Inference for a Population Mean — AP Statistics Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: t-distribution properties, one-sample t-interval for a population mean, one-sample t-test for a population mean, paired t-procedures for dependent samples, conditions for t-inference, and context interpretation of inference results for means.

You should already know: Basic confidence interval and hypothesis test logic, properties of the normal distribution, how to calculate sample mean and sample standard deviation.

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 Inference for a Population Mean?

Inference for a population mean is the process of using quantitative data from a random sample to draw evidence-based conclusions about the unknown true mean ($\mu$) of a full population. This topic is core to Unit 7 (Inference for Quantitative Data: Means), which makes up 12-15% of the total AP Statistics exam weight, and inference for a single population mean accounts for roughly 1/3 of that unit’s testing time. This topic appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a standalone question or embedded into larger problems about paired data.

Notation conventions are standard: $\mu$ = unknown population mean, $\bar{x}$ = sample mean (our point estimate for $\mu$), $s$ = sample standard deviation, $n$ = sample size, and $df$ = degrees of freedom. Unlike inference for proportions, which relies on the standard normal (z) distribution, inference for means almost always uses the t-distribution because we almost never know the true population standard deviation $\sigma$. The two core goals of inference for a population mean are to estimate $\mu$ with a confidence interval or test a claim about $\mu$ with a hypothesis test.

2. The t-Distribution and Conditions for Inference

When we calculate the standard deviation of the sampling distribution of $\bar{x}$, we get $\sigma /\sqrt{n}$. But since we almost never know $\sigma$, we estimate it with the sample standard deviation $s$, giving us the standard error $s/\sqrt{n}$. The ratio $\frac{\bar{x}-\mu }{s/\sqrt{n}}$ does not follow a normal distribution — it follows a t-distribution, a symmetric, bell-shaped distribution centered at 0, similar to the z-distribution but with fatter tails. The fatter tails account for the extra variability introduced by estimating $\sigma$ with $s$.

The shape of the t-distribution depends only on degrees of freedom ($df$), which for one-sample inference equals $n-1$. As $df$ (and thus sample size) increases, the t-distribution approaches the z-distribution, because $s$ becomes a more accurate estimate of $\sigma$ as $n$ grows. Before conducting any t-inference, we must check three required conditions:

1. Random: The data comes from a random sample or randomized experiment, to ensure unbiasedness.
2. Independence: Individual observations are independent. For sampling without replacement, this requires the 10% condition: $n<0.1N$ (sample is less than 10% of the population).
3. Normal/Large Sample: The sampling distribution of $\bar{x}$ is approximately normal. This is true if $n\ge 30$ (by the Central Limit Theorem) or, for smaller $n$, the sample data has no strong skewness or outliers.

Worked Example

A student wants to estimate the mean height of 10th graders at their high school. They collect a random sample of 18 10th graders. What are the degrees of freedom for t-inference, and what conditions must the student check?

1. Degrees of freedom for one-sample inference: $df=n-1=18-1=17$.
2. Random: The problem states the sample is random, so this condition is satisfied.
3. Independence: The population of 10th graders at a typical high school is more than $10\times 18=180$, so the 10% condition is satisfied and independence holds.
4. Normal/Large Sample: The sample size $n=18<30$, so the student must confirm the sample height data has no strong skewness or outliers to justify the normality assumption.

Exam tip: On the AP exam, you must explicitly name and justify each of the three conditions for full credit. Just writing "conditions are met" without explanation will cost you an easy point.

3. One-Sample t-Interval for a Population Mean

A one-sample t-interval is used to estimate the unknown value of a population mean $\mu$ from sample data. The interval follows the general confidence interval structure: $$\text{Point Estimate}\pm \text{Critical Value}\times \text{Standard Error}$$ For a population mean, the point estimate is $\bar{x}$, the critical value is $t_{df}^{\ast }$ (the t-value corresponding to the desired confidence level with $df=n-1$, found from a t-table or calculator), and the standard error is $s/\sqrt{n}$. The full formula for the interval is: $$\bar{x}\pm t_{df}^{\ast }\frac{s}{\sqrt{n}}$$ The correct interpretation of a C% confidence interval is: We are C% confident that the interval from [lower bound] to [upper bound] captures the true population mean [in context of the problem]. The confidence level itself refers to the long-run behavior of the method: if we repeated sampling many times, C% of all intervals constructed this way would capture the true mean.

