Confidence Intervals for a Population Mean — AP Statistics Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: One-sample and two-sample t-intervals for population means, t-distribution properties, inference conditions, margin of error calculation, confidence interval interpretation, and sample size determination for a desired margin of error.

You should already know: Basics of confidence interval construction for proportions, properties of the normal distribution, summary statistics for quantitative data.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Statistics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Confidence Intervals for a Population Mean?

A confidence interval for a population mean gives a range of plausible values for the true unknown population mean $\mu$, based on a random sample of quantitative data. Unlike confidence intervals for proportions (which use the normal distribution when conditions are met), most real-world cases for means use the t-distribution because we almost never know the true population standard deviation $\sigma$.

Per the College Board AP Statistics Course and Exam Description (CED), this topic falls within Unit 7 (Inference for Quantitative Data: Means), which accounts for 12-15% of the total AP exam score. It appears in both multiple-choice (MCQ) and free-response (FRQ) questions, often as a standalone short FRQ or as a component of a larger inference question.

Standard notation used here: $\mu$ = true population mean, $\bar{x}$ = sample mean, $s$ = sample standard deviation, $n$ = sample size, $df$ = degrees of freedom, $t^{\ast }$ = critical t-value, $z^{\ast }$ = critical z-value. Common synonyms include "t-interval for a mean" and "interval estimate for a population average." This procedure is used to estimate an unknown population average, rather than testing a specific claim about it (which is covered in hypothesis testing for means).

2. The t-Distribution for Inference on Means

When we estimate a population mean, we almost never have access to the true population standard deviation $\sigma$, so we estimate it with the sample standard deviation $s$. This extra step adds more variability to our sampling distribution than the normal distribution accounts for, so we use the t-distribution instead of the standard normal (z) distribution.

Like the normal distribution, the t-distribution is symmetric, bell-shaped, and centered at 0. However, it has fatter tails to accommodate the extra uncertainty from estimating $\sigma$ with $s$. The shape of the t-distribution depends on degrees of freedom: for a one-sample confidence interval, $df=n-1$. As sample size (and thus $df$) increases, the t-distribution converges to the normal distribution, because our estimate of $s$ becomes more accurate with larger samples.

The general formula for a one-sample confidence interval for $\mu$ is: $$\bar{x}\pm t^{\ast }\left(\frac{s}{\sqrt{n}}\right)$$ The term $t^{\ast }\left(\frac{s}{\sqrt{n}}\right)$ is called the margin of error ($ME$), which describes how far we expect our sample estimate to be from the true population mean, at the given confidence level.

Worked Example

An ecologist measures the mercury concentration (in ppm) of 15 randomly selected fish from a local lake. The sample mean concentration is 0.42 ppm, and the sample standard deviation is 0.11 ppm. What sampling distribution should be used to construct a confidence interval for the true mean mercury concentration, and what are its degrees of freedom?

1. We are estimating a population mean, and we only have the sample standard deviation ($s=0.11$), not the true population standard deviation $\sigma$.
2. The appropriate distribution for the critical value is a t-distribution, which accounts for extra variability from estimating $\sigma$.
3. For a one-sample procedure, degrees of freedom are calculated as $df=n-1=15-1=14$.
4. This t-distribution with 14 df is bell-shaped with fatter tails than the standard normal distribution.

Exam tip: On the AP exam, you will never lose points for using a t-interval when $\sigma$ is unknown, even for large sample sizes. The t-distribution converges to the z-distribution, so the interval will be nearly identical, and graders accept t-intervals as correct.

3. Conditions for Inference for a Population Mean

Before you can construct a valid confidence interval for a population mean, you must verify three core conditions to ensure your sampling distribution is well-behaved and your inference is reliable. On AP FRQs, you must explicitly state and check all three conditions for full credit.

The three conditions are:

1. Random: The data comes from a random sample from the population of interest, or a randomized experiment. This ensures your sample is representative and reduces selection bias.
2. Independent: Individual observations are independent of each other. When sampling without replacement from a finite population, this is verified via the 10% condition: the sample size $n$ must be less than 10% of the total population size. This ensures sampling without replacement does not meaningfully change the variability of our estimate.
3. Normal/Large Sample: The sampling distribution of $\bar{x}$ is approximately normal. This is true if either: (a) the sample size is $n\ge 30$ (by the Central Limit Theorem), or (b) for $n<30$, the sample data has no extreme outliers or strong skewness, indicating the population is approximately normal.

Worked Example

A university wants to estimate the average GPA of its current undergraduate students. It obtains a list of all 12,000 undergraduates, randomly selects 40 students, and finds the distribution of GPAs is slightly left-skewed with no extreme outliers. Check the conditions for constructing a confidence interval for the true mean undergraduate GPA.

1. Random: The problem explicitly states the sample of 40 students is randomly selected from the population of undergraduates, so the random condition is satisfied.
2. Independent: The sample size is 40, and 10% of the total population of 12,000 is 1200. 40 < 1200, so the 10% condition is met, and observations are independent.
3. Normal/Large Sample: The sample size $n=40\ge 30$, so by the Central Limit Theorem, the sampling distribution of $\bar{x}$ is approximately normal, even with slight skewness. All conditions are satisfied.

