ADA1: Class 14, Logistic regression

Advanced Data Analysis 1, Stat 427/527, Fall 2024, Prof. Erik Erhardt, UNM

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July 6, 2026

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AddHealth W4: Age at first intercourse vs pregancy

Data from the National Longitudinal Study of Adolescent Health (Add Health), Wave IV.

Research question

Research question: Is there a relationship between having at least one pregancy and the age at first vaginal intercourse for people in their fertile school years (12–22 years old)? We will define a pregnancy for females as “has been or is pregnant” and for males as “has impregnated partner”.

Intuition: The younger you are when you start having intercourse, the more opportunities you’ll have for pregancy so the more likely it will be that you’ve experienced at least one pregancy. Let’s see if this intuition is observed in the data.

Data

The AddHealth codebook variables used to address this question are below.

Note

If you get in the habit of providing a personal codebook for your work you can give an analysis to someone else to read without having to explain it to them. Everything is documented (also helpful for your future self who doesn’t have as good a memory as you may think you will have.

ADDHEALTH: WAVE4 IN-HOME INTERVIEW CODE BOOK
Wave IV Section 15: Suicide, Sexual Experiences, & Sexually Transmitted Diseases

H4SE7 (AgeAtInt)
7. How old were you the first time you ever had vaginal intercourse?
  1-30 years
  96 refused
  97 legitimate skip
  98 don't know

H4TR9 (NPreg)
9. Thinking about all the relationships and sexual encounters you have ever
  had, (how many times have you ever been pregnant/how many times have you ever
  made a partner pregnant)? Include all pregnancies, whether they resulted in
  babies born alive, stillbirth, abortion, miscarriage, or ectopic or tubal
  pregnancy. [If Q.7=1] say: Be sure to include your current pregnancy in your
  count.
  1-19 pregancies
  96 refused
  98 don't know


Created variables:
  EverPreg:   TRUE if NPreg > 0, FALSE if NPreg == 0
  EverPreg01:    0 if NPreg > 0,     1 if NPreg == 0

We read the data where missing values have already been coded and create a binary variable for “ever pregnant” (EverPreg).

library(erikmisc)
library(tidyverse)
# read data
dat_preg <-
  readr::read_csv(
    file = "ADA1_CL_14_Data-AH4_AgeAtInt-Preg.csv"
  )
Rows: 6504 Columns: 2
── Column specification ────────────────────────────────────────────────────────
Delimiter: ","
dbl (2): AgeAtInt, NPreg

ℹ Use `spec()` to retrieve the full column specification for this data.
ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.
dat_preg <-
  dat_preg |>
  mutate(
  # set as a binary variable, if ever pregnant, then 1, otherwise 0
    EverPreg   = (NPreg > 0)  # TRUE or FALSE (1 or 0)
    # alternatively, to code as 1 and 0
  , EverPreg01 = ifelse(NPreg > 0, 1, 0)  # 1 or 0
  ) |>
  # remove NAs (the NAs break the logi.hist.plot() below)
  # also, with only two variables in the dataset,
  #   no analysis can be done if either variable
  #   has a missing value for a row observation
  na.omit() |>
  # keep people between 12 and 22 years old at first vaginal intercourse
  filter(
    AgeAtInt >= 12
  , AgeAtInt <= 24
  )

Summarize observed probability of pregnancy for each age at first vaginal intercourse.

dat_preg_sum <-
  dat_preg |>
  group_by(
    AgeAtInt
  ) |>
  summarize(
    Success = sum(EverPreg)
  , Total   = n()
  # estimated proportion of preg for each age group
  , p_hat   = Success / Total
  , .groups = "drop_last"
  ) |>
  ungroup()

dat_preg_sum
# A tibble: 13 × 4
   AgeAtInt Success Total p_hat
      <dbl>   <int> <int> <dbl>
 1       12     131   161 0.814
 2       13     251   308 0.815
 3       14     357   462 0.773
 4       15     503   719 0.700
 5       16     581   844 0.688
 6       17     392   644 0.609
 7       18     342   596 0.574
 8       19     123   278 0.442
 9       20      87   190 0.458
10       21      65   179 0.363
11       22      37    96 0.385
12       23      25    56 0.446
13       24      15    46 0.326

Plot the data

Probability/proportion scale

Plots on the probability scale should follow a sigmoidal curve (a little difficult to see here).

