When do I use p versus p_0 in the standard error?

For a confidence interval, always use the sample proportion p in the standard error, because there is no hypothesized value. For a significance test (U6.4), you use the null value p_0 instead. This is the main computational difference between the two procedures.

Why do I round the sample size up instead of using normal rounding?

Rounding down would produce a sample too small to guarantee the required margin of error. To be sure the precision target is met, you always round the required sample size up to the next whole number, even if the decimal part is small.

Can I say there is a 95% chance the true proportion is in my interval?

No. Once the interval is computed, the true proportion is a fixed number that is either inside or outside it — no probability remains. The 95% refers to the long-run proportion of intervals from repeated samples that capture the parameter. This distinction is heavily tested.

What makes a confidence interval wider or narrower?

Three things: a higher confidence level widens it (larger z^*), a larger sample size narrows it (larger n reduces the standard error), and a sample proportion closer to 0.5 slightly widens it. To halve the margin of error you must quadruple the sample size because of the square root.

AP-STATS-6.1-6.3

U6.1 One-Proportion Confidence Interval

Master the one-proportion z-confidence interval for AP Statistics: check conditions, build the interval, and interpret confidence level correctly.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U6.1 One-Proportion Confidence Interval, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U6.1 One-Proportion Confidence Interval free

What this lesson covers

When a poll reports that 52% of voters favor a candidate "with a margin of error of 3 points," it is really reporting a confidence interval for a population proportion. In this lesson you will learn to build that interval yourself using the one-proportion z-procedure, verify the conditions that make it valid, and interpret the result in language that earns full credit on the AP exam.

This is one of the most heavily tested skills in Unit 6. The mechanics are short, but the interpretation trips up many students. We will separate the two clearly: how to compute the interval and how to talk about it precisely.

The structure of the interval

Every confidence interval on the AP exam follows the same template:

\text{statistic} \pm (\text{critical value})(\text{standard error})

. For a single proportion, the statistic is the sample proportion

\hat{p}

, the critical value is a z-score

z^*

determined by the confidence level, and the standard error is

\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

.

Put together, the one-proportion z-interval is:

\hat{p} \pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

Notice we use

\hat{p}

(the sample proportion) inside the standard error, not a hypothesized value. This is the key difference from the significance test in U6.4, which uses the null value

p_0

. The quantity

z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

is called the margin of error.

Common critical values you should memorize:

Confidence level	$z^*$
90%	1.645
95%	1.960
99%	2.576

Higher confidence means a larger

z^*

, which makes the interval wider. A wider interval is more likely to capture the true proportion but is less precise. This trade-off between confidence and precision appears constantly on the exam.

Checking the conditions

Before computing anything, you must verify three conditions. Skipping them costs points even when your arithmetic is perfect.

Random: The data must come from a random sample or randomized experiment. State this explicitly using the problem's wording ("a random sample of 200 adults").

Independence (10% condition): When sampling without replacement, the sample size must be less than 10% of the population,

n \le 0.10N

. This keeps observations approximately independent.

Normal (Large Counts): The sampling distribution of

\hat{p}

is approximately Normal when both

n\hat{p} \ge 10

and

n(1-\hat{p}) \ge 10

. In other words, you expect at least 10 successes and at least 10 failures. For a confidence interval, use

\hat{p}

(not

p_0

) in these checks because there is no hypothesized value.

Condition	What to check	Why it matters
Random	Random sample or randomized experiment	Allows generalizing to the population
10%	$n \le 0.10N$	Keeps observations independent
Large Counts	$n\hat{p}\ge10$ and $n(1-\hat{p})\ge10$	Makes $\hat{p}$ approximately Normal

A frequent mistake is checking counts with round numbers only. Always plug the actual sample counts (successes and failures) into the inequalities and show the numbers.

Interpreting the interval and the confidence level

AP graders distinguish sharply between interpreting an interval and interpreting a confidence level. You need both phrasings.

Interpreting the interval: "We are 95% confident that the interval from (lower) to (upper) captures the true proportion of [context]." Always name the parameter in context. Never say the interval captures

\hat{p}

— it already contains

\hat{p}

at its center; the interval is an estimate for the population proportion

p

.

