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Unit 6: Inference for Proportions

Confidence Intervals · Hypothesis Tests · p-values · Type I & II Errors · Two-Proportion Procedures

📊 12–15% of Exam ⏱ ~3–4 weeks
Section 6.1

What Is Statistical Inference?

Statistical inference uses sample data to draw conclusions about an unknown population parameter. There are two main tools:

Confidence Interval ESTIMATION "We estimate the true proportion is between 0.42 and 0.58" vs Hypothesis Test DECISION MAKING "Is there convincing evidence that p ≠ 0.5?" Yes / No decision
🔑 The Big Picture of Inference

Both procedures require the same three conditions to be verified first:

1. Random: Data came from a random sample (or randomized experiment).

2. Normal: The sampling distribution of \(\hat{p}\) is approximately Normal. Check: \(np \geq 10\) AND \(n(1-p) \geq 10\).

3. Independent: Individual observations are independent. Check: \(n \leq 10\%\) of the population (10% condition).

Section 6.2

Confidence Intervals for a Population Proportion \(p\)

A confidence interval gives a range of plausible values for the unknown population proportion \(p\), based on what we observed in our sample \(\hat{p}\).

One-Sample z-Interval for a Proportion
\[ \hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
\(\hat{p}\) = sample proportion  |  \(n\) = sample size
\(z^*\) = critical value (depends on confidence level)
\(\sqrt{\hat{p}(1-\hat{p})/n}\) = estimated standard error of \(\hat{p}\)
Common Critical Values z* 90% Confidence 1.645 α/2 = 0.05 95% Confidence 1.960 most common 99% Confidence 2.576 α/2 = 0.005
Anatomy of a Confidence Interval p̂ − z*SE lower bound p̂ sample proportion p̂ + z*SE upper bound margin of error = z* × SE
📌 Example: Constructing a CI

In a random sample of 120 voters, 78 say they support a local measure. Construct a 95% confidence interval for the true proportion of all voters who support it.

Step 1 — Check conditions:

Random ✓   |   \(n\hat{p} = 78 \geq 10\) ✓   |   \(n(1-\hat{p}) = 42 \geq 10\) ✓   |   Independent (120 < 10% of all voters) ✓

Step 2 — Calculate: \(\hat{p} = 78/120 = 0.65\)

\(SE = \sqrt{\frac{0.65(0.35)}{120}} = \sqrt{0.001896} \approx 0.04354\)

\(CI = 0.65 \pm 1.960(0.04354) = 0.65 \pm 0.0854\)

Interval: (0.565, 0.735)

✓ Always state conditions, show the formula, and write the final interval as (lower, upper).

Section 6.3

Interpreting Confidence Intervals

⚠️ Most Common AP Mistake: Wrong Interpretation

WRONG: "There is a 95% probability that the true proportion is between 0.565 and 0.735."

CORRECT: "We are 95% confident that the true proportion of all voters who support the measure is between 0.565 and 0.735."

95% Confidence — Meaning: 95 of 100 Intervals Capture the True p true p ← missed! Captures true p (95 out of 100) Misses true p (5 out of 100)
📐 The Correct Interpretation Template

"We are [C%] confident that the true [parameter in context] is between [lower bound] and [upper bound]."

💡 Effect of Confidence Level and Sample Size

Increase confidence level (e.g., 90% → 99%): Larger z*, wider interval — less precise but more confident.

Increase sample size n: Smaller SE, narrower interval — more precise.

Margin of error formula: \(ME = z^* \sqrt{\hat{p}(1-\hat{p})/n}\). To find required sample size: \(n = \left(\frac{z^*}{ME}\right)^2 \hat{p}(1-\hat{p})\). If no estimate of \(\hat{p}\) is available, use \(\hat{p} = 0.5\) (gives largest/most conservative n).

Section 6.4

Hypothesis Tests for a Population Proportion

A hypothesis test evaluates evidence against a specific claim about \(p\). The 4-step AP procedure must be memorized and applied in order every time.

1 2 3 4 Hypotheses State H₀ and Hₐ Conditions Random, Normal, Indep. Calculations Test statistic & p-value Conclusion In context with α

Step 1: State the Hypotheses

📐 Hypotheses for a Proportion Test

Null Hypothesis \(H_0\): \(p = p_0\)   (the "nothing is happening" claim — always has an equals sign)

Alternative Hypothesis \(H_a\): Choose based on the research question:

\(H_a: p > p_0\)   (one-sided, right — "greater than")

\(H_a: p < p_0\)   (one-sided, left — "less than")

\(H_a: p \neq p_0\)   (two-sided — "different from")

⚠️ H₀ Always Uses =

The null hypothesis always contains an equality (=). Never write H₀: p > p₀ or H₀: p ≠ p₀. Also, hypotheses are always about parameters (\(p\)), never about statistics (\(\hat{p}\)).

