6. Inference with Mathematical Models

b. Confidence Intervals

Yurk

Where We’ve Been

Yesterday (6a): the normal model gave us a math shortcut for hypothesis tests on a proportion.

  • Check conditions → the sampling distribution of \hat{p} is roughly normal
  • Standardize \hat{p} to a Z-score → read the p-value as a tail area

Today: the same normal model gives us a math shortcut for confidence intervals.

Recall: The Bootstrap CI

Back in 4 we built a 95% CI by:

  1. Resampling the original sample many times (with replacement)
  2. Computing \hat{p} for each resample → bootstrap distribution
  3. Reading off the 2.5th and 97.5th percentiles

Because the bootstrap distribution was approximately normal and centered at \hat{p}, those 2.5th and 97.5th percentiles fall close to \hat{p} \pm 1.96 \cdot SE — and that’s the formula we’ll use today.

CI Formula via the Normal Model

If the sampling distribution of \hat{p} is approximately normal, the confidence interval is:

\hat{p} \;\pm\; z^* \cdot SE

where:

  • \hat{p} — the observed proportion
  • z^* — critical value (multiplier) for the chosen confidence level
  • SE = \sqrt{\dfrac{\hat{p}\,(1-\hat{p})}{n}}

Conditions: CI for One Proportion

Before using the normal model for a CI, check:

  1. Independence — observations are independent (random sample, sample < 10% of population)

  2. Success-failure — using \hat{p} this time:

    • n \cdot \hat{p} \geq 10 (observed successes)
    • n \cdot (1 - \hat{p}) \geq 10 (observed failures)
    • At least 10 successes AND 10 failures observed

Key difference from H-test: a CI plugs in \hat{p}; an H-test plugs in p_0. A CI has no null value to assume.

Critical Values: z*

Confidence level z^*
90% 1.645
95% 1.960
99% 2.576

These come straight from the standard normal:

  • 95% confident → capture the middle 95% of the standard normal → 2.5% left in each tail → z^* is the value with 2.5% above it, 1.96.

The Model-Based Inference calculator in Jamovi’s Randomize module gives you z^* for any confidence level (90%, 92%, 99%, …) — not just the three above. It also computes the CI and p-value directly.

Margin of Error = Dart Radius

Back in 4a: each \hat{p} is a dart, p is the bullseye. The CI is a circle around our dart, sized to catch the bullseye some fraction of the time.

The margin of error is the radius of that circle:

\text{margin of error} \;=\; z^* \cdot SE

So the CI is just \hat{p} \;\pm\; margin of error.

Whatever makes the radius bigger makes the CI wider — next slide names what controls it.

What Drives the CI Width?

\hat{p} \;\pm\; \underbrace{z^* \cdot SE}_{\text{margin of error}} \;=\; \hat{p} \;\pm\; z^* \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Three things change the margin of error — all visible in the formula:

Change Effect Why
Confidence level goes up wider z^* gets bigger
Sample size goes up narrower SE gets smaller
\hat{p} closer to 0.5 wider \hat{p}(1-\hat{p}) gets bigger

Topic 4b discovered these patterns by experiment with the bootstrap. Now you can read them straight off the formula.

Recall: Payday Loan Study

From 6a: in a national random sample of n = 826 payday loan borrowers, 424 (51.3%) said they support additional regulation.

Yesterday’s H-test: H_0: p = 0.5 vs H_A: p > 0.5 → p-value = 0.222 → failed to reject H_0.

Today’s question: what range of values for p is plausible given our data? Build a 95% confidence interval.

95% CI for Payday

Same data as 6a: n = 826, \hat{p} = 0.513.

Conditions:

  • Random sample → ✓ independence
  • Observed successes: 424 ≥ 10 ✓    Observed failures: 402 ≥ 10 ✓

Standard error (uses \hat{p}): SE = \sqrt{\dfrac{0.513 \times 0.487}{826}} = 0.0174

95% CI: 0.513 \;\pm\; 1.96 \times 0.0174 \;=\; (0.479,\; 0.547)

Payday — Interpret the CI

95% CI: (0.479,\; 0.547)

We are 95% confident that the true proportion of payday borrowers who support the regulation is between 47.9% and 54.7%.

Notice: the interval includes 0.5 — the null value from yesterday’s hypothesis test. That’s consistent with our H-test result (we failed to reject H_0: p = 0.5). Topic 7 will tighten this “consistent with” into a precise equivalence.

Build CIs at Different Levels

A campus survey of n = 400 randomly sampled university students asks whether they’ve used a generative-AI tool for a class assignment. 272 say yes (\hat{p} = 0.68). Conditions are met (272 ≥ 10 successes, 128 ≥ 10 failures).

  1. Compute SE, then build the 95% CI. SE = \sqrt{\dfrac{0.68 \times 0.32}{400}} \approx 0.0233.   95% CI: 0.68 \pm 1.96 \times 0.0233 \approx (0.634,\; 0.726).

  2. Interpret the 95% CI in plain English. We are 95% confident that the true proportion of university students who have used a generative-AI tool for a class assignment is between 63.4% and 72.6%.

  3. Now build the 99% CI. (use z^* = 2.576) 0.68 \pm 2.576 \times 0.0233 \approx (0.620,\; 0.740) — wider, as the formula predicts.

  4. Without computing, will the 90% CI be wider or narrower than the 95%? Why? Narrower. z^* drops from 1.96 to 1.645, so the margin of error shrinks. Less confidence required → smaller circle around our dart.

When Does the Normal Model Work?

Blue bars = simulated sampling distribution. Red curve = normal approximation.

Left (n=100, p=0.5): success-failure passes (np = 50, n(1-p) = 50) — great fit.

Right (n=30, p=0.05): success-failure fails (np = 1.5 < 10) — true distribution is right-skewed, normal model misses it. That’s why we check the conditions.

Simulation or Normal Model?

For each scenario, you want a 95% CI for p. Which approach should you use — simulation or the normal model?

  1. Random sample of n = 1{,}200 likely voters; \hat{p} = 0.52. Normal model. Observed successes \approx 624 and failures \approx 576 — both far above 10. Conditions met, so use the formula.

  2. Pilot trial of n = 25 patients on a new pain medication; 5 report substantial relief (\hat{p} = 0.20). Simulation. Only 5 observed successes — fails success-failure. Use the bootstrap percentile CI from Topic 4.

  3. Quality-control sample of n = 200 widgets; 3 are defective (\hat{p} = 0.015). Simulation. Only 3 observed successes — fails success-failure (rare events even at moderate n). Use the bootstrap percentile CI from Topic 4.

Summary

CI from the normal model: \;\hat{p} \pm z^* \cdot SE, with SE = \sqrt{\hat{p}(1-\hat{p})/n}.

Conditions to check:

  • Independence
  • n\hat{p} \geq 10 and n(1-\hat{p}) \geq 10 — at least 10 successes and 10 failures observed

When both conditions hold: simulation and the normal model give similar answers, and the formula is faster. Success-failure fails (independence OK): fall back on the bootstrap. Independence fails: both methods break down — the data themselves need a closer look.

References

Course Home