Logistic regression

Prediction + Assessment

Prof. Maria Tackett

Mar 19, 2026

Announcements

Project EDA due March 24 at 11:59pm (on GitHub)
Statistics experience due April 15

Submit questions by the end of the day. Form at the end of today’s notes.

Topics

Interpreting logistic regression models
Relationship between odds and probabilities
Odds ratios

Computational setup

# load packages
library(tidyverse)
library(tidymodels)
library(knitr)
library(patchwork)
library(ggridges) # for ridgeline plots
library(RColorBrewer)

# set default theme in ggplot2
ggplot2::theme_set(ggplot2::theme_bw())

Data: Concern about rising AI

This data comes from the 2023 Pew Research Center’s American Trends Panel. The survey aims to capture public opinion about a variety of topics including politics, religion, and technology, among others. We will use data from 11201 respondents in Wave 132 of the survey conducted July 31 - August 6, 2023.

The goal of this analysis is to understand the relationship between age, time to complete the survey, and concern about the increased use of AI in daily life.

A more complete analysis on this topic can be found in the Pew Research Center article Growing public concern about the role of artificial intelligence in daily life by Alec Tyson and Emma Kikuchi.

Variables

ai_concern: Whether a respondent said they are “more concerned than excited” about in the increased use of AI in daily life (1: yes, 0: no)

Source: Pew Research

Variables

age_cat: Age category
- 18-29
- 30-49
- 50-64
- 65+
- Refused
survey_time: Time to complete the survey (in minutes)

Data prep

# change variable names and recode categories
pew_data <- pew_data |>
  mutate(ai_concern = if_else(CNCEXC_W132 == 2, 1, 0),
         age_cat = case_when(F_AGECAT == 1 ~ "18-29",
                             F_AGECAT == 2 ~ "30-49",
                             F_AGECAT == 3 ~ "50-64",
                             F_AGECAT == 4 ~ "65+",
                             TRUE ~ "Refused"
))

# Make factors and  relevel 
pew_data <- pew_data |>
  mutate(ai_concern = factor(ai_concern),
         age_cat = factor(age_cat))

# compute the time it took to do the interview
pew_data <- pew_data |>
  mutate(
    start_time = ymd_hms(INTERVIEW_START_W132),
    end_time = ymd_hms(INTERVIEW_END_W132),
    survey_time = as.numeric(difftime(end_time, start_time, units = "mins"))
)
  
pew_data <- 
  pew_data |> filter(survey_time <= 70)

Bivariate EDA

Code

ggplot(data = pew_data, aes(x = age_cat, fill = ai_concern)) + 
  geom_bar(position = "fill", color = "black") +
  labs(x = "",
       y = "Proportion",
       title = "AI Concern versus Age", 
       fill = "AI Concern") +
  scale_fill_brewer(palette = "Dark2")

Bivariate EDA

Code

ggplot(data = pew_data, aes(x = survey_time, y = ai_concern, 
                                  fill = ai_concern)) + 
  geom_density_ridges() +
  theme_ridges() + 
  labs(title = "AI concern versus survey time", 
         x = "Survey time in minutes", 
         y = "AI concern",
       fill = "AI concern") +
  theme(legend.position = "none") +
  scale_fill_brewer(palette = "Dark2")

Logistic regression model

Let \(\pi\) be the probability that \(Y = 1\). Then the population-level logistic regression model is of the form

\[ \log\big(\frac{\pi}{1-\pi}\big) = \beta_0 + \beta_1~X_1 + \beta_2X_2 + \dots + \beta_p X_p \]

where \(\log\big(\frac{\pi}{1-\pi}\big)\) = the log odds \(Y = 1\)

AI concern versus age

ai_concern_fit <- glm(ai_concern ~ age_cat, data = pew_data, 
                      family = "binomial")

tidy(ai_concern_fit) |> kable(digits = 3)

term	estimate	std.error	statistic	p.value
(Intercept)	-0.248	0.068	-3.625	0.000
age_cat30-49	0.126	0.077	1.630	0.103
age_cat50-64	0.538	0.078	6.904	0.000
age_cat65+	0.643	0.078	8.284	0.000
age_catRefused	0.493	0.322	1.531	0.126

