Lesson 20: Confidence Intervals for One Proportion

(a) Warm-Up

In Lesson 19 you met the big idea: a point estimate like a sample proportion p̂ is our single best guess for a population parameter, but it's almost never exactly right. So we wrap a margin of error around it and report a whole interval of plausible values — a confidence interval.

Here's the scenario for today. A national survey contacts 1,012 U.S. teens and finds that 567 of them say they use TikTok every day. That gives p̂ = 567/1012 ≈ 0.560 — about 56% of the sampled teens. But you weren't hired to describe 1,012 teens. You were hired to say something about all U.S. teens, the population proportion p you'll never get to measure directly.

Quick question to hold in your head: if you ran this exact survey again tomorrow with a different 1,012 teens, would you get 567 again? Almost certainly not. The estimate would wobble. Today you'll learn to quantify that wobble and turn "about 56%" into a defensible range like "somewhere between 53% and 59%, with 95% confidence." That's the one-proportion z-interval — the first real inference procedure of the course.

(b) Core Concept

What we're building

We want to estimate a single population proportion p — the true fraction of some population that has a characteristic (uses TikTok daily, supports a policy, has a defect). We can't see p. We can only see the sample proportion p̂ from one sample. The one-sample z-interval for a proportion captures p with a stated level of confidence:

p̂ ± z*·SE

That's the whole formula. The three jobs are: find p̂, find the critical value z*, and find the standard error SE. But before we touch numbers, we check conditions — because the formula only works when the sampling distribution of p̂ is approximately Normal.

The three conditions

1. Random. The data come from a random sample (or randomized experiment). This is what lets us generalize from the sample to the population. If the sample is convenience-based or voluntary, the interval is not trustworthy no matter how clean the arithmetic is.

2. 10% condition. When we sample without replacement, we need the sample to be small relative to the population: n ≤ 0.10N. This keeps the observations approximately independent so the standard error formula is valid. For our poll, 1,012 teens is far less than 10% of the millions of U.S. teens — fine.

3. Large Counts. The sampling distribution of p̂ is approximately Normal only when we expect enough successes and enough failures. We check np̂ ≥ 10 and n(1 − p̂) ≥ 10.

Critical detail for AP scoring: For a confidence interval, you do NOT have a hypothesized value p₀ — that only exists in a hypothesis test. So you check Large Counts with the sample proportion p̂, using the actual counts of successes and failures. In fact np̂ is just x (the number of successes) and n(1 − p̂) is the number of failures. For our poll: np̂ = 567 ≥ 10 and n(1 − p̂) = 445 ≥ 10. (Remember this — it's the #1 way students lose a condition point on interval problems.)

The critical value z*

The critical value z* is the number of standard errors you reach out on each side of p̂ to capture the middle C% of a Normal distribution. The three you must know cold:

Higher confidence ⇒ larger z ⇒ wider interval. There's no free lunch: to be more sure you've captured p, you have to cast a wider* net.

The standard error

For a one-proportion interval the standard error is:

SE = sqrt( p̂(1 − p̂) / n )

Note again: we use p̂, not p₀. We don't have a hypothesized value here — our best estimate of the standard deviation of p̂ uses the data we actually collected.

The margin of error

The margin of error is the part after the ±:

ME = z·SE = z·sqrt( p̂(1 − p̂) / n )

Read that formula like a control panel:

• Bigger n ⇒ smaller ME. n is under the square root, so quadrupling n only halves the margin. Precision is expensive.
• Higher confidence ⇒ bigger z* ⇒ bigger ME. More confidence costs width.
• p̂ near 0.5 ⇒ the product p̂(1 − p̂) is at its maximum (0.25), so the ME is largest. p̂ near 0 or 1 ⇒ smaller ME.

The full PANIC walkthrough

Inference at AP has a fixed structure. Memorize PANIC:

P — Parameter: define the parameter in context.

A — Assumptions/conditions: state and verify Random, 10%, Large Counts.

N — Name the procedure.

I — Interval: compute it.

C — Conclusion in context.

Let's run it on the TikTok poll: n = 1012, x = 567, 95% confidence.

P — Parameter. Let p = the true proportion of all U.S. teens who use TikTok daily.

A — Assumptions/conditions.

• Random: The 1,012 teens were a random sample of U.S. teens. ✓
• 10%: 1,012 ≤ 10% of all U.S. teens (there are tens of millions, so 10% is millions). ✓
• Large Counts: np̂ = 567 ≥ 10 and n(1 − p̂) = 445 ≥ 10. ✓

N — Name the procedure. A one-sample z-interval for a proportion.

