Math 115: Introduction to Statistics – 3. Introduction to Hypothesis Testing

The Big Picture

So far we’ve described data (e.g., center, spread, proportions)

Now we ask harder questions:

Does this drug actually work?
Are college graduation rates accurate?
Are MLB pitchers really more left-handed than the general population?

Each of these requires an inference — a statement about an unknown population, supported by data from a sample. Hypothesis testing is the framework; we’ll start with studies involving a single proportion.

One Proportion

These studies involve a single binary categorical variable — two levels, one labeled the success. The statistic of interest is a single proportion, denoted \(\hat{p}\).

Examples:

Variable	Levels (success in bold)	\(\hat{p}\) is the proportion of…
vote choice	A / B	voters choosing A
side effect	yes / no	patients with side effects
product condition	defective / OK	defective products

Note: “success” just means the outcome we’re counting — it doesn’t have to be a positive outcome.

These studies share one defining feature — a single binary categorical variable, with two levels, where we label one of them the success. The statistic of interest is the proportion of cases in the success category — that’s p hat. The table shows three quick examples. First, imagine a pre-election poll: each respondent is asked which candidate they support — A or B. We label candidate A the success, so p hat is the proportion of voters who chose candidate A. Second, picture a clinical trial in which we record whether each patient experiences a particular side effect — yes or no. Here we treat “yes” as the success, so p hat is the proportion of patients who did. Third, think of a quality-control inspection at a factory, where we check whether each item off the line is defective or OK. Defective is the success here, so p hat is the proportion that come out defective. And remember — success doesn’t have to mean a positive outcome; it just means the outcome we’re counting.

Statistical Notation: The “Hat”

The hat (as in \(\hat{p}\)) denotes an estimate from our sample:

\(p\) — population parameter

The TRUE proportion (unknown).

\(\hat{p}\) — sample statistic

Our ESTIMATE from data.

The true population proportion \(p\) is unknown — that’s what we’re trying to learn about.

Computing p̂

Count the successes (\(x\)) and divide by the sample size (\(n\)):

\[\hat{p} = \frac{x}{n}\]

Example: 12 of 50 students sampled prefer online classes:

\[\hat{p} = \frac{12}{50} = 0.24\]

Identify the Pieces

Scenario: a streaming service surveys 80 of its users. 56 of them watch more than 10 hours of programming per week.

Identify each piece — advance for answers:

Question	Answer
Cases?	the 80 surveyed users
Variable?	does the user watch more than 10 hours per week?
Two levels?	yes and no. Below we’ll treat “yes” as the success.
Single-proportion study?	Yes — one binary variable
Parameter \(p\)?	the true proportion of all users who watch more than 10 hours per week (unknown)
Sample size \(n\)?	80
Count of successes \(x\)?	56
Sample statistic \(\hat{p}\)?	\(\hat{p} = x/n = 56/80 = 0.70\)

Sampling Variability

Imagine we knew the true left-handedness rate was \(p = 0.10\). Three samples of 100 people might yield \(\hat{p}\) = 0.08, 0.13, 0.11. This natural sample-to-sample variation is sampling variability.

In real life we don’t know \(p\). We see one sample and have to reason backward: when our \(\hat{p}\) looks unusual,

is something real going on in the population we sampled from, or is it just due to sampling variability?

Medical Consultant

Setup: a consultant claims her liver-donor patients have a lower complication rate than the 10% national average.

Data: 3 complications in 62 surgeries.

\[\hat{p} = \frac{3}{62} = 0.048 \quad (\approx 4.8\%)\]

Is \(\hat{p} = 0.048\) evidence that the consultant’s true rate \(p < 0.10\)?

Source: Çetinkaya-Rundel & Hardin, Introduction to Modern Statistics.

Innocent Until Proven Guilty

We assume the consultant is no different from the national rate — and reject that assumption only if our data look very unusual.

Step	Courtroom	Hypothesis test
1.	Presume innocence	Assume \(p = 0.10\)
2.	Look for evidence of guilt	Compare data to chance
3.	Convict only if evidence is strong	Reject only if data look unusual

Null Hypothesis (\(H_0\))

The null hypothesis is the skeptical position — what we assume is true and look for evidence against.

\[H_0: p = 0.10\]

“The consultant’s rate equals the national average.”

Alternative Hypothesis (\(H_A\))

The alternative hypothesis is the claim we want to test:

\[H_A: p < 0.10\]

“The consultant’s rate is lower than the national average.”

Important: hypotheses are about the population parameter \(p\), not the sample statistic \(\hat{p}\). We already know \(\hat{p} = 0.048\) — the question is what \(p\) is.

State the Hypotheses

Scenario: a college claims its 4-year graduation rate is 70%, but a consumer-research organization suspects the actual rate is lower. They follow a sample of 80 students from a recent cohort and find that 48 graduated within 4 years.