The model

term	estimate	std.error	statistic	p.value
(Intercept)	-0.248	0.068	-3.625	0.000
age_cat30-49	0.126	0.077	1.630	0.103
age_cat50-64	0.538	0.078	6.904	0.000
age_cat65+	0.643	0.078	8.284	0.000
age_catRefused	0.493	0.322	1.531	0.126

\[\begin{aligned}\log\Big(\frac{\hat{\pi}}{1-\hat{\pi}}\Big) =& -0.248 + 0.126\times\text{age_cat30-49} + 0.538 \times \text{age_cat50-64}\\ &+ 0.643 \times \text{age_cat65+} + 0.493\times \text{age_catRefused} \end{aligned}\]

where \(\hat{\pi}\) is the predicted probability of being concerned about increased use of AI in daily life

Interpreting `age_cat30-49`: log-odds

term	estimate	std.error	statistic	p.value
(Intercept)	-0.248	0.068	-3.625	0.000
age_cat30-49	0.126	0.077	1.630	0.103
age_cat50-64	0.538	0.078	6.904	0.000
age_cat65+	0.643	0.078	8.284	0.000
age_catRefused	0.493	0.322	1.531	0.126

The predicted log-odds of being concerned about increased use of AI in daily life are 0.126 higher for individuals 30 - 49 years old compared to 18-29 year-olds (the baseline group).

Warning

We would not use the interpretation in terms of log-odds in practice.

Interpreting `age_cat30-49`: odds

The predicted odds of being concerned about increased use of AI in daily life for 30 - 49 year-olds are 1.134 ( \(e^{0.126}\) ) times the odds for 18-29 year-olds.

Coefficients & odds ratios

The model coefficient, 0.126, is the expected difference in the log-odds when comparing 30 - 49 year olds to 18 - 29 year olds.

Therefore, \(e^{0.126}\) = 1.134 is the expected change in the odds when comparing 30 - 49 year olds to 18-29 year olds.

\[ OR = e^{\hat{\beta}_j} = \exp\{\hat{\beta}_j\} \]

Interpret in terms of percent change

You can also interpret the change in the odds in terms of a percent change. The percent change in the odds can be computed as the following

\[\% \text{ change } = (e^{\hat{\beta}_j} - 1) \times 100\]

Interpret the coefficient of age_cat30-49 (0.126) in terms of the percent change in the odds.

Quantitative predictor

Now let’s look at the relationship between survey_time and ai_concern

ai_time_fit <- glm(ai_concern ~ survey_time, data = pew_data,
family = "binomial")

term	estimate	std.error	statistic	p.value
(Intercept)	0.069	0.037	1.853	0.064
survey_time	0.005	0.002	2.434	0.015

For each additional minute of taking the survey, the predicted odds of being concerned about increased AI in daily life multiply by a factor of 1.005 ( \(e^{0.005}\)).

Multiple predictors

Now let’s consider a model that takes into account age and survey_time

ai_concern_full_fit <- glm(ai_concern ~ age_cat  + survey_time, data = pew_data, family = "binomial")

Multiple predictors

term	estimate	std.error	statistic	p.value
(Intercept)	-0.240	0.073	-3.281	0.001
age_cat30-49	0.127	0.077	1.640	0.101
age_cat50-64	0.541	0.078	6.904	0.000
age_cat65+	0.647	0.079	8.217	0.000
age_catRefused	0.497	0.322	1.543	0.123
survey_time	-0.001	0.002	-0.330	0.742

Describe the type of respondent represented by the intercept.
Interpret the coefficient of survey_time in the context of the data.

Displaying logistic model in R

By default, tidy() displays the model in terms of \(\log\big(\frac{\hat{\pi}}{1 - \hat{\pi}}\big)\) . Use argument exponentiate = TRUE to display the model in terms of the odds.

tidy(ai_concern_full_fit, exponentiate = TRUE) |>
  kable(digits = 3)

term	estimate	std.error	statistic	p.value
(Intercept)	0.787	0.073	-3.281	0.001
age_cat30-49	1.135	0.077	1.640	0.101
age_cat50-64	1.717	0.078	6.904	0.000
age_cat65+	1.910	0.079	8.217	0.000
age_catRefused	1.644	0.322	1.543	0.123
survey_time	0.999	0.002	-0.330	0.742

Prediction

Data: Risk of coronary heart disease

This data set is from an ongoing cardiovascular study on residents of the town of Framingham, Massachusetts.

high_risk: 1 = High risk of having heart disease in next 10 years, 0 = Not high risk of having heart disease in next 10 years
age: Age at exam time (in years)
totChol: Total cholesterol (in mg/dL)
currentSmoker: 0 = nonsmoker; 1 = smoker

The goal is to use logistic regression to identify patients who are high risk of heart disease.

