(Normal) Linear Regression Response Variables
Birth weight of Babies (55 - 176 ounces)
Sale Prices ($12789 - $755,000)
Number of Species (6 - 129 mammals)
Logistic Regression Response Variables
Will it rain tomorrow? (Yes/No)
Is email spam? (Yes/No)
Does the candidate receive a callback? (Yes/No)
When the response variable is binary a logistic regression model can be utilized.

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 × 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.

\(\text{transformation}(\pi_i) = \beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}\)

\(logit(\pi_i) = \beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}\)

\(logit(\pi_i) = log(\frac{\pi_i}{1-\pi_i})\)

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

function in R.

Probability \(\pi_i\) Probability of receiving a callback.

Odds \(\frac{\pi_i}{1-\pi_i}\) Odds of receiving a callback.

Logit \(log(\frac{\pi_i}{1-\pi_i})\) Logit of receiving a callback.

When race is black (0)
resume %>%
filter (race == "black" ) %>%
count (received_callback) %>%
mutate (prop = n / sum (n))

```
# A tibble: 2 × 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.

## Odds
odds_b <- p_b / (1 - p_b)
odds_b

## Logit
logit_b <- log (odds_b)
logit_b

## Odds
odds_w <- p_w / (1 - p_w)
odds_w

## Logit
logit_w <- log (odds_w)
logit_w

This is THE LINE of the linear model. As x increases by 1 unit, the expected change in the logit of receiving call back is 0.4381802. In this case, this is just the difference between logit for the white group and the black group.

The slope of the line is:

The intercept is

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

```
# A tibble: 2 × 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
```

\(log(\frac{\hat \pi_i}{1-\hat \pi_i}) = -2.67 + 0.438\times racewhite_i\)

Probability
0 to 1
Odds
0 to \(\infty\)
Logit
- \(\infty\) to \(\infty\)

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)

```
# A tibble: 2 × 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

\(odds = e^{logit}\)

\(\frac{\pi_i}{1-\pi_i} = e^{log(\frac{\pi_i}{1-\pi_i})}\)

\(\frac{\hat\pi_i}{1-\hat\pi_i} = e^{-2.76 + 0.0391 \times 1}\)

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.

Logistic Regression model
Logit form:

\(log(\frac{\pi_i}{1-\pi_i}) = \beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}\)

Probability form:

\(\large{\pi_i = \frac{e^{\beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}}}{1+e^{\beta_0 + \beta_1x_{1i}+\beta_2x_{2i} +.... \beta_kx_{ki}}}}\)

Estimated probability of a candidate with 0 years of experience receiving a callback

\(\hat\pi_i = \frac{e^{-2.76 + 0.0391 \times 0}}{1+e^{-2.76 + 0.0391 \times 0}} = 0.0595\)

Estimated probability of a candidate with 1 year of experience receiving a callback

\(\hat\pi_i = \frac{e^{-2.76 + 0.0391 \times 1}}{1+e^{-2.76 + 0.0391 \times 1}} = 0.0618\)

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

```
# A tibble: 4 × 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.

\(\large{\hat\pi_i = \frac{e^{-2.78 + (0.0440 \times 0) + (0.0332\times10) + (-0.0329\times 0)}}{1+e^{-2.78 + (0.0440 \times 0) + (0.0332\times10) + (-0.0329\times 0)}} = 0.0796}\)

Model Evaluation
babies <- babies %>%
mutate (low_bwt = case_when (bwt < 88 ~ TRUE ,
bwt >= 88 ~ FALSE )) %>%
drop_na (gestation)

model_g <- glm (low_bwt ~ gestation,
data = babies,
family = "binomial" )

```
# A tibble: 2 × 5
term estimate std.error statistic p.value
<chr> <dbl> <dbl> <dbl> <dbl>
1 (Intercept) 17.5 2.24 7.82 5.11e-15
2 gestation -0.0758 0.00846 -8.96 3.27e-19
```

\(\hat p = \frac{\exp(b_0 +b_1x)}{1+\exp(b_0 + b_1x)}\)

\(\hat p\) when gestation is 284 = \(\frac{\exp(17.5 -0.0758 \cdot 284)}{1+\exp(17.5 -0.0758 \cdot 284)} = \frac{\exp(-4.0272)}{1+ \exp(-4.0272)} = 0.01751203\)

babies <- babies %>%
add_predictions (model_g)
select (babies, pred)

```
# A tibble: 1,223 × 1
pred
<dbl>
1 -4.02
2 -3.87
3 -3.64
4 -3.87
5 -4.17
6 -0.986
7 -1.06
8 -4.40
9 -5.15
10 -9.10
# ℹ 1,213 more rows
```

babies <- babies %>%
mutate (pred_p = exp (pred)/ (1 + exp (pred)))
select (babies, pred, pred_p)

```
# A tibble: 1,223 × 2
pred pred_p
<dbl> <dbl>
1 -4.02 0.0177
2 -3.87 0.0205
3 -3.64 0.0256
4 -3.87 0.0205
5 -4.17 0.0152
6 -0.986 0.272
7 -1.06 0.257
8 -4.40 0.0122
9 -5.15 0.00574
10 -9.10 0.000112
# ℹ 1,213 more rows
```

Cutoff
babies <- babies %>%
mutate (pred_y = case_when (pred_p < 0.5 ~ FALSE ,
pred_p >= 0.5 ~ TRUE ))
select (babies, low_bwt, pred, pred_p, pred_y)

```
# A tibble: 1,223 × 4
low_bwt pred pred_p pred_y
<lgl> <dbl> <dbl> <lgl>
1 FALSE -4.02 0.0177 FALSE
2 FALSE -3.87 0.0205 FALSE
3 FALSE -3.64 0.0256 FALSE
4 FALSE -3.87 0.0205 FALSE
5 FALSE -4.17 0.0152 FALSE
6 FALSE -0.986 0.272 FALSE
7 FALSE -1.06 0.257 FALSE
8 FALSE -4.40 0.0122 FALSE
9 FALSE -5.15 0.00574 FALSE
10 FALSE -9.10 0.000112 FALSE
# ℹ 1,213 more rows
```

Confusion Matrix
janitor:: tabyl (babies, low_bwt, pred_y) %>%
janitor:: adorn_totals (c ("row" , "col" ))

```
low_bwt FALSE TRUE Total
FALSE 1161 5 1166
TRUE 53 4 57
Total 1214 9 1223
```

Sensitivity (true-positive rate): 4/57 = 0.0702
Specificity (true-negative rate): 1161/1166 = 0.9957
Accuracy : (1161+4)/1223 = 0.9526