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

Where We’ve Been

So far we’ve done inference with simulation:

Hypothesis tests — null distribution by simulating \(H_0\) (Topic 3)
Confidence intervals — bootstrap distribution by resampling (Topic 4)

Both are examples of a sampling distribution — a distribution showing how a statistic like \(\hat{p}\) varies across many possible samples.

These methods work, but each needs thousands of simulated samples. Today’s question: Is there a math shortcut?

Sampling Distributions Look Bell-Shaped

Across very different \(n\) and \(p\), the simulated sampling distribution of \(\hat{p}\) is bell-shaped. This isn’t a coincidence — it’s a general pattern (with one important caveat we’ll see soon).

The Normal Distribution

The bell-shaped curve is called a normal distribution.

Properties:

Symmetric and unimodal
Fully described by two parameters:
- Mean (\(\mu\)) — center
- Standard deviation (\(\sigma\)) — spread

Notation: \(N(\mu, \sigma)\)

Same Family, Different Parameters

Mean shifts the curve left/right; SD controls how wide it is.

Z-Scores

A Z-score measures how many standard deviations an observation sits from the mean:

\[Z = \frac{\text{observation} - \text{mean}}{\text{SD}}\]

\(Z = 2\) → 2 SDs above the mean
\(Z = -1.5\) → 1.5 SDs below the mean
\(Z = 0\) → exactly at the mean

Z-scores let us compare values across very different normal distributions.

Quick Z-Score Check

A class’s exam scores are approximately normal with \(\mu = 75\), \(\sigma = 8\).

Compute the Z-score for a student who scored 91. \(Z = \dfrac{91 - 75}{8} = \dfrac{16}{8} = 2.0\) — two SDs above the mean.
Compute the Z-score for a student who scored 63. \(Z = \dfrac{63 - 75}{8} = \dfrac{-12}{8} = -1.5\) — one and a half SDs below the mean.
What’s the Z-score of a student who scored exactly 75? What does that tell you? \(Z = (75 - 75)/8 = 0\). The score sits at the mean — neither above nor below. \(Z = 0\) always means “equal to the mean.”

Your turn. A class’s exam scores are approximately normal, with a mean of seventy-five and a standard deviation of eight. Work through all three Z-scores yourself, then advance when you are ready to see the answers. The first question asks for the Z-score of a student who scored ninety-one. Ninety-one minus seventy-five is sixteen; sixteen over eight is two. So Z equals two, meaning the student is two standard deviations above the class mean — well above average. The second question asks for the Z-score of a student who scored sixty-three. Sixty-three minus seventy-five is negative twelve; negative twelve over eight is negative one point five. That puts the student one and a half standard deviations below the mean — below average, though not extreme. The third question asks about a student who scored exactly seventy-five — the class mean itself. Seventy-five minus seventy-five is zero, and zero divided by eight is still zero. A Z-score of zero always means the observation sits right at the mean, neither above nor below.

The Standard Normal

Converting to Z-scores standardizes any normal distribution to \(N(0, 1)\).

Why this matters: one table (or one Jamovi calculator) handles every normal distribution.

68-95-99.7 Rule

68% of values fall within 1 SD of the mean
95% within 2 SDs (more precisely, 1.96)
99.7% within 3 SDs

Probabilities Are Areas Under the Curve

Key idea: the probability of getting a Z-score in some range is the area under the normal curve in that range.

The shaded area between \(Z = -1\) and \(Z = 1\) is ≈ 68% — that’s exactly the 68 in the 68-95-99.7 rule, now reframed as a probability.

SAT Scores

SAT scores are approximately normal with \(\mu = 1050\), \(\sigma = 210\).

Question: What proportion of test-takers score above 1470?

Step 1 — Z-score: \[Z = \dfrac{1470 - 1050}{210} = \dfrac{420}{210} = 2\]

So 1470 sits 2 SDs above the mean. The next slide turns that Z-score into a probability.

SAT — The Top 2.5%

Step 2 — Use the 68-95-99.7 rule.

About 95% of scores fall between \(Z = -2\) and \(Z = 2\) (within 2 SDs of the mean). That leaves about 5% outside — split between the two tails.

The normal curve is symmetric, so each tail gets half: ≈ 2.5% above \(Z = +2\), ≈ 2.5% below \(Z = -2\).