Worked Example

A local bakery wants to estimate the true mean weight of its sourdough loaves. A random sample of 25 loaves gives a sample mean weight of 802 grams and a sample standard deviation of 15 grams. Construct and interpret a 95% confidence interval for the true mean weight of all sourdough loaves from this bakery.

1. Check conditions: Random sample stated, population of loaves is far more than 250, $n=25<30$ so we assume no extreme skewness in weights (reasonable for this context), so all conditions are met.
2. Calculate degrees of freedom: $df=25-1=24$. For 95% confidence, $t_{24}^{\ast }=2.064$.
3. Calculate the interval: $$802\pm 2.064\times \frac{15}{\sqrt{25}}=802\pm 6.19=(795.81,808.19)$$
4. Interpret: We are 95% confident that the interval from 795.81 grams to 808.19 grams captures the true mean weight of all sourdough loaves from this bakery.

Exam tip: If you use a calculator to compute the interval, you still need to write the formula and plug in your values for full credit on FRQ. The AP exam requires you to show your work, not just give a final number.

4. One-Sample t-Test for a Population Mean

A one-sample t-test is used to test a claim about the value of a population mean $\mu$. The null hypothesis is always $H_0:\mu ={\mu }_0$, where ${\mu }_0$ is the hypothesized value from the claim. The alternative hypothesis is two-sided ($H_a:\mu \ne {\mu }_0$), left-tailed ($H_a:\mu <{\mu }_0$), or right-tailed ($H_a:\mu >{\mu }_0$), depending on the research question.

The test statistic for a one-sample t-test is: $$t=\frac{\bar{x}-{\mu }_0}{s/\sqrt{n}}$$ This statistic measures how far our sample mean is from the hypothesized mean, measured in standard error units. The p-value is the probability of observing a t-statistic as extreme or more extreme than the one we calculated, assuming $H_0$ is true. We compare the p-value to the significance level $\alpha$ (usually 0.05, unless stated otherwise): if $p<\alpha$, we reject $H_0$; otherwise, we fail to reject $H_0$.

Worked Example

A coffee shop chain claims that the mean caffeine content of its 12-oz lattes is 100 mg. An independent tester collects a random sample of 12 lattes and gets a sample mean caffeine content of 108 mg with a sample standard deviation of 10 mg. Test whether the true mean caffeine content differs from the chain’s claim at the $\alpha =0.05$ significance level.

1. State hypotheses: Let $\mu$ = true mean caffeine content of all 12-oz lattes from this chain. $H_0:\mu =100$, $H_a:\mu \ne 100$.
2. Check conditions: Random sample stated, population of lattes more than 120, n=12 so we assume no extreme outliers in caffeine content, conditions met.
3. Calculate test statistic: $df=12-1=11$, $$t=\frac{108-100}{10/\sqrt{12}}\approx 2.77$$
4. Find p-value: For a two-tailed test with $df=11$ and $t=2.77$, $p$-value $\approx 0.018$.
5. Conclusion: Since $0.018<0.05$, we reject $H_0$. There is convincing evidence at the 0.05 significance level that the true mean caffeine content of the chain’s 12-oz lattes differs from the claimed 100 mg.

Exam tip: Always state your conclusion in the context of the problem, not just "reject $H_0$". Failing to add context will cost you a point on FRQ.

5. Paired t-Procedures for Dependent Samples

Paired data occurs when we have two dependent measurements (e.g., before/after treatment on the same subject, matched pairs of subjects with similar characteristics). Because the two measurements are not independent, we cannot use two-sample t-procedures. Instead, we calculate the difference $d_i$ for each pair (e.g., $d_i=\text{before }i-\text{after }i$), then conduct one-sample inference on the true mean difference ${\mu }_d$. All rules for one-sample t-intervals and t-tests apply directly to paired data: $df=n-1$ where $n$ is the number of pairs, and we use the mean difference $\bar{d}$ and standard deviation of differences $s_d$ in all calculations.