Step 3: The Test Statistic

One-Sample z-Test Statistic for a Proportion
\[ z = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}} \]
\(p_0\) = the hypothesized value from \(H_0\)
Note: Use \(p_0\) (not \(\hat{p}\)) in the standard error — this is different from the CI formula!
⚠️ Key Formula Difference: CI vs Hypothesis Test

Confidence Interval SE: Uses \(\hat{p}\) — because we're estimating \(p\).

Hypothesis Test SE: Uses \(p_0\) — because we're assuming \(H_0\) is true and \(p = p_0\).

Section 6.5

p-values and Drawing Conclusions

🔑 What Is a p-value?

The p-value is the probability of obtaining a test statistic as extreme as (or more extreme than) the observed value, assuming \(H_0\) is true.

A small p-value means: "If \(H_0\) were true, this result would be very unlikely." → Evidence against \(H_0\).

Right-Tailed Hₐ: p > p₀ Left-Tailed Hₐ: p < p₀ Two-Tailed Hₐ: p ≠ p₀ μ₀ z=1.5 p ≈0.067 μ₀ z=−1.5 p ≈0.067 μ₀ −1.5 +1.5 p/2 p/2 total p ≈ 0.134 p-value = P(getting result this extreme or more | H₀ true)
📐 Decision Rule

Compare the p-value to the significance level \(\alpha\) (usually 0.05):

p-value \(\leq \alpha\): Reject \(H_0\). We have convincing evidence for \(H_a\).

p-value \(> \alpha\): Fail to reject \(H_0\). We do NOT have convincing evidence for \(H_a\).

💡 "Fail to Reject" ≠ "Accept H₀"

Never say "we accept \(H_0\)" or "we proved \(H_0\) is true." We simply don't have enough evidence to reject it. The absence of evidence is not evidence of absence.

AP conclusion template:
"Because the p-value [value] is [less than / greater than] α = [value], we [reject / fail to reject] \(H_0\). There [is / is not] convincing evidence that [Hₐ in context]."

📌 Full Worked Example: One-Sample z-Test

A coin is flipped 100 times and lands heads 62 times. Test whether the coin is biased at α = 0.05.

Step 1 — Hypotheses: \(H_0: p = 0.5\) vs \(H_a: p \neq 0.5\) (two-sided — "biased" means either direction)

Step 2 — Conditions: Random (stated) ✓ | \(np_0 = 50 \geq 10\) ✓ | \(n(1-p_0) = 50 \geq 10\) ✓ | Independent ✓

Step 3 — Calculate: \(\hat{p} = 62/100 = 0.62\)

\(z = \frac{0.62 - 0.50}{\sqrt{0.50(0.50)/100}} = \frac{0.12}{0.05} = 2.40\)

Two-sided p-value = \(2 \times P(Z > 2.40) = 2(0.0082) = \mathbf{0.0164}\)

Step 4 — Conclude: Since p-value = 0.0164 < α = 0.05, we reject \(H_0\). There is convincing evidence that the coin is biased (the true proportion of heads differs from 0.5).

Section 6.6

Type I and Type II Errors

Any time we make a decision based on sample data, we might be wrong. There are exactly two ways to make an error.

REALITY H₀ is TRUE H₀ is FALSE OUR DECISION Reject H₀ Fail to Reject H₀ Type I Error P = α Correct! Power = 1−β Correct! P = 1−α Type II Error P = β
Error TypeWhat HappenedProbabilityExample
Type I Error (α) Rejected H₀ when H₀ was actually TRUE \(\alpha\) (significance level) Convicting an innocent person; alarm goes off but there's no fire
Type II Error (β) Failed to reject H₀ when H₀ was actually FALSE \(\beta\) Acquitting a guilty person; no alarm when there IS a fire
Power (1 − β) Correctly rejected a false H₀ \(1 - \beta\) Correctly detecting a real effect
💡 Increasing Power

Power increases when:

• Sample size \(n\) increases (most effective)

• The true \(p\) is far from \(p_0\) (effect size is large)

• \(\alpha\) increases (but this also increases Type I error risk)

There is always a tradeoff: decreasing α (being more strict) increases β (more Type II errors).

Section 6.7

Two-Proportion Procedures

When comparing proportions from two independent groups, we use two-proportion procedures.

Two-Sample z-Interval for \(p_1 - p_2\)
\[ (\hat{p}_1 - \hat{p}_2) \pm z^* \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \]
Two-Sample z-Test Statistic for \(p_1 - p_2\)
\[ z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}_c(1-\hat{p}_c)\left(\dfrac{1}{n_1} + \dfrac{1}{n_2}\right)}} \]
\(\hat{p}_c = \dfrac{X_1 + X_2}{n_1 + n_2}\) = pooled proportion (used only in the hypothesis test, not CI)
\(X_1, X_2\) = number of successes in each group
🔑 Pooled vs Unpooled

Hypothesis test \(H_0: p_1 = p_2\): Use the pooled proportion \(\hat{p}_c\) because we assume both populations have the same proportion under \(H_0\).