Univariate EDA

Bivariate EDA

Modeling risk of coronary heart disease

heart_disease_fit <- glm(high_risk ~ age + totChol + currentSmoker, 
              data = heart_disease, family = "binomial")

tidy(heart_disease_fit) |> 
  kable(digits = 3)

term	estimate	std.error	statistic	p.value
(Intercept)	-6.638	0.372	-17.860	0.000
age	0.082	0.006	14.430	0.000
totChol	0.002	0.001	2.001	0.045
currentSmoker1	0.457	0.092	4.951	0.000

Prediction and classification

We are often interested in using the model to classify observations, i.e., predict whether a given observation will have a 1 or 0 response
For each observation
- Use the logistic regression model to calculate the predicted log-odds the response for the \(i^{th}\) observation is 1
- Use the log-odds to calculate the predicted probability the \(i^{th}\) observation is 1
- Then, use the predicted probability to classify the observation as having a 1 or 0 response using some predefined threshold

Augmented data frame

heart_disease_aug <- augment(heart_disease_fit)

# A tibble: 4,190 × 10
   high_risk   age totChol currentSmoker .fitted .resid     .hat .sigma  .cooksd
   <fct>     <dbl>   <dbl> <fct>           <dbl>  <dbl>    <dbl>  <dbl>    <dbl>
 1 0            39     195 0              -3.06  -0.302 0.000594  0.890  6.94e-6
 2 0            46     250 0              -2.38  -0.420 0.000543  0.890  1.25e-5
 3 0            48     245 1              -1.77  -0.560 0.000527  0.890  2.24e-5
 4 1            61     225 1              -0.751  1.51  0.00164   0.889  8.70e-4
 5 0            46     285 1              -1.86  -0.539 0.000830  0.890  3.25e-5
 6 0            43     228 0              -2.67  -0.366 0.000546  0.890  9.43e-6
 7 1            63     205 0              -1.08   1.66  0.00154   0.889  1.15e-3
 8 0            45     313 1              -1.88  -0.532 0.00127   0.890  4.86e-5
 9 0            52     260 0              -1.87  -0.535 0.000542  0.890  2.08e-5
10 0            43     225 1              -2.22  -0.454 0.000532  0.890  1.44e-5
# ℹ 4,180 more rows
# ℹ 1 more variable: .std.resid <dbl>

Predicted log-odds

# A tibble: 5 × 1
  .fitted
    <dbl>
1  -3.06 
2  -2.38 
3  -1.77 
4  -0.751
5  -1.86

Observation 1

\[ \log(\hat{\text{odds}}) = \log\Big(\frac{\hat{\pi}}{1- \hat{\pi}}\Big) = -3.06 \]

Predicted odds

# A tibble: 5 × 1
  .fitted
    <dbl>
1  -3.06 
2  -2.38 
3  -1.77 
4  -0.751
5  -1.86

Observation 1

\[ \hat{\text{odds}} = \frac{\hat{\pi}}{1- \hat{\pi}} = \exp\{-3.06\} = 0.0469 \]

Predicted probability

# A tibble: 5 × 1
  .fitted
    <dbl>
1  -3.06 
2  -2.38 
3  -1.77 
4  -0.751
5  -1.86

Observation 1

\[ \text{predicted prob.} = \hat{\pi} = \frac{\hat{\text{odds}}}{1+\hat{\text{odds}}} = \frac{\exp\{-3.06\}}{1 + \exp\{-3.06\}}= 0.045 \]

An individual has a predicted probability of 0.045. Would you consider this person high risk for heart disease?
An individual has a predicted probability of 0.321. Would you consider this person high risk for heart disease?

🔗 https://forms.office.com/r/VPjCMfrsLy

Predicted probabilities in R

Compute predicted probabilities by adding type.predict = "response" argument in augment()

heart_disease_aug <- augment(heart_disease_fit, type.predict = "response")

Predicted probabilities for Observations 1 -5

# A tibble: 6 × 1
  .fitted
    <dbl>
1  0.0446
2  0.0845
3  0.145 
4  0.321 
5  0.135 
6  0.0646

Classifying observations

You would like to determine a threshold for classifying individuals as high risk or not high risk.

What considerations would you make in determining the threshold?

Classify using 0.5 as threshold

We can use a threshold of 0.5 to classify observations.