I — Interval. First the estimate:

p̂ = 567 / 1012 = 0.5603

Standard error:

SE = sqrt( 0.5603 × (1 − 0.5603) / 1012 )
   = sqrt( 0.5603 × 0.4397 / 1012 )
   = sqrt( 0.24636 / 1012 )
   = sqrt( 0.00024344 )
   = 0.01560

Critical value for 95%: z* = 1.96. Margin of error:

ME = 1.96 × 0.01560 = 0.03058

Interval:

0.5603 ± 0.03058  →  (0.5297, 0.5909)

C — Conclusion in context. We are 95% confident that the interval from 0.530 to 0.591 captures the true proportion of all U.S. teens who use TikTok daily.

Confirming with the calculator

TI-84: STAT → TESTS → 1-PropZInt
Input: x = 567, n = 1012, C-Level = 0.95
Output: (0.52969, 0.59086)
        p̂ = 0.5603162055
        n = 1012

Rounded, the calculator gives (0.530, 0.591) — an exact match to our by-hand interval. The TI-84 uses the same sqrt(p̂(1 − p̂)/n) standard error, so it should always match (any tiny difference is rounding). One non-negotiable: x must be a whole number of successes. If you only know p̂ = 0.56, you must convert: x = 0.56 × 1012 = 566.72 → round to the nearest whole number of successes (567) before entering. Entering a decimal for x gives an error.

(c) Worked Examples

Example 1 — A clean PANIC interval (medium)

A polling firm randomly samples 400 likely voters in a city; 152 say they will vote for a transit bond. Construct a 90% confidence interval for the proportion of all likely voters who support the bond.

Strategy / Solution — PANIC.

P: Let p = the true proportion of all likely voters in the city who support the transit bond.

A:

• Random: 400 voters randomly sampled. ✓
• 10%: 400 ≤ 10% of all likely voters in the city (a city has far more than 4,000 voters). ✓
• Large Counts: np̂ = 152 ≥ 10, n(1 − p̂) = 248 ≥ 10. ✓

N: One-sample z-interval for a proportion.

I:

p̂ = 152/400 = 0.38
SE = sqrt(0.38 × 0.62 / 400) = sqrt(0.2356/400) = sqrt(0.000589) = 0.02427
z* = 1.645
ME = 1.645 × 0.02427 = 0.03992
Interval: 0.38 ± 0.0399 → (0.340, 0.420)

Calculator check — 1-PropZInt, x = 152, n = 400, C-Level = 0.90 → (0.34008, 0.41992). ✓

C: We are 90% confident that between 34.0% and 42.0% of all likely voters in the city support the transit bond. (Note: the entire interval is below 0.50, so we have evidence the bond would fail a majority vote.)

Example 2 — A condition violation (medium)

A quality engineer inspects a random sample of 25 circuit boards from a large production run and finds 4 defective. Can she build a one-proportion z-interval for the true defect rate?

Strategy / Solution. Check the conditions before computing anything.

• Random: random sample. ✓
• 10%: 25 ≤ 10% of a large production run. ✓
• Large Counts: p̂ = 4/25 = 0.16. np̂ = 25 × 0.16 = 4 and n(1 − p̂) = 25 × 0.84 = 21. Since np̂ = 4 < 10, the Large Counts condition fails.

Interpretation. Because we expect fewer than 10 successes, the sampling distribution of p̂ is not approximately Normal — it's right-skewed when p̂ is small and n is modest. The z-interval is not appropriate here; its stated 95% confidence would be inaccurate. The fix: collect a larger sample (enough that np̂ ≥ 10 and n(1 − p̂) ≥ 10). On the AP exam, noticing and naming the violation earns the point — proceeding to compute an invalid interval loses it.

Example 3 — Choosing a sample size (AP-style)

A campaign wants to estimate the proportion of voters who support a candidate to within a margin of error of 0.03 at 95% confidence. How many voters must they survey?

Strategy. We need the smallest n with ME ≤ 0.03. Solve the margin-of-error formula for n:

ME = z*·sqrt( p̂(1 − p̂)/n )   ⇒   n = (z*/ME)² · p̂(1 − p̂)

We don't know p̂ in advance, so use the conservative guess p̂ = 0.5, which maximizes p̂(1 − p̂) = 0.25 and therefore the required n (guarantees ME ≤ target no matter the true p).

n = (1.96 / 0.03)² × 0.25
  = (65.333...)² × 0.25
  = 4268.44 × 0.25
  = 1067.11

Always round UP to guarantee the margin is met: n = 1,068 voters. (Rounding down to 1,067 would leave the margin just over 0.03.)

Quick extension — what if they wanted ME ≤ 0.02 at 99% confidence?

n = (2.576 / 0.02)² × 0.25 = (128.8)² × 0.25 = 16589.44 × 0.25 = 4147.36 → round up to 4,148.