→ About 2.5% of test-takers score above 1470.

Heights and the 68-95-99.7 Rule

US adult female heights are approximately normal with \(\mu = 64\) in, \(\sigma = 3\) in.

About what percent of women are between 61 and 67 inches tall? That range is \(\mu \pm 1\,\text{SD}\) (\(Z = -1\) to \(Z = 1\)). By 68-95-99.7, about 68%.
About what percent are between 58 and 70 inches tall? That’s \(\mu \pm 2\,\text{SD}\) (\(Z = -2\) to \(Z = 2\)) — about 95%.
A woman is 58 inches tall. Is this unusual? \(Z = (58 - 64)/3 = -2\). Only about 2.5% of women are this short or shorter — yes, unusual.
About what percent are shorter than 61 inches? \(Z = (61-64)/3 = -1\). About 16% — the 32% outside \(\mu \pm 1\,\text{SD}\) split symmetrically into 16% per tail.

Your turn. Heights of US adult women are approximately normal, with a mean of sixty-four inches and a standard deviation of three inches. Use the 68-95-99.7 rule to answer four questions — advance when you are ready to see the answers. The first question asks roughly what percent of women are between sixty-one and sixty-seven inches tall. Sixty-one is one standard deviation below sixty-four, and sixty-seven is one above — so this range is the central one-SD region. By the 68-95-99.7 rule, that’s about sixty-eight percent. The second question widens the range — between fifty-eight and seventy inches. That’s two SDs on each side of the mean, so by the rule, about ninety-five percent of women fall in there. The third question asks whether a woman who is fifty-eight inches tall is unusually short. Her Z-score is negative two, putting her two SDs below the mean. The 95 percent inside plus-or-minus two SDs leaves five percent in the two tails combined — about two and a half percent in each. So only about two and a half percent of women are this short or shorter, which does qualify as unusual. The fourth question asks what percent of women are shorter than sixty-one inches. Z equals negative one. The thirty-two percent outside the central sixty-eight percent splits evenly into the two tails, so about sixteen percent of women are shorter than sixty-one inches.

When Does the Normal Model Apply?

Now back to our problem. When does the sampling distribution of \(\hat{p}\) behave like a normal distribution?

For a single proportion, \(\hat{p}\) is approximately normal when both of these technical conditions (also called validity conditions) hold:

Independence — observations are independent (random sample, sample < 10% of population)
Success-failure — at least 10 expected successes and 10 expected failures
- For an H-test, “expected” is computed under \(H_0\): check \(n \cdot p_0 \geq 10\) and \(n \cdot (1-p_0) \geq 10\)

We’ve spent the last stretch on the normal distribution itself — but the opening promise of this lecture was a math shortcut for inference. The catch is that the shortcut isn’t always available. Sometimes the sampling distribution of p hat really does look normal, and we can swap simulation for a formula. Other times it doesn’t, and we have to stick with simulation. Two technical conditions tell us which case we’re in. First, independence: each observation has to be independent of the others. A random sample gives us that, provided the sample is less than 10 percent of the population. Second, the success-failure condition: at least 10 expected successes and 10 expected failures. For a hypothesis test, “expected” is computed under the null — that’s where p naught comes in. We plug in p naught and check that n times p naught is at least 10, and n times one minus p naught is at least 10.

The Normal Model for p̂

When both technical conditions hold:

\[\hat{p} \;\sim\; N\!\left(p,\; \sqrt{\tfrac{p(1-p)}{n}}\right)\]

A special case of the Central Limit Theorem (CLT) — sample proportions and averages become approximately normal as \(n\) grows.

Both hold → normal model. Success-failure fails (independence OK) → simulation. Independence fails → both break down.

When both conditions hold, we have a clean mathematical statement: the sampling distribution of p hat is approximately normal, centered at the true proportion p, with standard deviation equal to the square root of p times one minus p, divided by n. This is the normal model. It’s a special case of one of the most important results in statistics — the Central Limit Theorem, often abbreviated CLT — which says that sample proportions and averages tend toward a normal distribution as the sample grows. The bell shape we kept seeing earlier in this lecture? That’s the CLT in action. When both technical conditions hold, the formula gives us the sampling distribution — no simulation required. If success-failure fails but independence holds, simulation still works. If independence itself fails, both methods break down — the issue isn’t the math, it’s the data.