If \(\hat{\pi} > 0.5\), classify as 1
If \(\hat{\pi} \leq 0.5\), classify as 0

heart_disease_aug <- heart_disease_aug |>
  mutate(pred_class = factor(if_else(.fitted > 0.5, 1, 0)))

# A tibble: 5 × 3
  high_risk .fitted pred_class
  <fct>       <dbl> <fct>     
1 0          0.0446 0         
2 0          0.0845 0         
3 0          0.145  0         
4 1          0.321  0         
5 0          0.135  0

Confusion matrix

A confusion matrix is a \(2 \times 2\) table that compares the predicted and actual classes. We can produce this matrix using the conf_mat() function in the yardstick package (part of tidymodels).

heart_disease_aug |>
  conf_mat(high_risk, pred_class)

          Truth
Prediction    0    1
         0 3553  635
         1    2    0

Visualize confusion matrix

heart_conf_mat <- heart_disease_aug |>
  conf_mat(high_risk, pred_class)

autoplot(heart_conf_mat, type = "heatmap")

Using the confusion matrix

          Truth
Prediction    0    1
         0 3553  635
         1    2    0

The accuracy of this model with a classification threshold of 0.5 is

\[ \text{accuracy} = \frac{3553 + 0}{3553 + 635 + 2 + 0} = 0.848 \]

Using the confusion matrix

          Truth
Prediction    0    1
         0 3553  635
         1    2    0

The misclassification rate of this model with a threshold of 0.5 is

\[ \text{misclassification} = \frac{635 + 2}{3553 + 635 + 2 + 0} = 0.152 \]

Using the confusion matrix

          Truth
Prediction    0    1
         0 3553  635
         1    2    0

Accuracy is 0.848 and the misclassification rate is 0.152.

What is the limitation of solely relying on accuracy and misclassification to assess the model performance?
What is the limitation of using a single confusion matrix to assess the model performance?

Changing threshold

What is the limitation of using a single confusion matrix to assess the model performance?

Sensitivity and specificity

True/false positive/negative

	Not high risk \((y_i = 0)\)	High risk \((y_i = 1)\)
Classified not high risk \((\hat{\pi}_i \leq \text{threshold})\)	True negative (TN)	False negative (FN)
Classified high risk \((\hat{\pi}_i > \text{threshold})\)	False positive (FP)	True positive (TP)

\(\text{accuracy} = \frac{TN + TP}{TN + TP + FN + FP}\)
\(\text{misclassification} = \frac{FN + FP}{TN+ TP + FN + FP}\)

False negative rate

	Not high risk \((y_i = 0)\)	High risk \((y_i = 1)\)
Classified not high risk \((\hat{\pi}_i \leq \text{threshold})\)	True negative (TN)	False negative (FN)
Classified high risk \((\hat{\pi}_i > \text{threshold})\)	False positive (FP)	True positive (TP)

False negative rate: Proportion of actual positives that were classified as negatives

P(classified not high risk | high risk) = \(\frac{FN}{TP + FN}\)

False positive rate

	Not high risk \((y_i = 0)\)	High risk \((y_i = 1)\)
Classified not high risk \((\hat{\pi}_i \leq \text{threshold})\)	True negative (TN)	False negative (FN)
Classified high risk \((\hat{\pi}_i > \text{threshold})\)	False positive (FP)	True positive (TP)

False positive rate: Proportion of actual negatives that were classified as positives

P(classified high risk | not high risk) = \(\frac{FP}{TN + FP}\)

Sensitivity

	Not high risk \((y_i = 0)\)	High risk \((y_i = 1)\)
Classified not high risk \((\hat{\pi}_i \leq \text{threshold})\)	True negative (TN)	False negative (FN)
Classified high risk \((\hat{\pi}_i > \text{threshold})\)	False positive (FP)	True positive (TP)

Sensitivity: Proportion of actual positives that were correctly classified as positive

Also known as true positive rate (TPR) and recall
P(classified high risk | high risk) = 1 − False negative rate

Specificity

	Not high risk \((y_i = 0)\)	High risk \((y_i = 1)\)
Classified not high risk \((\hat{\pi}_i \leq \text{threshold})\)	True negative (TN)	False negative (FN)
Classified high risk \((\hat{\pi}_i > \text{threshold})\)	False positive (FP)	True positive (TP)

Specificity: Proportion of actual negatives that were correctly classified as negative

P(classified not high risk | not high risk) = 1 − False positive rate

Practice

          Truth
Prediction    0    1
         0 3553  635
         1    2    0

Calculate the

False negative rate
False positive rate
Sensitivity
Specificity

Using metrics to select model and threshold

Metric	Guidance for use
Accuracy	For balanced data, use only in combination with other metrics. Avoid using for imbalanced data.
Sensitivity (true positive rate)	Use when false negatives are more “expensive” than false positives.
False positive rate	Use when false positives are more “expensive” than false negatives.
Precision = \(\frac{TP}{TP + FP}\)	Use when it’s important for positive predictions to be accurate.