Worked Example

A physical therapist tests whether a new stretching routine reduces resting hamstring flexibility. She measures flexibility (in cm) for 8 subjects before and after the 4-week routine, and calculates a mean increase in flexibility of $\bar{d}=1.2$ cm (after - before) with a standard deviation of differences $s_d=1.1$ cm. Test the therapist’s hypothesis that the routine increases flexibility at $\alpha =0.05$.

1. Hypotheses: Let ${\mu }_d$ = true mean difference (after - before) in flexibility. $H_0:{\mu }_d=0$ (no change), $H_a:{\mu }_d>0$ (flexibility increases).
2. Conditions: Random assignment assumed, 8 subjects less than 10% of all potential patients, no extreme outliers in differences, conditions met.
3. Test statistic: $df=8-1=7$, $$t=\frac{1.2-0}{1.1/\sqrt{8}}\approx 3.09$$
4. P-value: For a right-tailed test, $p$-value $\approx 0.009$.
5. Conclusion: Since $0.009<0.05$, we reject $H_0$. There is convincing evidence that the stretching routine increases mean hamstring flexibility.

Exam tip: Always calculate differences first and do inference on the differences. Never treat paired data as two independent samples, this is one of the most commonly tested errors on the AP exam.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Using a z-distribution for inference on a mean when the population standard deviation $\sigma$ is unknown. Why: Students confuse inference for means (almost always use t) with inference for proportions (always use z). Correct move: Unless the problem explicitly gives you the true population standard deviation $\sigma$, always use t-procedures for inference on a mean.
• Wrong move: Failing to check normality for small samples, and automatically concluding "conditions are not met" when $n<30$. Why: Students memorize the $n\ge 30$ rule and forget that normality can still be assumed for small samples with roughly symmetric data. Correct move: For $n<30$, explicitly state that you check for no strong skewness or outliers in the sample data to justify the normality assumption.
• Wrong move: Interpreting a confidence interval as "C% of sample means fall in this interval" or "C% of population data is in this interval". Why: Students confuse the location of the true mean with the behavior of the sampling method. Correct move: Always interpret the interval as "We are C% confident that the true population mean is between [lower] and [upper]".
• Wrong move: Saying "we accept $H_0$" when the p-value is greater than $\alpha$. Why: Students think a large p-value proves the null hypothesis is true. Correct move: Always say "we fail to reject $H_0$" — we only lack sufficient evidence to reject the null, we cannot prove it is true.
• Wrong move: Treating paired dependent data as two independent samples. Why: Students see two groups of data and automatically jump to a two-sample test, without noticing the pairing. Correct move: Always check if there is a natural matching of observations (same subject before/after, matched pairs) — if yes, use a paired t-procedure on differences.
• Wrong move: Forgetting that degrees of freedom is $n-1$, not $n$, for one-sample and paired procedures. Why: Students mix up sample size and degrees of freedom, leading to incorrect critical values and p-values. Correct move: Always subtract 1 from the number of observations (or number of pairs for paired data) when calculating df.

7. Practice Questions (AP Statistics Style)

Question 1 (Multiple Choice)

A botanist is estimating the mean height of 1-year-old pine seedlings in a large forest. She takes a random sample of 15 seedlings, and gets a sample mean height of 28 cm with a sample standard deviation of 4 cm. Assuming all conditions for inference are met, what is the margin of error for a 95% confidence interval for the true mean height? A) 2.22 cm B) 2.03 cm C) 1.03 cm D) 2.14 cm

Worked Solution: The margin of error for a t-interval is $ME=t_{df}^{\ast }\frac{s}{\sqrt{n}}$. Degrees of freedom $df=15-1=14$. For 95% confidence, the critical $t_{14}^{\ast }=2.145$. Plugging in values: $ME=2.145\times \frac{4}{\sqrt{15}}=2.145\times 1.033\approx 2.22$. Option B uses $z^{\ast }=1.96$ instead of $t^{\ast }$, which is incorrect. Option C is just the standard error without the critical value. Option D is just the critical value, not scaled by standard error. The correct answer is A.