Confidence interval: Do NOT pool — use separate \(\hat{p}_1\) and \(\hat{p}_2\) in the SE formula because we're not assuming they're equal.

📌 Example: Two-Proportion Test

Group 1 (new drug): 45 out of 100 improved. Group 2 (placebo): 30 out of 100 improved. Test \(H_0: p_1 = p_2\) at α = 0.05.

\(\hat{p}_1 = 0.45\), \(\hat{p}_2 = 0.30\), \(\hat{p}_c = (45+30)/(100+100) = 75/200 = 0.375\)

\(z = \frac{0.45-0.30}{\sqrt{0.375(0.625)(1/100+1/100)}} = \frac{0.15}{\sqrt{0.004688}} = \frac{0.15}{0.0685} \approx 2.19\)

p-value = \(2 \times P(Z > 2.19) \approx 2(0.0143) = 0.0286 < 0.05\)

Reject \(H_0\). There is convincing evidence that the new drug produces a higher improvement rate than the placebo.

Exam Practice

Multiple Choice Questions

Try each question, then reveal the answer and explanation.

MCQ · Q1 Interpreting a CI

A 90% confidence interval for the proportion of adults who exercise regularly is (0.41, 0.53). Which of the following is a correct interpretation?

  • A There is a 90% probability that the true proportion is between 0.41 and 0.53.
  • B 90% of adults exercise between 41% and 53% of the time.
  • C We are 90% confident that the true proportion of adults who exercise regularly is between 0.41 and 0.53.
  • D If we took 100 samples, exactly 90 would give proportions between 0.41 and 0.53.
  • E 90% of the time, the sample proportion will be 0.47.
✓ Correct Answer: C

(A) is the most common wrong answer — the true proportion is fixed, not random, so we can't assign probability to it after the interval is computed. The correct language is "we are 90% confident." (D) is close but says "exactly 90" — it should say "approximately 90 in repeated sampling."

MCQ · Q2 Hypothesis Test Setup

A researcher wants to test whether more than 60% of college students use social media daily. Which are the correct hypotheses?

  • A \(H_0: \hat{p} = 0.60\) vs \(H_a: \hat{p} > 0.60\)
  • B \(H_0: p = 0.60\) vs \(H_a: p > 0.60\)
  • C \(H_0: p > 0.60\) vs \(H_a: p = 0.60\)
  • D \(H_0: p = 0.60\) vs \(H_a: p \neq 0.60\)
  • E \(H_0: p < 0.60\) vs \(H_a: p > 0.60\)
✓ Correct Answer: B

Hypotheses use the parameter \(p\), never the statistic \(\hat{p}\) — eliminates (A). \(H_0\) always has an equality — eliminates (C) and (E). "More than 60%" means one-sided right — eliminates (D). Answer is \(H_0: p = 0.60\) vs \(H_a: p > 0.60\).

MCQ · Q3 p-value Interpretation

A hypothesis test yields a p-value of 0.03. The significance level is α = 0.05. Which conclusion is correct?

  • A There is a 3% probability that \(H_0\) is true.
  • B There is a 97% probability that \(H_a\) is true.
  • C We fail to reject \(H_0\); there is insufficient evidence.
  • D We reject \(H_0\); there is convincing evidence for \(H_a\).
  • E We accept \(H_0\) because the p-value is very small.
✓ Correct Answer: D

p-value (0.03) < α (0.05), so we reject \(H_0\). (A) and (B) misinterpret p-value as a probability about the hypotheses being true — it is not. (C) is wrong because we DO reject. (E) is never correct language — we never "accept" \(H_0\).

MCQ · Q4 Type I and Type II Errors

A medical test is used to detect a disease. The null hypothesis is H₀: the patient does NOT have the disease. A Type I error in this context would be:

  • A Concluding the patient has the disease when they actually do not.
  • B Concluding the patient does not have the disease when they actually do.
  • C Correctly diagnosing a healthy patient as healthy.
  • D The test failing to run properly.
  • E Using too small a sample size.
✓ Correct Answer: A

Type I error = rejecting \(H_0\) when \(H_0\) is true. Here \(H_0\) is "patient is healthy." Rejecting it means concluding they have the disease when they actually don't — a false positive. Type II error (B) would be concluding healthy when they actually do have the disease — a false negative.

MCQ · Q5 Margin of Error

A pollster wants to estimate a proportion with a margin of error of no more than 0.03 at 95% confidence. Using \(\hat{p} = 0.5\) as a conservative estimate, what is the minimum sample size needed?