This table is a modification of work created and shared by Google in the Google Machine Learning Crash Course.

Choosing a classification threshold

A doctor plans to use your model to determine which patients are high risk for heart disease. The doctor will recommend a treatment plan for high risk patients.

Would you want sensitivity to be high or low? What about specificity?
What are the trade-offs associated with each decision?

ROC curve

So far the model assessment has depended on the model and selected threshold. The receiver operating characteristic (ROC) curve allows us to assess the model performance across a range of thresholds.

x-axis: 1 - Specificity (False positive rate)
y-axis: Sensitivity (True positive rate)

Which corner of the plot indicates the best model performance?

ROC curve

ROC curve in R

# calculate sensitivity and specificity at each threshold
roc_curve_data <- heart_disease_aug |>
  roc_curve(high_risk, .fitted, 
            event_level = "second") 

# plot roc curve
autoplot(roc_curve_data)

ROC curve in R

# A tibble: 20 × 3
   .threshold specificity sensitivity
        <dbl>       <dbl>       <dbl>
 1  -Inf         0                  1
 2     0.0259    0                  1
 3     0.0262    0.000281           1
 4     0.0283    0.000844           1
 5     0.0287    0.00113            1
 6     0.0290    0.00141            1
 7     0.0297    0.00169            1
 8     0.0302    0.00197            1
 9     0.0307    0.00225            1
10     0.0310    0.00253            1
11     0.0313    0.00309            1
12     0.0320    0.00338            1
13     0.0320    0.00366            1
14     0.0323    0.00422            1
15     0.0325    0.00450            1
16     0.0327    0.00478            1
17     0.0331    0.00506            1
18     0.0333    0.00534            1
19     0.0334    0.00591            1
20     0.0334    0.00619            1

Area under the curve

The area under the curve (AUC) can be used to assess how well the logistic model fits the data

AUC=0.5: model is a very bad fit (no better than a coin flip)
AUC close to 1: model is a good fit

heart_disease_aug |>
  roc_auc(high_risk, .fitted,
    event_level = "second"
  )

# A tibble: 1 × 3
  .metric .estimator .estimate
  <chr>   <chr>          <dbl>
1 roc_auc binary         0.696

Recap

Interpreted coefficients of logistic regression
Calculated predicted probabilities from the logistic regression model
Used predicted probabilities to classify observations
Made decisions and assessed model performance using
- Confusion matrix
- ROC curve

Questions about content?

🔗 https://forms.office.com/r/QvBy4yvV8N

Next class

Inference for logistic regression
Complete Lecture 18 prepare

Logistic regression

Announcements

Topics

Computational setup

Data: Concern about rising AI

Variables

Variables

Data prep

Bivariate EDA

Bivariate EDA

Logistic regression model

AI concern versus age

The model

Interpreting age_cat30-49: log-odds

Interpreting age_cat30-49: odds

Coefficients & odds ratios

Interpret in terms of percent change

Quantitative predictor

Multiple predictors

Multiple predictors

Displaying logistic model in R

Prediction

Data: Risk of coronary heart disease

Univariate EDA

Bivariate EDA

Modeling risk of coronary heart disease

Prediction and classification

Augmented data frame

Predicted log-odds

Predicted odds

Predicted probability

Classifying observations

Predicted probabilities in R

Classifying observations

Classify using 0.5 as threshold

Confusion matrix

Visualize confusion matrix

Using the confusion matrix

Using the confusion matrix

Using the confusion matrix

Changing threshold

Sensitivity and specificity

True/false positive/negative

False negative rate

False positive rate

Sensitivity

Specificity

Practice

Using metrics to select model and threshold

Choosing a classification threshold

ROC curve

ROC curve

ROC curve in R

ROC curve in R

Area under the curve

Recap

Questions about content?

Next class

Interpreting `age_cat30-49`: log-odds

Interpreting `age_cat30-49`: odds