Notice how much higher n jumps when you demand both a tighter margin and more confidence — precision compounds in cost.

Example 4 — Interpreting and critiquing an interval (AP-style)

A study reports: "Based on a random sample, we are 95% confident that between 0.41 and 0.49 of adults exercise daily." A student writes three interpretations. Which is correct, and what's wrong with the others?

1. "There is a 95% probability that p is between 0.41 and 0.49."
2. "95% of adults exercise between 0.41 and 0.49 of the time."
3. "We are 95% confident that the true proportion of all adults who exercise daily is between 0.41 and 0.49."

Solution.

• #3 is correct. It names the parameter (true proportion of all adults), states the confidence level, and gives the interval — in context.
• #1 is wrong. Once the interval is computed, p either is or isn't inside it; there's no probability about p. The 95% describes the long-run success rate of the method, not the chance for this one interval.
• #2 is wrong. It misreads a proportion of people as a proportion of time per person. The parameter is the fraction of adults, not how often each adult exercises.

Why this matters: AP graders reward the interpretation that (a) is about the population parameter, (b) states the confidence level, and (c) is in context. Miss any one and you can lose the point.

(d) Common Mistakes

1. Using p₀ instead of p̂ in the standard error. Students who memorized the hypothesis-test SE formula sqrt(p₀(1 − p₀)/n) try to use it here. There is no p₀ in a confidence interval — you have no hypothesized value. Always use the sample proportion: SE = sqrt(p̂(1 − p̂)/n). (The test SE uses p₀ because it assumes the null is true; the interval has no such assumption.)

2. Skipping or fudging the conditions. Writing "conditions met" with no evidence earns nothing. You must state all three — Random, 10%, Large Counts — and back them up. For Large Counts on an interval, show the numbers: np̂ = (successes) ≥ 10 and n(1 − p̂) = (failures) ≥ 10.

3. Misinterpreting the confidence level. "There's a 95% chance p is in this interval" is wrong. The correct idea: if we repeated this sampling process many times, about 95% of the intervals produced would capture the true p. The confidence is in the method, not in any single interval.

4. Interpreting with a count instead of a proportion. The parameter p is a proportion (between 0 and 1), not a number of people. An interval like "(0.530, 0.591)" means 53.0%–59.1% of the population, not "530 to 591 teens." Define p as a proportion and keep the interpretation in proportion (or percent) terms.

5. Forgetting to round sample size UP. Sample-size answers must be whole numbers and must guarantee the margin, so always round up — even 1067.1 becomes 1,068. Rounding down leaves the margin too large.

(e) Practice Problems

1. (MC) Which formula gives the standard error for a one-proportion z-interval?

(A) sqrt(p₀(1 − p₀)/n)

(B) sqrt(p̂(1 − p̂)/n)

(C) sqrt(p̂(1 − p̂)/n) · z*

(D) p̂(1 − p̂)/n

2. (MC) A 95% confidence interval for a proportion is (0.42, 0.50). What is the sample proportion p̂ and the margin of error?

(A) p̂ = 0.42, ME = 0.50

(B) p̂ = 0.46, ME = 0.04

(C) p̂ = 0.46, ME = 0.08

(D) p̂ = 0.50, ME = 0.04

3. (MC) Holding everything else fixed, which change makes a confidence interval wider?

(A) Increasing the sample size

(B) Decreasing the confidence level from 95% to 90%

(C) Increasing the confidence level from 95% to 99%

(D) A sample proportion closer to 0 or 1

4. (MC) A random sample of 50 customers finds 6 are dissatisfied. Before building a 95% z-interval for the true dissatisfaction rate, which condition fails?

(A) Random

(B) 10% condition

(C) Large Counts (np̂ ≥ 10)

(D) No condition fails

5. (MC) To cut the margin of error of a proportion interval in half (same confidence, same p̂), the sample size should be multiplied by about:

(A) 2

(B) 4

(C) 1/2

(D) 1/4

6. (MC) A researcher will estimate a proportion to within 0.04 at 95% confidence and uses the conservative p̂ = 0.5. The required sample size is closest to:

(A) 423

(B) 601

(C) 1,068

(D) 2,401

7. (MC, in context) A 90% confidence interval for the proportion of students who skip breakfast is (0.61, 0.69). Which statement is the correct interpretation?

(A) 90% of students skip breakfast between 61% and 69% of the time.

(B) There is a 90% probability the true proportion is between 0.61 and 0.69.

(C) We are 90% confident that the true proportion of all students who skip breakfast is between 0.61 and 0.69.

(D) 90% of all samples will have a sample proportion between 0.61 and 0.69.