  • A 752
  • B 1068
  • C 1537
  • D 1111
  • E 2401
✓ Correct Answer: B — 1068

\(n = \left(\frac{z^*}{ME}\right)^2 \hat{p}(1-\hat{p}) = \left(\frac{1.960}{0.03}\right)^2 (0.5)(0.5)\)
\(= (65.33)^2 (0.25) = 4268.4 \times 0.25 = 1067.1 \approx \mathbf{1068}\)
Always round UP to the next whole number to ensure the margin of error is satisfied.

Exam Practice

Free Response Questions

Follow the 4-step procedure every time. Use proper notation and always conclude in context.

FRQ 1 — One-Proportion z-Test

~15 minutes
A school administrator claims that at least 70% of students at the school eat breakfast before school. To investigate, a random sample of 80 students is selected, and 50 report eating breakfast. Test the administrator's claim at a significance level of α = 0.05.
(a)
State the hypotheses for this test.
(b)
Check the conditions for the test.
(c)
Calculate the test statistic and p-value.
(d)
State your conclusion in context.
✓ Model Solution

(a) Hypotheses:

\(H_0: p = 0.70\)   (the true proportion who eat breakfast is 70%)

\(H_a: p < 0.70\)   (the true proportion is less than 70% — testing if the claim is too high)

(b) Conditions:

Random: The 80 students were randomly selected. ✓

Normal (Large Counts): \(np_0 = 80(0.70) = 56 \geq 10\) ✓   \(n(1-p_0) = 80(0.30) = 24 \geq 10\) ✓

Independent: 80 students is less than 10% of the school's population. ✓

(c) Test Statistic:

\(\hat{p} = 50/80 = 0.625\)

\(z = \frac{0.625 - 0.70}{\sqrt{0.70(0.30)/80}} = \frac{-0.075}{\sqrt{0.002625}} = \frac{-0.075}{0.05123} \approx -1.464\)

p-value: One-sided left: \(P(Z < -1.464) \approx \mathbf{0.0716}\)

(d) Conclusion:

Since the p-value (0.0716) > α (0.05), we fail to reject \(H_0\). There is not convincing evidence that the true proportion of students who eat breakfast is less than 70%. The data are consistent with the administrator's claim.

✓ Key AP points: (a) use p not p̂; one-sided left because claim says "at least 70%." (b) use p₀ in Normal condition check. (c) use p₀ in the SE denominator. (d) say "fail to reject" not "accept" — and always include context.

FRQ 2 — Two-Proportion z-Test & CI

~15 minutes
A researcher compares the proportion of smokers in two cities. In City A, 85 of 250 randomly sampled adults smoke. In City B, 60 of 200 randomly sampled adults smoke.
(a)
Construct a 95% confidence interval for the difference in proportions \(p_A - p_B\). Interpret your interval.
(b)
Based on your confidence interval, is there convincing evidence that the smoking rates differ between the two cities? Explain.
(c)
If you were to perform a two-sided hypothesis test at α = 0.05, would you expect to reject \(H_0: p_A = p_B\)? Explain without performing the test.
✓ Model Solution

(a) 95% Confidence Interval:

\(\hat{p}_A = 85/250 = 0.340\)  |  \(\hat{p}_B = 60/200 = 0.300\)

Conditions: Random ✓ | Large Counts: \(n_A\hat{p}_A=85\geq10\), \(n_A(1-\hat{p}_A)=165\geq10\), \(n_B\hat{p}_B=60\geq10\), \(n_B(1-\hat{p}_B)=140\geq10\) ✓ | Independent ✓

\(SE = \sqrt{\frac{0.340(0.660)}{250} + \frac{0.300(0.700)}{200}} = \sqrt{0.000898 + 0.001050} = \sqrt{0.001948} \approx 0.04414\)

\(CI = (0.340 - 0.300) \pm 1.960(0.04414) = 0.040 \pm 0.0865\)

Interval: (−0.046, 0.126)

Interpretation: We are 95% confident that the true difference in smoking rates (City A minus City B) is between −0.046 and 0.126.

(b) Is there convincing evidence of a difference?

No. The confidence interval (−0.046, 0.126) contains 0. This means 0 is a plausible value for \(p_A - p_B\), so we do not have convincing evidence that the smoking rates differ between the two cities.

(c) Would we reject H₀?

No. Because the 95% confidence interval contains 0, a two-sided test at α = 0.05 would fail to reject \(H_0: p_A = p_B\). A confidence interval and a two-sided hypothesis test at the corresponding α level always lead to the same conclusion: if 0 is in the CI, we fail to reject; if 0 is outside the CI, we reject.

✓ AP tip: The CI-test duality (part c) is frequently tested. For two-proportion problems, if the CI for p₁−p₂ contains 0, we fail to reject H₀: p₁=p₂ at that α level.

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