Exam tip: If the problem does not explicitly state the sample is random, you must write "we assume the sample is random" to earn credit for the condition. Never skip checking the random condition just because it is not given.

4. Two-Sample Confidence Intervals for the Difference in Means

We use a two-sample t-interval when we have two independent samples from two different populations, and we want to estimate the difference between their true population means (${\mu }_1-{\mu }_2$). This is commonly used to compare the average of a quantitative variable across two groups (e.g., average test scores for two teaching methods, average commute times for two cities).

The formula for a 95% two-sample t-interval for ${\mu }_1-{\mu }_2$ is: $$({\bar{x}}_1-{\bar{x}}_2)\pm t^{\ast }\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$$ For the AP exam, the conservative degrees of freedom (always accepted for full credit) is $df=\min (n_1-1,n_2-1)$. Most calculators use the Welch-Satterthwaite approximation to get a larger df, which gives a slightly narrower interval, but both methods are acceptable.

Worked Example

A bakery wants to estimate the difference in average rise height of two brands of yeast. They test 12 batches of dough with brand A (mean rise 6.2 cm, s = 0.8 cm) and 10 batches with brand B (mean rise 5.7 cm, s = 0.7 cm). Construct a 95% confidence interval for the difference in true mean rise (${\mu }_A-{\mu }_B$).

1. Check conditions: Both samples are independent and random, total batches of dough are far more than 10 times each sample size, and we assume no extreme outliers, so all conditions are met.
2. Calculate the difference in sample means: ${\bar{x}}_A-{\bar{x}}_B=6.2-5.7=0.5$ cm.
3. Conservative degrees of freedom: $\min (12-1,10-1)=9$, so the critical $t^{\ast }=2.262$.
4. Calculate the standard error: $SE=\sqrt{\frac{0.8^2}{12}+\frac{0.7^2}{10}}=\sqrt{0.0533+0.049}\approx 0.32$.
5. Margin of error: $ME=2.262\ast 0.32\approx 0.72$.
6. Final interval: $0.5\pm 0.72=(-0.22,1.22)$ cm.

Exam tip: If 0 is inside your two-sample confidence interval for the difference in means, that means no difference between the population means is plausible, which is a key result for inference that AP examiners test frequently.

5. Sample Size Determination for a Desired Margin of Error

Before collecting data, researchers often calculate the minimum sample size needed to achieve a maximum desired margin of error for their confidence interval. Since we do not have a sample standard deviation $s$ before collecting data, we use a prior estimate of $\sigma$ (from a pilot study or previous research) and use the z-distribution for calculation. The difference between $t^{\ast }$ and $z^{\ast }$ is negligible for the large sample sizes produced by this calculation.

The formula for minimum sample size is: $$n={\left(\frac{z^{\ast }{\sigma }^{\ast }}{ME}\right)}^2$$ where ${\sigma }^{\ast }$ is the prior estimate of $\sigma$, $z^{\ast }$ is the critical z-value for the desired confidence level, and we always round up to the next whole number, regardless of the decimal part. Rounding down would produce a margin of error larger than the maximum allowed.

Worked Example

A market researcher wants to estimate the average amount spent by customers at a new coffee shop, with a 95% confidence interval and a margin of error no larger than $\mathrm{1.50.}Apilotstudyfoundthestandarddeviationofpurchaseamountsisapproximately$6.20. What is the minimum number of customers the researcher needs to sample?

1. For 95% confidence, $z^{\ast }=1.96$, $ME=1.50$, ${\sigma }^{\ast }=6.20$.
2. Plug into the formula: $n={\left(\frac{1.96\ast 6.20}{1.50}\right)}^2={\left(\frac{12.152}{1.50}\right)}^2=(8.1013{)}^2\approx 65.63$.
3. We round up to the next whole number, so the minimum sample size is 66 customers.
4. Check: For $n=65$, $ME=1.96\ast 6.20/\sqrt{65}\approx 1.51$, which exceeds the maximum allowed $ME$ of 1.50, so 65 is too small.

Exam tip: Never follow standard rounding rules for sample size calculation. Even 65.1 rounds up to 66, because any decimal means you need one additional observation to meet the margin of error requirement.