Note

The overlayed reference curve (red) is a weighted smoothed curve (loess), not the model fit.

library(ggplot2)
p1 <- ggplot(dat_preg, aes(x = AgeAtInt, y = EverPreg01))
p1 <- p1 + theme_bw()
# data
p1 <- p1 + geom_jitter(width = 0.25, height = 0.025, size = 0.5, alpha = 1/10, colour = "green")
# summaries
p1 <- p1 + geom_point(data = dat_preg_sum, aes(x = AgeAtInt, y = p_hat, size = Total))
p1 <- p1 + geom_smooth(data = dat_preg_sum, aes(x = AgeAtInt, y = p_hat,weight = Total), se = FALSE, colour = "red")  # just for reference
# axes
p1 <- p1 + scale_y_continuous(breaks = seq(0, 1, by = 0.2))
p1 <- p1 + expand_limits(y = c(0, 1))
p1 <- p1 + scale_x_continuous(breaks = seq(0, 100, by = 2))
# labels
p1 <- p1 + labs(
    title   = "Proportion of at least one pregnancy by Age at first intercourse"
  , subtitle= "Observed data"
  , x       = "Age at first vaginal intercourse"
  , y       = "At least one pregnancy (0/1)\nObserved probability of at least one pregnancy"
  , caption = paste(
                "Green = Indicator points of at least one pregnancy (1) or not (0)."
              , "Black = Observed proportions of at least one pregnancy given age at first intercourse"
              , "Red = Smoothed curve to proportions"
              , sep = "\n" # separate by new lines
              )
  )
print(p1)
`geom_smooth()` using method = 'loess' and formula = 'y ~ x'

Logit scale

On the logit scale, if points follow a straight line, then we can fit a simple logistic regression model. :::{.callout-note} The overlayed reference straight line (blue) is from weighted least squares (not the official fitted line), and the curve (red) is a weighted smoothed curve (loess). ::: If these two lines are similar, that suggests a straight line on the logit scale is a good fit. Both lines are weighted by the total number of observations that each point represents, so that points representing few observations don’t contribute as much as points representing many observations, thus our decision should not be heavily influcenced by random deviations where there is little data.

# emperical logits
dat_preg_sum <-
  dat_preg_sum |>
  mutate(
    emp_logit = log((p_hat + 0.5/Total) / (1 - p_hat + 0.5/Total))
  )

# plot on logit scale
library(ggplot2)
p1 <- ggplot(dat_preg_sum, aes(x = AgeAtInt, y = emp_logit))
p1 <- p1 + theme_bw()

# summaries
p1 <- p1 + geom_point(aes(size = Total))
p1 <- p1 + stat_smooth(aes(weight = Total), method = "lm", se = FALSE, linetype = 3)  # just for reference
p1 <- p1 + geom_smooth(aes(weight = Total), se = FALSE, colour = "red")  # just for reference

# axes
p1 <- p1 + scale_y_continuous(breaks = seq(-10, 10, by = 0.5))
p1 <- p1 + scale_x_continuous(breaks = seq(0, 100, by = 2))
# labels
p1 <- p1 + labs(
    title   = "Logit of at least one pregnancy by Age at first intercourse"
  , subtitle= "Observed data"
  , x       = "Age at first vaginal intercourse"
  , y       = "Empirical logit of the probability of at least one pregnancy"
  , caption = paste(
                "Black = Observed logit proportions of at least one pregnancy given age at first intercourse"
              , "Blue = Naive LM fit of logit proportions"
              , "Red = Loess smooth curve of empirical logits"
              , sep = "\n" # separate by new lines
              )
  )
print(p1)
`geom_smooth()` using formula = 'y ~ x'
`geom_smooth()` using method = 'loess' and formula = 'y ~ x'

(1 p) For the plot above, on the logit scale, does it appear that a straight line fits the data well? If not, what are the features in the data that aren’t being captured by the model, and how concerned are you about any observed deviations from the straight line?

  • ???

Simple logistic regression model

The simple logistic regression model expresses the population proportion \(p\) of individuals with a given attribute (called the probability of success) as a function of a single predictor variable \(X\). The model assumes that \(p\) is related to \(X\) through \[ \log \left( \frac{p}{1-p} \right) = \beta_0 + \beta_1 X \] or, equivalently, as \[ p = \frac{ \exp( \beta_0 + \beta_1 X ) }{ 1 + \exp( \beta_0 + \beta_1 X ) }. \] The logistic regression model is a binary response model, where the response for each case falls into one of two exclusive and exhaustive categories, success (cases with the attribute of interest) and failure (cases without the attribute of interest).

Fit the model.