Standard Error

The standard error (SE) is the SD of the sampling distribution of \(\hat{p}\):

\[SE = \sqrt{\frac{p(1-p)}{n}}\]

Which \(p\) do we plug in?

Hypothesis tests: use \(p_0\) (the null value) — we’re asking “if \(H_0\) were true…”
Confidence intervals: use \(\hat{p}\) (next time, in 6b)

The Test Statistic

For an H-test, convert \(\hat{p}\) to a Z-score measuring how many SEs \(\hat{p}\) sits from \(p_0\):

\[Z = \frac{\hat{p} - p_0}{SE}\]

The p-value is the tail area of \(N(0,1)\) beyond \(Z\) — the side that matches \(H_A\).

Payday Loan Study

Background: a payday loan is a small, short-term loan due on the borrower’s next paycheck, typically with extremely high interest (often 300%+ APR). Critics worry borrowers get trapped in cycles of debt. Should the industry be more tightly regulated? The borrowers themselves are a key voice.

The data: in a national random sample of \(n = 826\) payday loan borrowers, 424 (51.3%) said they would support additional regulation of the industry.

We’ll use this single study for both 6a (hypothesis test, today) and 6b (confidence interval, next class) — two methods, two angles on the same question.

H-Test on Payday

Research question: Do a majority of payday borrowers support regulation?

Hypotheses:

\(H_0: p = 0.5\) (no majority — exactly half support)
\(H_A: p > 0.5\) (majority supports)

We’ll use \(\alpha = 0.05\) as our significance level.

Payday — Check Conditions

Independence: Random sample of borrowers. ✓

Success-failure (under \(H_0\), so use \(p_0 = 0.5\)):

Expected successes: \(n \cdot p_0 = 826 \times 0.5 = 413\) ✓
Expected failures: \(n \cdot (1-p_0) = 826 \times 0.5 = 413\) ✓

Both well above 10 → conditions met. We can use the normal model.

Payday — Compute SE and Z

Standard error: \(SE = \sqrt{\dfrac{0.5 \times 0.5}{826}} \approx 0.0174\)

Z-score: \(Z = \dfrac{0.5133 - 0.5}{0.0174} \approx 0.77\)

So \(\hat{p}\) sits less than 1 SE above \(p_0\). Before crunching the precise tail area, let’s get a rough sense from the 68-95-99.7 rule.

Payday — Rough P-value from 68-95-99.7

By 68-95-99.7: the right tail beyond \(Z = 1\) is ≈ 16%.

The area we want includes that 16% region — so our p-value > 16%, well above \(\alpha = 0.05\). Already pointing to fail to reject \(H_0\).

Payday — Precise P-value

Precise p-value (Jamovi’s Model-Based Inference calculator, or any standard-normal table): \(0.222\).

Conclusion: \(0.222 > 0.05\). We fail to reject \(H_0\). There’s not convincing evidence that a majority of payday borrowers support the regulation. (The national random sample lets us generalize this to US payday borrowers.)

Run a Hypothesis Test

A national report places the rate of seniors with full-time jobs by graduation at 75%. Your university suspects its own rate is higher. In a random sample of \(n = 200\) recent grads, 164 report having a job by graduation (\(\hat{p} = 0.82\)). Is there evidence the true rate at your university exceeds 75%? Use \(\alpha = 0.05\).

State \(H_0\) and \(H_A\). \(H_0: p = 0.75\) vs \(H_A: p > 0.75\)
Conditions met? Random sample → ✓ independence. \(200 \times 0.75 = 150 \geq 10\) and \(200 \times 0.25 = 50 \geq 10\) → ✓ success-failure.
Compute SE and \(Z\). \(SE = \sqrt{\frac{0.75 \times 0.25}{200}} \approx 0.0306\). \(Z = \frac{0.82 - 0.75}{0.0306} \approx 2.29\).
P-value and decision. Right-tail area beyond \(Z = 2.29\) is about 0.011. Since \(0.011 < 0.05\), reject \(H_0\) — evidence the true employment rate is above 75%. (The random sample of recent grads lets us generalize to your university’s recent grads.)

