Confidence interval practice problems

By Jude Wallis · Published July 28, 2026

These eight problems build and interpret confidence intervals for a proportion and a mean, then test how the margin of error responds to sample size and confidence level. Work each one before opening the solution, then match your bounds to the steps.

This set spans Unit 3 (topics 3.3 to 3.4, proportion intervals) and Unit 4 (topics 4.2 to 4.3, mean intervals) in the Fall 2026 AP Statistics course.

What these problems build

These eight problems drill the two confidence intervals you meet in AP inference: the one-proportion $z$ -interval for a categorical variable and the one-sample $t$ -interval for a quantitative variable. You will construct each interval from a sample, write what it means in context, and judge claims against it. Two problems isolate how the margin of error responds when you change the sample size or the confidence level.

Every interval uses the same form from the formula sheet, statistic $\pm$ (critical value)(standard error), so one recipe carries across both types. For the full walkthrough first, read how to calculate a confidence interval, and check any bounds you compute with the confidence interval calculator.

Try each problem before opening its solution. Round critical values to three decimals and keep the standard error to at least four decimals until the final step, the way the worked solutions do.

Problem 1

A community garden club plants 150 randomly chosen wildflower seeds and records that 90 of them sprout. Treat the 150 seeds as fewer than 10% of all seeds of this type. Build a 90% confidence interval for the true proportion of these seeds that sprout.

Show the worked solution

Find the sample proportion (call it p-hat, written $\hat{p}$ ): $\hat{p} = \frac{90}{150} = 0.6$ .
Check conditions. The seeds are random and are under 10% of all such seeds. Large counts: $n\hat{p} = 150(0.6) = 90 \geq 10$ and $n(1 - \hat{p}) = 150(0.4) = 60 \geq 10$ .
Critical value for 90% confidence: $z^* = 1.645$ .
Standard error (the typical sample-to-sample spread of $\hat{p}$ ): $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6(0.4)}{150}} = \sqrt{\frac{0.24}{150}} = \sqrt{0.0016} = 0.04$ .
Margin of error: $z^* \cdot SE = 1.645 \times 0.04 = 0.0658$ .
Interval: $0.6 \pm 0.0658 = (0.5342,\ 0.6658)$ .

The 90% confidence interval is about $(0.534,\ 0.666)$ . You are 90% confident that between 53.4% and 66.6% of these seeds sprout.

Problem 2

A tea company checks moisture content in a random sample of 10 batches of dried leaves. The sample mean is 12.0% moisture with a sample standard deviation of 1.5%. Moisture content is roughly normal. Build a 95% confidence interval for the true mean moisture content.

Show the worked solution

The sample mean (call it x-bar, written $\bar{x}$ ) is $\bar{x} = 12.0$ and the sample standard deviation is $s = 1.5$ , with $n = 10$ .
Check conditions: the sample is random, the population is roughly normal, and 10 batches is under 10% of all batches.
Because the population standard deviation is unknown, use a $t$ critical value with degrees of freedom $df = n - 1 = 10 - 1 = 9$ . For 95% confidence, $t^* = 2.262$ .
Standard error: $SE = \frac{s}{\sqrt{n}} = \frac{1.5}{\sqrt{10}} = \frac{1.5}{3.1623} = 0.4743$ .
Margin of error: $t^* \cdot SE = 2.262 \times 0.4743 = 1.073$ .
Interval: $12.0 \pm 1.073 = (10.927,\ 13.073)$ .

The 95% confidence interval is about $(10.93,\ 13.07)$ percent. You are 95% confident that the true mean moisture content is between 10.93% and 13.07%.

Problem 3

A microbrewery pulls 5 cans off the line and measures their fill volumes in milliliters: 350, 352, 348, 351, 349. Fill volume is roughly normal. Build a 90% confidence interval for the true mean fill volume.

Show the worked solution

Sample mean: $\bar{x} = \frac{350 + 352 + 348 + 351 + 349}{5} = \frac{1750}{5} = 350$ mL.
Deviations from the mean: $0, 2, -2, 1, -1$ . Squared: $0, 4, 4, 1, 1$ , which sum to $10$ .
Sample standard deviation (divide the squared deviations by $n - 1$ under the root): $s = \sqrt{\frac{10}{5 - 1}} = \sqrt{\frac{10}{4}} = \sqrt{2.5} = 1.5811$ mL.
Degrees of freedom $df = 5 - 1 = 4$ , so for 90% confidence $t^* = 2.132$ .
Standard error: $SE = \frac{s}{\sqrt{n}} = \frac{1.5811}{\sqrt{5}} = \frac{1.5811}{2.2361} = 0.7071$ mL.
Margin of error: $t^* \cdot SE = 2.132 \times 0.7071 = 1.508$ mL.
Interval: $350 \pm 1.508 = (348.492,\ 351.508)$ .

The 90% confidence interval is about $(348.49,\ 351.51)$ mL. You are 90% confident that the true mean fill volume falls in that range.

Problem 4

A city planning office surveys a random sample of 400 residents about a proposed bike lane, and 216 say they support it. Treat the sample as under 10% of all residents. Build a 95% confidence interval for the proportion of all residents who support the bike lane, then decide whether there is convincing evidence that a majority (more than 50%) support it.