# For our summarized data (with frequencies and totals for each age)
# The left-hand side of our formula binds two columns together with cbind():
#   the columns are the number of "successes" and "failures".
# For logistic regression with logit link we specify family = binomial,
#   where logit is the default link function for the binomial family.

glm_p_a <-
  glm(
    cbind(Success, Total - Success) ~ AgeAtInt
  , family = binomial
  , data   = dat_preg_sum
  )

Deviance statistic for lack-of-fit

Unfortunately, there aren’t many model diagnostics for logistic regression. Visual inspection of the empirical logits is good when you can summarize the data as we’ve done.

One simple test for lack-of-fit uses the deviance statistic. Under the null hypothesis (that you’ll state below), the residual deviance follows a \(\chi^2\) distribution with the associated degrees-of-freedom.

Below, we calculate the deviance p-value and perform the test for lack-of-fit.

# Test residual deviance for lack-of-fit (if > 0.10, little-to-no lack-of-fit)
glm_p_a$deviance
[1] 17.36055
glm_p_a$df.residual
[1] 11
dev_p_val <- 1 - pchisq(glm_p_a$deviance, glm_p_a$df.residual)
dev_p_val
[1] 0.09765373

(1 p) State the null and alternative hypotheses for lack-of-fit.

  • \(H_0:\) ???

  • \(H_A:\) ???

(1 p) For your preferred model, the deviance statistic is

  • D = ??? with
  • ??? df, giving
  • p-value = ???.

(1 p) What is your conclusion for the model fit?

  • ???

Model visualization and interpretion

The glm() statement creates an object which we can use to create the fitted probabilities and 95% CIs for the population proportions at the ages at first vaginal intercourse. The fitted probabilities and the limits are stored in columns labeled fit_p, fit_p_lower, and fit_p_upper, respectively.

Below I create confidence bands for the model and plot the fit against the data, first on the logit scale to assess model fit and then on the probability scale for interpretation.

# predict() uses all the Load values in dataset, including appended values
fit_logit_pred <-
  predict(
    glm_p_a
  , data.frame(AgeAtInt = dat_preg_sum$AgeAtInt)
  , type   = "link"
  , se.fit = TRUE
  ) |>
  as_tibble()

# put the fitted values in the data.frame
dat_preg_sum <-
  dat_preg_sum |>
  mutate(
    # logit scale values
    fit_logit       = fit_logit_pred$fit
  , fit_logit_se    = fit_logit_pred$se.fit
  , fit_logit_lower = fit_logit - 1.96 * fit_logit_se
  , fit_logit_upper = fit_logit + 1.96 * fit_logit_se
  # proportion scale values
  , fit_p           = exp(fit_logit) / (1 + exp(fit_logit))
  , fit_p_lower     = exp(fit_logit_lower) / (1 + exp(fit_logit_lower))
  , fit_p_upper     = exp(fit_logit_upper) / (1 + exp(fit_logit_upper))
  )

Logit scale

# plot on logit scale
library(ggplot2)
p1 <- ggplot(dat_preg_sum, aes(x = AgeAtInt, y = emp_logit))
p1 <- p1 + theme_bw()

# MODEL FIT
# predicted curve and point-wise 95% CI
p1 <- p1 + geom_ribbon(aes(x = AgeAtInt, ymin = fit_logit_lower, ymax = fit_logit_upper), fill = "blue", linetype = 1, alpha = 0.2)
p1 <- p1 + geom_line(aes(x = AgeAtInt, y = fit_logit), colour = "blue", linewidth = 1)
# fitted values
p1 <- p1 + geom_point(aes(y = fit_logit), colour = "blue", size = 2)

# summaries
p1 <- p1 + geom_point(aes(size = Total))

# axes
p1 <- p1 + scale_y_continuous(breaks = seq(-10, 10, by = 0.5))
p1 <- p1 + scale_x_continuous(breaks = seq(0, 100, by = 2))
# labels
p1 <- p1 + labs(
    title   = "Logit of at least one pregnancy by Age at first intercourse"
  , subtitle= "Logistic model fit"
  , x       = "Age at first vaginal intercourse"
  , y       = "Logit scale of the probability of at least one pregnancy"
  , caption = paste(
                "Black = Observed logit proportions of at least one pregnancy given age at first intercourse"
              , "Blue = Logistic model fitted logit proportions"
              , sep = "\n" # separate by new lines
              )
  )
print(p1)

(1 p) For the plot above, comment on the model fit.

  • ???