Your turn for a full one-sided hypothesis test. A national report says seventy-five percent of seniors have full-time jobs by graduation. Your university suspects its rate is higher. In a random sample of two hundred recent grads, one hundred sixty-four reported having a job by graduation — that’s eighty-two percent. Use alpha equals zero point zero five. Work through all four steps yourself, then advance when you are ready to see the answers. The first step is the hypotheses. H naught says p equals zero point seven five — the national rate. The alternative is p greater than zero point seven five, matching the university’s suspicion that its rate is higher. The second step is checking conditions. The random sample gives independence. Under the null, n times p naught is two hundred times zero point seven five, which is one hundred fifty expected employed, and n times one minus p naught is two hundred times zero point two five, which is fifty expected unemployed. Both far above ten, so success-failure is met. The third step computes the standard error and Z-score. SE is the square root of zero point seven five times zero point two five divided by two hundred, which works out to about zero point zero three zero six. Z is zero point eight two minus zero point seven five divided by that SE, which gives about two point two nine. So the observed p hat sits roughly two point three standard errors above the null value. The fourth step turns Z equals two point two nine into a decision. Even before computing the precise p-value, the 68-95-99.7 rule pins this down: Z equals plus two marks the right two and a half percent tail of the standard normal, and our Z is even further out, so the right tail beyond it is less than two and a half percent. The p-value is under zero point zero two five — already below alpha equals zero point zero five — so we reject H naught. A precise calculation gives a p-value of about zero point zero one one, which confirms that call: there’s evidence the university’s true employment rate exceeds seventy-five percent.

When Conditions Are NOT Met

Scenario (back to Topic 3): A medical consultant had 3 complications in 62 surgeries. We want to test:

\[H_0: p = 0.1 \quad\text{vs}\quad H_A: p < 0.1\]

Check success-failure under \(p_0 = 0.1\):

Expected successes: \(62 \times 0.1 = 6.2\) ✗

Only 6.2 expected complications — less than 10. Normal model not appropriate.

→ Fall back on simulation (our Topic 3 approach). The math shortcut isn’t always available.

Conditions Met?

For each scenario, decide whether the normal model is appropriate for a hypothesis test on \(p\).

Random sample of \(n = 50\) adults screened for a rare condition; testing whether local prevalence differs from the national 10% (\(H_0: p = 0.10\)). No. Expected positives: \(50 \times 0.10 = 5\) — fewer than 10. Fall back on simulation.
Random sample of \(n = 150\) voters polled on a ballot measure; testing whether a majority support it (\(H_0: p = 0.50\)). Yes. Expected supporters: \(150 \times 0.5 = 75\); expected opposers: 75. Both well above 10.
Convenience sample of \(n = 500\) mall shoppers asked whether they used contactless payment last visit; testing \(H_0: p = 0.30\). Watch out. Success-failure is fine — expected users: \(500 \times 0.3 = 150\), non-users: 350. But independence is questionable: shoppers often arrive in groups (couples, families) and influence each other’s payment habits. Independence is required for both the normal model and simulation, so neither gives reliable inference here.

Three quick condition checks. For each scenario, decide whether the normal model is appropriate for a hypothesis test on p — advance when you are ready to see the answers. The first scenario tests fifty adults randomly screened for a rare condition, with national prevalence ten percent. Under the null, the expected number of positives is fifty times zero point one, which is only five — fewer than ten. Success-failure fails, so the normal model isn’t appropriate here, and we’d fall back on simulation. The second scenario polls one hundred fifty voters on a ballot measure with H naught: p equals zero point five. Expected supporters under the null are seventy-five, and expected opposers are also seventy-five. Both are well above ten, so both conditions are met and the normal model is fine to use. The third scenario surveys five hundred mall shoppers about contactless payment, testing H naught: p equals zero point three. The success-failure numbers look fine — one hundred fifty expected users and three hundred fifty expected non-users. But the catch is independence: shoppers often arrive in groups, and people in couples or families tend to influence each other’s payment habits, so the responses aren’t truly independent. And here’s the part worth holding onto — independence is required for both the normal model and simulation. When independence fails, neither approach gives reliable inference, and the fix has to come from the data, not from a different method.

Looking Ahead

Today: normal model gives us a math shortcut for hypothesis tests on a proportion.

Next time (6b): the same normal model gives us a math shortcut for confidence intervals — and we’ll see when the two methods agree (and when they don’t).

References

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