Set up the test — advance for answers:

Question	Answer
Parameter of interest?	\(p\) = the college’s true 4-year graduation rate
\(H_0\)?	\(H_0: p = 0.70\)
\(H_A\)?	\(H_A: p < 0.70\)
\(\hat{p}\)?	\(48 / 80 = 0.60\)

Try this one yourself. A college claims its four-year graduation rate is seventy percent. Consumer-research organizations sometimes audit these self-reported numbers — and one of them suspects this college’s actual rate is lower. They follow a sample of eighty students from a recent cohort, and find that forty-eight of them graduated within four years. Pause and identify the parameter, both hypotheses, and p hat. Advance when you’re ready to check your answers. The parameter of interest is p — the college’s true four-year graduation rate. The null hypothesis H naught says p equals zero point seven zero — that’s the college’s claim. The alternative H A says p is less than zero point seven zero — because the organization suspects the rate is lower. And p hat from this sample is forty-eight over eighty, which is zero point six.

The Testing Strategy

Assume \(H_0\) is true (consultant rate = 0.10).
Simulate what we’d expect to see under that assumption.
Compare our observed \(\hat{p}\) to the simulations.
Decide:
- If our data look very unusual under \(H_0\) → reject \(H_0\).
- If they look plausible → cannot reject \(H_0\).

We’ll spend the rest of this lecture seeing what “simulate” looks like. Making “unusual” precise comes next class.

Simulating One Sample

To test the hypothesis, we simulate samples we’d expect to see if \(H_0\) (\(p = 0.10\)) were true. Two ways to simulate one sample:

Spinner

10% orange (complication), 90% blue (no complication). Spin 62 times, count the complications, and divide by 62 to get a simulated \(\hat{p}\).

Random number generator

Draw 62 numbers from 1 to 10 (e.g. via random.org). Each “1” = complication; “2–10” = no complication. Count the 1’s and divide by 62 to get a simulated \(\hat{p}\).

What does “simulate a sample under H naught” actually mean in practice? It means setting up a physical or digital mechanism that produces ten-percent successes by design, running it sixty-two times to mimic the consultant’s sample size, and recording the resulting p hat. The slide shows two ways to build that mechanism — a spinner ten-percent orange, or a random number generator pulling integers one through ten with the ones counted as complications. The mechanism doesn’t matter; what matters is that the success rate is exactly ten percent — that’s the under H naught part — and the sample size is exactly sixty-two — that’s the like the consultant’s sample part. Either way, each run gives us one possible p hat under H naught — one realization of what the consultant’s data could have looked like if the null were really true.

Try It Yourself

Activity: simulate one sample under \(H_0\). Go to random.org/integers and generate 62 random integers from 1 to 10 (each “1” = complication).

Identify each piece — advance for answers:

Question	Answer
Count of successes \(x\)?	the number of 1’s you observed
Sample size \(n\)?	62
Simulated \(\hat{p}\)?	\(\hat{p} = x/n\)

Bonus: try the interactive spinner demo for an animated version (Run 1 Sample).

Repeat 100 Times

How did your simulated \(\hat{p}\) compare to the consultant’s observed \(\hat{p} = 0.048\)?
Each simulated sample (62 spins, or 62 random integers) gives one \(\hat{p}\) — the proportion of times the spinner landed on orange, or the proportion of 1’s in the draw.
Imagine doing this 100 times. The collection of those 100 simulated \(\hat{p}\) values is the null distribution — what we’d expect to see when \(H_0\) is true.

100 Simulations: The Null Distribution

This collection of 100 simulated \(\hat{p}\) values is the null distribution — what’s typical when \(H_0\) is true. It’s centered around \(p = 0.10\) (the rate we assumed), not around our observed \(\hat{p} = 0.048\) — because each simulation came from a population where the true rate really is 0.10.

Where Does Our p̂ Land?

The red dashed line marks the consultant’s observed \(\hat{p} = 0.048\). The 9 pink dots are simulations at or below it — putting our observed value in the left tail. Unusual enough to reject \(H_0\)? With only 100 simulations, hard to tell — let’s do more.

1000 Simulations: A Sharper Picture

Same null distribution, sharper picture — still centered around \(p = 0.10\). With 1000 simulations the dots overlap, so we switch to a histogram. 111 of 1000 simulated proportions are at or below 0.048.

Surprising Enough?

So if \(H_0\) is really true, we’d only expect to see a result like the consultant’s — or more extreme — about 11.1% of the time.

Surprising enough to reject \(H_0\)? Next class, we’ll dig in.

References

Introduction to Modern Statistics (2e) textbook by Mine Cetinkaya-Rundel and Johanna Hardin
Chapter 11, Section 16.1