Probability/proportion scale

# plot on probability scale
library(ggplot2)
p1 <- ggplot(dat_preg, aes(x = AgeAtInt, y = EverPreg01))
p1 <- p1 + theme_bw()
# data
p1 <- p1 + geom_jitter(width = 0.25, height = 0.025, size = 0.5, alpha = 1/10, colour = "green")
# summaries
p1 <- p1 + geom_point(data = dat_preg_sum, aes(x = AgeAtInt, y = p_hat, size = Total))

# MODEL FIT
# predicted curve and point-wise 95% CI
p1 <- p1 + geom_ribbon(data = dat_preg_sum, aes(x = AgeAtInt, y = fit_p, ymin = fit_p_lower, ymax = fit_p_upper), fill = "blue", linetype = 1, alpha = 0.2)
p1 <- p1 + geom_line(data = dat_preg_sum, aes(x = AgeAtInt, y = fit_p), colour = "blue", linewidth = 1)
# fitted values
p1 <- p1 + geom_point(data = dat_preg_sum, aes(y = fit_p), colour = "blue", size = 2)

# axes
p1 <- p1 + scale_y_continuous(breaks = seq(0, 1, by = 0.2))
p1 <- p1 + expand_limits(y = c(0, 1))
p1 <- p1 + scale_x_continuous(breaks = seq(0, 100, by = 2))
# labels
p1 <- p1 + labs(
    title   = "Proportion of at least one pregnancy by Age at first intercourse"
  , subtitle= "Logistic model fit"
  , x       = "Age at first vaginal intercourse"
  , y       = "At least one pregnancy (0/1)\nObserved probability of at least one pregnancy"
  , caption = paste(
                "Green = Indicator points of at least one pregnancy (1) or not (0)."
              , "Black = Observed proportions of at least one pregnancy given age at first intercourse"
              , "Blue = Logistic model fitted proportions"
              , sep = "\n" # separate by new lines
              )
  )
print(p1)

## ## Alternative plot.

 # plot histograms for each condition and fit a logistic curve
 #library(popbio)
 popbio::logi.hist.plot(
     dat_preg$AgeAtInt
   , dat_preg$EverPreg
   , logi.mod = 1 # logistic fit
   , type     = "hist", boxp = FALSE, rug = FALSE
   , col      = "gray"
   , ylabel   = "Probability at least one pregnancy (red)"
   , ylabel2  = "Frequency"
   , xlabel   = "Age at first vaginal intercourse"
   )

(1 p) For the plot above, describe the general pattern of the probability of pregnancy given the age at first vaginal intercourse.

  • ???

Interpret a model prediction

Let’s consider those people who first had vaginal intercourse at age 15.

dat_preg_sum |>
  filter(
    AgeAtInt == 15
  ) |>
  print(width = Inf)
# A tibble: 1 × 12
  AgeAtInt Success Total p_hat emp_logit fit_logit fit_logit_se fit_logit_lower
     <dbl>   <int> <int> <dbl>     <dbl>     <dbl>        <dbl>           <dbl>
1       15     503   719 0.700     0.844     0.924       0.0395           0.847
  fit_logit_upper fit_p fit_p_lower fit_p_upper
            <dbl> <dbl>       <dbl>       <dbl>
1            1.00 0.716       0.700       0.731

(1 p) Complete the statement below. Use R code in the sentence to automatically print the values, using the code for the first value to replace the question marks (???).

  • “Using the model, the estimated population proportion of pregnancy when age at first intercourse was 15 is ???. In repeated sampling, 95% of confidence intervals will contain the true population proportion. Our data gave us the confidence interval of (???, ???).”

The summary table gives MLEs and standard errors for the regression parameters. The z-value column is the parameter estimate divided by its standard error. The p-values are used to test whether the corresponding parameters of the logistic model are zero.

summary(glm_p_a)

Call:
glm(formula = cbind(Success, Total - Success) ~ AgeAtInt, family = binomial, 
    data = dat_preg_sum)

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
(Intercept)  4.21268    0.22268   18.92   <2e-16 ***
AgeAtInt    -0.21923    0.01313  -16.70   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 320.450  on 12  degrees of freedom
Residual deviance:  17.361  on 11  degrees of freedom
AIC: 96.833

Number of Fisher Scoring iterations: 3
  # see names(summary(glm_p_a)) to find the object that has the coefficients.
  # can also use coef(glm_p_a)

(1 p) Complete the equation below with the MLEs of the regression coefficients.

  • The MLE of the predicted probabilities satisfy the equation

    \[ \log \left( \frac{\hat{p}}{1-\hat{p}} \right) = ??? + ??? \textrm{ AgeAtInt} \].

Interpreting the model coefficients is tricky because they’re on the logit scale. We’d prefer to think of them on the probability scale. We’ll cover other ways of interpretating these coefficients next semester. For now, I want you to interpret qualities of the slope and intercept.

(1 p) Interpret the sign (\(+\) or \(-\)) of the slope.

  • ???

(1 p) What is the interpretation of the intercept? Is it meaningful?

  • ???
  • ???