Logistic Regression

STAT 20: Introduction to Probability and Statistics

Concept Questions

m1 <- glm(sex ~ body_mass_g, data = penguins, family = "binomial")

 (Intercept)  body_mass_g 
-5.162541644  0.001239819 

Which of the expressions given in the poll (math or code) will correctly calculate the predicted probability that a penguin that weighs 4000 g is a female? Select all that apply.

01:00

To answer this one correctly, they’ll need to know (or remember from the RQ) that the reference level is female, so this model is predicting the probability male.

You can show that they can either use R as a calculator or use the predict function.

1/(1 + exp(-(-5.15 + .00124 * 4000))) predict(m1, newdata = data.frame(body_mass_g = 4000), type = “response”)

The latter is more precise, which is useful here since p-hat (prob of male) is .449 therefore prob of female is .55.

As a bonus, try sketching this function on a scatterplot!

When sketching this, the positive slope means the s-curve goes up and to the right. The negative intercept shifts the whole curve a tiny bit to the left, reflecting the base rate of a couple more males than females (168 vs 165).

What is the misclassification rate of this model?

# A tibble: 4 × 3
# Groups:   sex [2]
  sex    sex_hat     n
  <fct>  <chr>   <int>
1 female female    109
2 female male       56
3 male   female     74
4 male   male       94

01:00

(56+74)/(109+56+74+94) = 130/333

m2 <- glm(sex ~ body_mass_g + bill_length_mm, 
          data = penguins, family = "binomial")

   (Intercept)    body_mass_g bill_length_mm 
   -6.91208086     0.00101530     0.06112808 

Open up RStudio and fit the model here in the slides. What are the predicted sexes of these two penguins, using a probability threshold of 0.5?

1. body mass = 3900 g, bill length = 50
2. body mass = 4100 g, bill length = 35

01:00

You can have students work on their laptops for this question.

This one extends the task of the previous one by having them include another covariate and also to do the thresholding procedure to go from p-hat to y-hat. This doesn’t specify the threshold, so using .5 is probably a good default.

predict(m2, newdata = data.frame(body_mass_g = 3900, bill_length_mm = 50), type = “response”) # male predict(m2, newdata = data.frame(body_mass_g = 4100, bill_length_mm = 35), type = “response”) # female

Or they can do it “by hand”, in which case, the first one uses \(z = -6.91208086 + 0.00101530 \times 3900 + 0.06112808\times 50 = 0.1039931\) so \(\hat{p} = \frac{1}{1+exp(-0.1039931)} = 0.5259749\) and we would predict male. For the second, we have that \(z = -6.91208086 + 0.00101530 \times 4100 + 0.06112808 \times 35 = -0.6098681\). Plugging this into the logistic function gives \(\hat{p} = \frac{1}{1+exp(0.6098681)} = 0.3520893\), so we would predict female.

Break

05:00

Problem Set

Fill out this end-of-class survey here and see how your answers differ from before!

Lab: Cancer Diagnosis

Lab: Cancer Diagnosis

45:00