 0)}}{1+e^{-2.78+(0.0440\times 0)+(0.0332\times 10)+(-0.0329\times 0)}}=0.0796$

We have used the data for educational purposes. The original study considers many other variables that may influence whether someone receives a callback or not. Read the original study for other considerations.

Are Emily and Greg More Employable than Lakisha and Jamal? A Field Experiment on Labor Market Discrimination.