Common Pitfalls (and how to avoid them)

• Wrong move: Using a z-distribution (z*) instead of t-distribution (t*) when the population standard deviation $\sigma$ is unknown. Why: Students confuse means intervals with proportion intervals, or assume $z$ is fine for large $n$. Correct move: Always use t* for confidence intervals for means unless the problem explicitly gives you the true population $\sigma$.
• Wrong move: Interpreting a confidence interval as "there is a 95% probability the true mean is in this interval." Why: Students confuse the confidence level (which describes the method, not a single interval) with probability; the true mean is fixed, not random. Correct move: Always interpret as "We are 95% confident that the true [population mean in context] is between [lower] and [upper]."
• Wrong move: Forgetting to check or state the 10% condition for independence. Why: Students remember random and normal conditions, but skip independence. Correct move: Always check the 10% condition (n < 10% of population) for independence when sampling without replacement.
• Wrong move: Rounding down the final sample size when the calculation gives a decimal. Why: Students use standard rounding rules instead of the special rule for sample size. Correct move: Always round up to the next integer, regardless of the decimal part, to ensure the margin of error requirement is met.
• Wrong move: For a two-sample interval, claiming "0 is in the interval so the population means are equal." Why: Students overstate the conclusion from the interval. Correct move: Say "0 is a plausible value for the difference, so we do not have convincing evidence of a difference between the population means."
• Wrong move: Skipping the random condition on FRQs because it is not stated in the problem. Why: Students assume randomness if it is not mentioned. Correct move: Explicitly write "we assume the sample is random" to earn credit for the condition.

Practice Questions (AP Statistics Style)

Question 1 (Multiple Choice)

A biologist takes a random sample of 25 hummingbirds and constructs a 90% confidence interval for the mean weight of this species: (3.1 g, 4.3 g). What is the margin of error for this interval? A) 0.6 g B) 1.2 g C) 3.7 g D) Cannot be calculated without the critical t-value

Worked Solution: A confidence interval is always structured as $\bar{x}\pm ME$, so the width of the interval is twice the margin of error. First calculate the interval width: $4.3-3.1=1.2$ g. The margin of error is half the width, so $ME=1.2/2=0.6$ g. We do not need the critical t-value to find ME when the interval bounds are already given. The correct answer is A.

Question 2 (Free Response)

A sports scientist wants to compare the average vertical jump height of high school football players and high school basketball players. (a) A random sample of 60 football players has a mean vertical jump of 26.2 inches, standard deviation 3.1 inches. A random sample of 50 basketball players has a mean vertical jump of 28.1 inches, standard deviation 3.5 inches. Check all conditions for constructing a 95% confidence interval for the difference in mean jump height (${\mu }_{basketball}-{\mu }_{football}$). (b) Construct and interpret the 95% confidence interval. (c) Based on the interval, is there convincing evidence that the true mean vertical jump differs between the two groups? Justify your answer.

Worked Solution: (a) 1. Random: Both samples are stated to be random, so the random condition is satisfied. 2. Independent: The population of high school football and basketball players is far larger than 1060=600 and 1050=500, so the 10% condition is met, and the samples are independent. 3. Normal/Large Sample: Both sample sizes (60 and 50) are greater than 30, so the Central Limit Theorem applies, and the sampling distribution is approximately normal. All conditions are satisfied. (b) Difference in sample means: $28.1-26.2=1.9$ inches. Conservative $df=\min (50-1,60-1)=49$, $t^{\ast }\approx 2.009$. Standard error: $\sqrt{\frac{3.5^2}{50}+\frac{3.1^2}{60}}=\sqrt{0.245+0.160}\approx 0.636$. Margin of error: $2.009\ast 0.636\approx 1.28$. Interval: $1.9\pm 1.28=(0.62,3.18)$ inches. Interpretation: We are 95% confident that the true difference in mean vertical jump height between basketball players and football players is between 0.62 inches and 3.18 inches. (c) 0 is not inside the 95% confidence interval, so 0 is not a plausible value for the true difference in means. We have convincing evidence at the 5% significance level that the true mean vertical jump differs between the two groups.

Question 3 (Application / Real-World Style)

A civil engineer wants to estimate the average daily water usage per household in a new subdivision, with a 99% confidence interval and a margin of error no larger than 12 gallons. A previous study in a similar subdivision found the standard deviation of daily water usage is 48 gallons. What is the minimum number of households the engineer needs to sample? Interpret your result in context.

Worked Solution: For 99% confidence, $z^{\ast }=2.576$. We have $ME=12$ gallons, ${\sigma }^{\ast }=48$ gallons. Plug into the sample size formula: $n={\left(\frac{2.576\ast 48}{12}\right)}^2=(10.304{)}^2\approx 106.17$. We round up to the next whole number, so $n=107$. The engineer needs to sample at least 107 households in the subdivision to achieve a 99% confidence interval for mean daily water usage with a margin of error no larger than 12 gallons.

Quick Reference Cheatsheet

What's Next

Confidence intervals for population means are the foundation for all inference on quantitative means, and the next topic you will study is hypothesis testing for population means. Without mastering the conditions for inference, t-distribution properties, and confidence interval interpretation from this chapter, interpreting hypothesis test results for means will be impossible. This topic builds directly on confidence intervals for proportions you learned earlier, extending that reasoning to account for the extra variability introduced by estimating the unknown population standard deviation. In the bigger picture, all inferential statistics for quantitative data relies on the core ideas of interval estimation for means introduced here. For follow-up study, explore these connected topics: Hypothesis Tests for a Population Mean Inference for Paired Data Comparing Two Population Means Confidence Intervals for a Population Proportion