Interpreting the confidence level: "If we took many random samples of this size and constructed a confidence interval from each, about 95% of those intervals would capture the true proportion." The 95% describes the long-run success rate of the method, not the probability that any one interval is correct.

The single most tested misconception: 95% confidence does NOT mean "there is a 95% probability the true proportion is in this interval." Once the interval is computed, the true proportion either is or isn't inside it — there is no probability left. The randomness is in the sampling process, not in the fixed parameter.

Also avoid saying "95% of the data" or "95% of samples fall in the interval." The confidence level is about intervals capturing a parameter, not about individual data points or sample proportions.

Sample size and margin of error

The margin of error

ME = z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

controls the width of the interval. Three factors change it:

Increasing the sample size

n

decreases the margin of error, because

n

is in the denominator under the square root. To cut the margin of error in half, you must quadruple the sample size, since

ME

depends on

\sqrt{n}

.

Increasing the confidence level increases

z^*

and therefore widens the interval.

The standard error is largest when

\hat{p} = 0.5

, so planners often use

\hat{p} = 0.5

as a conservative guess when designing a study.

A classic exam problem gives a target margin of error and asks for the required sample size. Set up the inequality

z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \le ME

and solve for

n

n \ge \left(\frac{z^*}{ME}\right)^2 \hat{p}(1-\hat{p})

Always round the final sample size UP to the next whole number, because rounding down would fail to meet the required precision. If no prior estimate of

\hat{p}

is given, use

\hat{p} = 0.5

to guarantee a large enough sample.

How the exam tests this topic

On multiple-choice questions, expect items that ask you to identify the correct interpretation of a confidence level, compute a margin of error, or predict how the interval width changes when

n

or the confidence level changes. Memorizing the three critical values saves time.

On free-response questions, the four-step template earns the points: State the parameter and interval type; Plan by checking all three conditions with numbers; Do the calculation showing the formula with values substituted; Conclude with a contextual interpretation.

Graders look for the parameter defined in context, all conditions verified with actual counts, correct substitution into the formula, and a conclusion that mentions the confidence level, the interval, and the population parameter in context. A common point loss is a conclusion that talks about the sample instead of the population, or that treats the confidence level as a probability about the specific interval.

If a question asks whether a particular value (like 0.50) is plausible, check whether it lies inside the interval. If the entire interval is above 0.50, you have evidence the true proportion exceeds 0.50 — this connects directly to the significance testing you'll do in U6.4.

Key terms

Sample proportion ( $\hat{p}$ ).: The fraction of successes in the sample, $\hat{p} = \frac{x}{n}$ , used as the point estimate for the population proportion.
Population proportion ( $p$ ).: The true, usually unknown, fraction of the entire population with the characteristic of interest; the parameter the interval estimates.
Standard error.: An estimate of the standard deviation of the sampling distribution of $\hat{p}$ , computed as $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ .
Critical value ( $z^*$ ).: The z-score corresponding to the chosen confidence level; it sets how many standard errors wide the interval is.
Margin of error.: The half-width of the interval, $z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ , expressing the maximum expected estimation error at the given confidence level.
Confidence level.: The long-run percentage of intervals, built by this method from repeated random samples, that would capture the true parameter.
Large Counts condition.: The requirement that $n\hat{p}\ge10$ and $n(1-\hat{p})\ge10$ so the sampling distribution of $\hat{p}$ is approximately Normal.
10% condition.: When sampling without replacement, the rule $n \le 0.10N$ that keeps observations approximately independent.

Worked example

A random sample of 250 registered voters is surveyed, and 140 say they support a proposed transit measure. Construct and interpret a 95% confidence interval for the proportion of all registered voters who support the measure. Assume there are far more than 2,500 registered voters.

Step 1 — Parameter and procedure. Let

p

be the true proportion of all registered voters who support the measure. We will construct a one-proportion z-interval.

Step 2 — Conditions. Random: the problem states a random sample of voters. 10%: the sample of 250 is less than 10% of the population, since there are far more than 2,500 voters. Large Counts: successes

= 140 \ge 10

and failures

= 250 - 140 = 110 \ge 10

. All conditions are met.