8. (Short answer) A random sample of 240 households finds 90 own a pet. Construct and interpret a 95% confidence interval for the true proportion of households that own a pet. Show the full PANIC sequence.

9. (Short answer) Explain, in one or two sentences, why a confidence interval uses p̂ in its standard error while a hypothesis test for a proportion uses p₀.

10. (Short answer, in context) A nutrition company claims "more than 60% of users lose weight." A random sample of 350 users yields a 99% confidence interval of (0.631, 0.758) for the true proportion who lose weight. Does the interval support the company's claim? Explain using the interval.

11. (Short answer) A pollster wants a margin of error of at most 0.025 at 95% confidence. Using the conservative guess, how many people must be surveyed?

12. (Short answer) A student computes a 95% interval as (0.30, 0.46) but reports it as "We are 95% confident that between 30 and 46 people in the population own a smartphone." Identify and correct the error.

(f) FRQ Practice (10 points)

A regional health department wants to know what proportion of adults in the county have received a flu vaccine this season. Public health officials take a random sample of 350 adults from the county (which has roughly 180,000 adults) and find that 243 of them report having received a flu vaccine.

(a) Construct and interpret a 99% confidence interval for the proportion of all adults in the county who have received a flu vaccine this season. Be sure to check the necessary conditions. (7 points)

(b) A health official states, "This proves that exactly 69.4% of county adults got the flu vaccine." Explain why this statement is incorrect, using your interval. (1 point)

(c) The department plans next year's survey and wants the margin of error to be no more than 0.04 at the 99% confidence level. Assuming they have no prior estimate of the proportion, how many adults should they sample? Show your work. (2 points)

Model Response

(a) — PANIC.

P — Parameter. Let p = the true proportion of all adults in the county who have received a flu vaccine this season.

A — Assumptions/conditions.

• Random: The 350 adults were a random sample of county adults. ✓
• 10% condition: n = 350 ≤ 0.10N = 0.10 × 180,000 = 18,000. Since 350 ≤ 18,000, ✓.
• Large Counts: p̂ = 243/350 = 0.6943. np̂ = 243 ≥ 10 and n(1 − p̂) = 107 ≥ 10. ✓

N — Name the procedure. A one-sample z-interval for a proportion.

I — Interval.

p̂ = 243/350 = 0.6943
SE = sqrt(0.6943 × 0.3057 / 350) = sqrt(0.21225/350) = sqrt(0.00060644) = 0.02463
z* = 2.576  (99% confidence)
ME = 2.576 × 0.02463 = 0.06344
Interval: 0.6943 ± 0.0634 → (0.631, 0.758)

Calculator check — 1-PropZInt, x = 243, n = 350, C-Level = 0.99 → (0.63085, 0.75772). ✓

C — Conclusion in context. We are 99% confident that the interval from 0.631 to 0.758 captures the true proportion of all adults in the county who have received a flu vaccine this season.

(b) The interval gives a range of plausible values (0.631 to 0.758), not a single exact value. p̂ = 0.694 is only our point estimate from one sample; we are not certain the true proportion equals 0.694 — it is simply our best guess, and the interval shows the true value could plausibly be anywhere from about 63% to 76%. We never "prove" an exact parameter value from a sample.

(c) Use the conservative guess p̂ = 0.5 (no prior estimate), z* = 2.576, ME = 0.04:

n = (z*/ME)² · p̂(1 − p̂) = (2.576/0.04)² × 0.25 = (64.4)² × 0.25 = 4147.36 × 0.25... 

Carefully: (2.576/0.04)² = 64.4² = 4147.36; times 0.25 = 1036.84. Round up ⇒ the department should sample 1,037 adults.

Rubric (10 points)

Part (a) — 7 points:

Part (b) — 1 point: Explains that the interval is a range of plausible values / 0.694 is just an estimate, so we cannot claim an exact proportion.

Part (c) — 2 points: 1 pt: correct setup using conservative p̂ = 0.5, z* = 2.576, ME = 0.04. 1 pt: correct value with round-up to 1,037 (1,036 unrounded shown).

Total: 10 points.

Where students lose points

• P: Defining p as the sample proportion or forgetting "all county adults" / context. The parameter is about the population.
• A: Writing "Large Counts met" without the numbers, or using a hypothesized p₀ instead of p̂. You must show np̂ = 243 and n(1 − p̂) = 107.
• I: Using z* = 1.96 (95%) instead of 2.576 (99%) — a very common slip. Or computing SE with the wrong proportion.
• C: Saying "95% confident" out of habit, omitting the interval, or interpreting it as a probability about p.
• (c): Forgetting to round up (writing 1,036), or using p̂ = 0.694 instead of the conservative 0.5 when "no prior estimate" is specified.