Show the worked solution

Sample proportion: $\hat{p} = \frac{216}{400} = 0.54$ .
Conditions: random and under 10% are given; large counts $n\hat{p} = 400(0.54) = 216 \geq 10$ and $n(1 - \hat{p}) = 400(0.46) = 184 \geq 10$ .
Critical value for 95% confidence: $z^* = 1.960$ .
Standard error: $SE = \sqrt{\frac{0.54(0.46)}{400}} = \sqrt{\frac{0.2484}{400}} = \sqrt{0.000621} = 0.02492$ .
Margin of error: $1.960 \times 0.02492 = 0.04884$ .
Interval: $0.54 \pm 0.04884 = (0.4912,\ 0.5888)$ .
The value $0.50$ lies inside this interval, so proportions at or below one-half are still plausible. There is not convincing evidence of a majority.

The 95% interval is about $(0.491,\ 0.589)$ . Because 0.50 is inside it, the data do not give convincing evidence that a majority support the bike lane.

Problem 5

A pollster's 90% confidence interval for a proportion had a margin of error of 0.06 from a random sample of 200 people. She wants to cut the margin of error to 0.03 while keeping the same confidence level and about the same sample proportion. About how large a sample does she need?

Show the worked solution

The margin of error is $z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$ . With the critical value $z^*$ and the sample proportion $\hat{p}$ held fixed, the margin of error is proportional to $\frac{1}{\sqrt{n}}$ .
Comparing two sample sizes gives $\frac{\text{new margin}}{\text{old margin}} = \sqrt{\frac{n_{\text{old}}}{n_{\text{new}}}}$ .
Set the ratio to the target: $\frac{0.03}{0.06} = 0.5 = \sqrt{\frac{200}{n_{\text{new}}}}$ .
Square both sides: $(0.5)^2 = 0.25 = \frac{200}{n_{\text{new}}}$ .
Solve: $n_{\text{new}} = \frac{200}{0.25} = 800$ .

She needs about 800 people. Halving the margin of error takes roughly four times the sample size, because the margin shrinks like $\frac{1}{\sqrt{n}}$ .

Problem 6

In a random sample of 100 store visitors, 50 use the self-checkout, so the sample proportion is 0.50. Find the margin of error for a 90% confidence interval and for a 99% confidence interval, and describe how raising the confidence level changes the interval width.

Show the worked solution

Sample proportion: $\hat{p} = \frac{50}{100} = 0.50$ , with $n = 100$ .
Standard error: $SE = \sqrt{\frac{0.50(0.50)}{100}} = \sqrt{\frac{0.25}{100}} = \sqrt{0.0025} = 0.05$ .
At 90% confidence, $z^* = 1.645$ , so the margin of error is $1.645 \times 0.05 = 0.08225$ .
At 99% confidence, $z^* = 2.576$ , so the margin of error is $2.576 \times 0.05 = 0.1288$ .
The standard error did not change, but the larger critical value at 99% makes the margin of error, and therefore the interval width, larger.

The margin of error grows from about 0.082 at 90% confidence to about 0.129 at 99% confidence. Higher confidence uses a bigger critical value, so the interval gets wider.

Problem 7

A sports scientist records the resting heart rate of a random sample of 25 college rowers. The sample mean is 58 beats per minute with a sample standard deviation of 5 beats per minute. Resting heart rate is roughly normal.

(a) Construct a 95% confidence interval for the true mean resting heart rate. (b) Interpret the interval in context. (c) Give two changes that would produce a narrower interval.

Show the worked solution

Given $\bar{x} = 58$ , $s = 5$ , and $n = 25$ . Use a $t$ interval because the population standard deviation is unknown.
Degrees of freedom $df = 25 - 1 = 24$ , so for 95% confidence $t^* = 2.064$ .
Standard error: $SE = \frac{s}{\sqrt{n}} = \frac{5}{\sqrt{25}} = \frac{5}{5} = 1$ beat per minute.
Margin of error: $t^* \cdot SE = 2.064 \times 1 = 2.064$ .
(a) Interval: $58 \pm 2.064 = (55.936,\ 60.064)$ .
(b) You are 95% confident that the true mean resting heart rate of all rowers in this population is between about 55.94 and 60.06 beats per minute.
(c) A larger sample size shrinks the standard error, and a lower confidence level such as 90% uses a smaller critical value. Either change narrows the interval.

The 95% interval is about $(55.94,\ 60.06)$ beats per minute. A larger sample or a lower confidence level would make it narrower.

Problem 8

A quality manager inspects a random sample of 250 LED bulbs from a large shipment and finds that 235 pass inspection.

(a) Check the conditions for a one-proportion z-interval. (b) Construct a 99% confidence interval for the proportion of bulbs in the shipment that pass. (c) The supplier guarantees that at least 90% of bulbs pass. Does the interval support the guarantee?

Show the worked solution

Sample proportion: $\hat{p} = \frac{235}{250} = 0.94$ .
(a) Conditions: the sample is random; 250 bulbs is under 10% of a large shipment; large counts $n\hat{p} = 250(0.94) = 235 \geq 10$ and $n(1 - \hat{p}) = 250(0.06) = 15 \geq 10$ . All hold.
Critical value for 99% confidence: $z^* = 2.576$ .
Standard error: $SE = \sqrt{\frac{0.94(0.06)}{250}} = \sqrt{\frac{0.0564}{250}} = \sqrt{0.0002256} = 0.01502$ .
Margin of error: $2.576 \times 0.01502 = 0.03869$ .
(b) Interval: $0.94 \pm 0.03869 = (0.9013,\ 0.9787)$ .
(c) The entire interval lies above 0.90, since the lower bound 0.9013 is greater than 0.90, so every plausible value clears the guarantee.

The 99% interval is about $(0.901,\ 0.979)$ . Because the whole interval sits above 0.90, it supports the supplier's guarantee that at least 90% of bulbs pass.