Question 2 (Free Response)

A fast-food chain advertises that the mean calorie count of its grilled chicken sandwich is less than 400 calories. A consumer advocacy group takes a random sample of 30 sandwiches and finds a mean calorie count of 392 calories with a standard deviation of 25 calories. (a) What type of inference procedure is appropriate here? Justify your answer. (b) State the null and alternative hypotheses in terms of the parameter of interest, check conditions for inference, and calculate the test statistic and degrees of freedom. (c) The p-value for the test is 0.041. Interpret the p-value in context, and state your conclusion at the $\alpha =0.05$ significance level.

Worked Solution: (a) A one-sample t-test for a population mean is appropriate. We are testing a claim about one single population mean (true mean calorie count of the sandwiches), and the population standard deviation is unknown. (b) Let $\mu$ = the true mean calorie count of the chain’s grilled chicken sandwiches. Hypotheses: $H_0:\mu =400$, $H_a:\mu <400$. Conditions: 1) Random: the sample is stated to be random, so met. 2) Independence: The population of sandwiches is far more than $10\times 30=300$, so the 10% condition is met. 3) Normal/Large Sample: $n=30\ge 30$, so the Central Limit Theorem guarantees an approximately normal sampling distribution. All conditions are satisfied. Degrees of freedom $df=30-1=29$. Test statistic: $t=\frac{392-400}{25/\sqrt{30}}\approx -1.753$. (c) P-value interpretation: If the true mean calorie count of the sandwiches is actually 400 calories, the probability of getting a sample mean as low or lower than 392 calories is 0.041. Since $0.041<0.05$, we reject the null hypothesis. There is convincing evidence at the $\alpha =0.05$ level that the true mean calorie count of the grilled chicken sandwiches is less than 400 calories, supporting the chain’s advertisement.

Question 3 (Application / Real-World Style)

A civil engineer wants to estimate the mean breaking strength of a new batch of steel support bolts for a bridge. The required mean breaking strength is at least 10,000 psi. A random sample of 40 bolts from the batch gives a mean breaking strength of 10,215 psi with a standard deviation of 580 psi. Construct a 90% confidence interval for the true mean breaking strength of all bolts in the batch. Is it plausible that the batch meets the requirement of a mean breaking strength of at least 10,000 psi?

Worked Solution: Conditions are met: random sample, $n=40\ge 30$ so CLT applies, 40 bolts is less than 10% of a large production batch. Degrees of freedom $df=40-1=39$. For 90% confidence, $t_{39}^{\ast }\approx 1.685$. Standard error $=\frac{580}{\sqrt{40}}\approx 91.7$. Margin of error $=1.685\times 91.7\approx 154.5$. The 90% confidence interval is $10215\pm 154.5=(10060.5,10369.5)$. We are 90% confident that the true mean breaking strength of all bolts in the batch is between 10060.5 psi and 10369.5 psi. The entire interval is above 10,000 psi, so it is plausible that the batch meets the required mean breaking strength.

8. Quick Reference Cheatsheet

9. What's Next

Inference for a population mean is the foundation for all inference on quantitative means, and the next topic in Unit 7 is inference for two population means from independent samples. Without mastering the t-distribution, conditions for inference, and core t-procedures covered here, two-sample t-procedures and later ANOVA for multiple means will be very difficult to master, as they build directly on the same logic. This topic also reinforces the general inference framework that applies to all inference problems across the AP Statistics course, from proportions to chi-square to regression inference. Mastering the common pitfalls here will help you avoid similar mistakes in more complex problems later.

• Inference for Two Population Means
• Confidence Interval General Framework
• Hypothesis Testing General Framework