Math 115: Introduction to Statistics – 6. Inference with Mathematical Models

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:

Resampling the original sample many times (with replacement)
Computing \(\hat{p}\) for each resample → bootstrap distribution
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:

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

The multiplier — z star — in our confidence interval formula depends on the confidence level. Three common ones: ninety percent, ninety-five percent, and ninety-nine percent — with z stars of one point six four five, one point nine six, and two point five seven six. Where do those numbers come from? For a ninety-five percent confidence interval, we want to capture the middle ninety-five percent of the standard normal — so two and a half percent gets left in each tail. z star is the value with two and a half percent above it: one point nine six. … Going from ninety-five to ninety-nine percent jumps z-star from one point nine six to two point five seven six — roughly thirty percent bigger, which means a roughly thirty percent wider interval at the same sample size. The Jamovi Randomize calculator handles any confidence level you want, not just these three.

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).

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

Your turn. A campus survey of four hundred randomly sampled university students — two hundred seventy-two say they’ve used a generative AI tool for a class assignment. p hat is zero point six eight. Conditions are met. Work through all four parts, then advance when you are ready to see the answers. First, the SE and the ninety-five percent confidence interval. Standard error is the square root of zero point six eight times zero point three two, divided by four hundred — about zero point zero two three three. The confidence interval is zero point six eight plus or minus one point nine six times that — roughly zero point six three four to zero point seven two six. Now interpret in plain English. We are ninety-five percent confident that the true proportion of university students who have used a generative AI tool for a class assignment is between sixty three point four percent and seventy two point six percent. Next, the ninety-nine percent confidence interval, using z-star equals two point five seven six. Zero point six eight plus or minus two point five seven six times zero point zero two three three — roughly zero point six two zero to zero point seven four zero. Wider than the ninety-five percent, exactly as the formula predicts. Last one — without computing, will the ninety percent confidence interval be wider or narrower than the ninety-five? Narrower. z-star drops from one point nine six to one point six four five, so the margin of error shrinks. Less confidence required means a 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.

We’ve been building confidence intervals assuming the conditions hold — let’s see what those conditions are actually buying us. Both blue histograms come from simulation — for each panel, we drew ten thousand samples of size n from a population with the listed true proportion p, computed p hat each time, and plotted those ten thousand p hats. So the histogram shows the actual shape of the sampling distribution in that scenario. The red curve is what the normal model predicts for the same scenario. The question is whether the red curve matches the simulated truth. On the left, n equals one hundred and p equals one half — both observed-counts checks pass, with fifty each. The histogram is beautifully bell-shaped, and the red normal curve sits right on top of it — a great fit. On the right, n equals thirty and p equals zero point zero five — only one and a half expected successes, far below ten. The true distribution is right-skewed, and the normal curve doesn’t fit the skew. So the normal model is a shortcut, not a free pass — when conditions hold, it gets you the right answer fast; when they don’t, you fall back on simulation.

Simulation or Normal Model?

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

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.
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.
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.

Your turn — for each scenario you want a ninety-five percent confidence interval for p. Which approach is right: simulation or the normal model? Advance when you are ready to see the answers. First — random sample of twelve hundred likely voters with p hat zero point five two. Observed successes are about six hundred twenty-four; observed failures about five hundred seventy-six. Both far above ten — conditions met, so use the normal model formula. Next — pilot trial of twenty-five patients, five report substantial relief, p hat zero point two zero. Only five observed successes — that fails the success-failure check since it’s less than ten. Fall back on the bootstrap percentile confidence interval from topic 4. Last — quality-control sample of two hundred widgets, three are defective, p hat zero point zero one five. Even though n is reasonable here, only three observed successes — still fails success-failure. Same fix: bootstrap percentile confidence interval.

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

Introduction to Modern Statistics (2e) textbook by Mine Çetinkaya-Rundel and Johanna Hardin
Chapter 13, Section 16.2