Step 3 — Compute. The sample proportion is

\hat{p} = \frac{140}{250} = 0.56

. The standard error is

\sqrt{\frac{0.56(0.44)}{250}} = \sqrt{\frac{0.2464}{250}} = \sqrt{0.0009856} \approx 0.0314

. With

z^* = 1.960

for 95% confidence:

0.56 \pm 1.960(0.0314)

The margin of error is

1.960 \times 0.0314 \approx 0.0615

, giving the interval

0.56 \pm 0.0615

, or approximately

(0.499, 0.622)

.

Step 4 — Interpret. We are 95% confident that the interval from 0.499 to 0.622 captures the true proportion of all registered voters who support the transit measure. Because the interval barely includes values below 0.50, we cannot be fully confident that a majority supports the measure.

Practice questions

A 95% confidence interval for a population proportion is calculated to be (0.42, 0.48). Which statement is the correct interpretation of the confidence level?

There is a 95% probability that the true proportion lies between 0.42 and 0.48.
If many random samples of the same size were taken and an interval computed from each, about 95% of those intervals would capture the true proportion.
About 95% of the sample proportions fall between 0.42 and 0.48.
We are 95% confident that the sample proportion lies between 0.42 and 0.48.

Answer: If many random samples of the same size were taken and an interval computed from each, about 95% of those intervals would capture the true proportion.

The confidence level describes the long-run capture rate of the method across repeated sampling, not a probability about one fixed interval. Once computed, the interval either contains

p

or it does not. Choice 1 wrongly assigns probability to a fixed interval; choice 3 confuses the level with the data distribution; choice 4 refers to

\hat{p}

, which is always the center of the interval by construction.

A researcher wants to estimate the proportion of students who skip breakfast, with a margin of error no larger than 0.03 at 95% confidence. No prior estimate is available. How large a sample is required?

Answer: At least 1068 students.

With no prior estimate, use the conservative value

\hat{p} = 0.5

, which maximizes the standard error. Set up

n \ge \left(\frac{z^*}{ME}\right)^2 \hat{p}(1-\hat{p}) = \left(\frac{1.960}{0.03}\right)^2 (0.5)(0.5)

. Compute

\frac{1.960}{0.03} \approx 65.33

, square it to get about 4268.4, then multiply by 0.25 to get about 1067.1. Always round up to guarantee the margin of error is met, so

n = 1068

A 90% confidence interval for the proportion of defective parts is (0.04, 0.10). Explain what would happen to the width of the interval if the confidence level were increased to 99%, assuming the same sample.

Answer: The interval would become wider.

Increasing the confidence level raises the critical value from

z^* = 1.645

z^* = 2.576

. Since the margin of error is

z^*

times a fixed standard error (the sample is unchanged), a larger

z^*

produces a larger margin of error and therefore a wider interval. This reflects the trade-off: greater confidence in capturing the true proportion comes at the cost of less precision.

FAQ

When do I use $\hat{p}$ versus $p_0$ in the standard error?: For a confidence interval, always use the sample proportion $\hat{p}$ in the standard error, because there is no hypothesized value. For a significance test (U6.4), you use the null value $p_0$ instead. This is the main computational difference between the two procedures.
Why do I round the sample size up instead of using normal rounding?: Rounding down would produce a sample too small to guarantee the required margin of error. To be sure the precision target is met, you always round the required sample size up to the next whole number, even if the decimal part is small.
Can I say there is a 95% chance the true proportion is in my interval?: No. Once the interval is computed, the true proportion is a fixed number that is either inside or outside it — no probability remains. The 95% refers to the long-run proportion of intervals from repeated samples that capture the parameter. This distinction is heavily tested.
What makes a confidence interval wider or narrower?: Three things: a higher confidence level widens it (larger $z^*$ ), a larger sample size narrows it (larger $n$ reduces the standard error), and a sample proportion closer to 0.5 slightly widens it. To halve the margin of error you must quadruple the sample size because of the square root.

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The Crimsora tutor teaches U6.1 One-Proportion Confidence Interval live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

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