Source: Principles of Statistics, University of Central Florida
Tags: inference for means, t-distribution, confidence intervals, hypothesis testing, t-test, two-sample t-test, paired t-test, margin of error, degrees of freedom, p-value, AP Stats Unit 7
Difficulty: Intermediate Prerequisites: Basic understanding of sampling distributions, standard deviation, and the normal distribution (Units 1–6). You should be comfortable with z-scores and the idea of a sampling distribution of a sample mean before starting here.
Big picture: This unit is where statistics stops being descriptive and starts being inferential. You are learning to make claims about an entire population using only a sample. Specifically, you are estimating population means and testing hypotheses about them using the t-distribution (because in real life you almost never know the true population standard deviation). If the earlier units taught you how to summarise data, this one teaches you how to draw conclusions from it. This is the heart of what shows up on the AP exam's free-response section.
When you do not know the population standard deviation (which is nearly always), you use the t-distribution instead of the normal distribution to build confidence intervals and run hypothesis tests for means. The key mechanics are: pick the right t* critical value using degrees of freedom, check your conditions, and interpret the results in context.
Point estimate
The single best guess for a population parameter based on your sample. For a population mean, the point estimate is the sample mean (x̄). Think of it as your starting number before you add or subtract a margin of error.
Margin of error
The amount added and subtracted from the point estimate to create a confidence interval. Calculated as t* × (s / √n). In simple terms, this is the "wiggle room" around your estimate that accounts for sampling variability.
t-distribution
A family of bell-shaped curves used in place of the normal distribution when the population standard deviation is unknown and estimated by the sample standard deviation. It is wider and has heavier tails than the normal curve, especially for small sample sizes. Think of it as the normal distribution's cautious cousin: it gives you wider intervals to compensate for the extra uncertainty.
Degrees of freedom (df)
For a one-sample t-procedure, df = n − 1. This value determines which specific t-distribution you use. As df increases, the t-distribution gets closer to the standard normal distribution.
t (t-critical value)*
The number of standard errors you go out from the mean to capture a given confidence level. Found in a t-distribution table by matching your df (row) to your tail probability (column). For a 95% confidence interval, the tail probability is 0.025.
Confidence interval
A range of plausible values for a population parameter, constructed as: point estimate ± margin of error. A 95% confidence interval means that if you repeated the sampling process many times, about 95% of the resulting intervals would contain the true parameter.
Null hypothesis (H₀)
The default claim about a population parameter, typically that nothing interesting is happening (e.g., μ = some value). This is the statement you assume to be true unless your data provides strong enough evidence against it.
Alternative hypothesis (H₁ or Hₐ)
The claim you are testing for, which contradicts the null. It can be one-sided (μ < value or μ > value) or two-sided (μ ≠ value).
p-value
The probability of obtaining a test statistic as extreme as (or more extreme than) the one observed, assuming the null hypothesis is true. A small p-value means the observed data would be unusual under H₀. In simple terms, it answers: "If nothing special were going on, how surprised should I be by this result?"
Significance level (α)
The threshold you set before collecting data. If your p-value falls below α, you reject the null hypothesis. Common values are 0.05 and 0.01.
One-sample t-test
A hypothesis test used to compare a sample mean to a claimed population mean when the population standard deviation is unknown. Uses the test statistic t = (x̄ − μ₀) / (s / √n).
Two-sample t-test
A hypothesis test used to compare the means of two independent groups. The key word is "independent": the individuals in one group have no natural connection to the individuals in the other.
Paired t-test
A hypothesis test for the mean difference within paired observations (e.g., before/after measurements on the same subjects, or matched pairs). You reduce the two measurements to a single set of differences and then run a one-sample t-test on those differences.
The general formula for a one-sample t-interval is:
x̄ ± t* × (s / √n)
where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and t* comes from the t-distribution table with df = n − 1.
Finding t from the table:*
Calculate your degrees of freedom: df = n − 1.
Determine the tail probability. For a 95% confidence level, each tail has 0.025 (since 5% is split across two tails). For 98% confidence, each tail has 0.01.
Look up the intersection of your df row and your tail probability column.
Worked example (coffee cups):
A sample of n = 10 cups has x̄ = 12.5 oz and s = 0.5 oz.
df = 10 − 1 = 9.
For 95% confidence, tail probability = 0.025. From the t-table, t* = 2.262.
Margin of error = 2.262 × (0.5 / √10).
Confidence interval: 12.5 ± 2.262 × (0.5 / √10).
When a confidence interval for μ₁ − μ₂ does not contain zero, there is evidence of a real difference between the two population means.
If the entire interval is negative, μ₁ is plausibly lower than μ₂ for every value in the interval.
If the entire interval is positive, μ₁ is plausibly higher than μ₂.
If the interval straddles zero, you cannot conclude that either mean is larger.
Worked example (blood pressure diet):
A 98% confidence interval for μ_treatment − μ_control is (−27, −4).
The interval is entirely negative, meaning for all plausible values of the difference, the treatment group's mean blood pressure is lower.
This supports the claim that the diet reduces blood pressure.
Step 1: State hypotheses.
H₀: μ = [claimed value]
H₁: μ < , > , or ≠ [claimed value], depending on the research question.
Step 2: Check conditions.
Random: the sample must be randomly selected or randomly assigned.
Independence (10% condition): the sample size must be less than 10% of the population.
Normality: the sampling distribution of x̄ should be approximately normal. Check via a graph of the data for strong skew or outliers, or rely on a large enough sample (n ≥ 30 by the Central Limit Theorem).
Step 3: Calculate the test statistic and p-value.
t = (x̄ − μ₀) / (s / √n)
Find the p-value using the t-distribution with df = n − 1.
Step 4: Conclude in context.
If p-value < α, reject H₀. State the conclusion in the context of the problem.
If p-value ≥ α, fail to reject H₀.
Worked example (TV watching):
A father claims his son watches 2 hours of TV per day and wants to test whether the son's peers watch less.
H₀: μ = 2, H₁: μ < 2, α = 0.05.
n = 20 TV times, data shows no strong skew or outliers, sample is random, and 20 is less than 10% of all TV times.
Test statistic: t = 1.81, p-value = 0.0431.
Since 0.0431 < 0.05, reject H₀. There is evidence at the 5% significance level that the son's peers watch less TV than he does.
This distinction is one of the most commonly tested conceptual points on the AP exam.
Use a two-sample t-test when:
The two groups are independently selected.
There is no natural one-to-one link between an observation in one group and an observation in the other.
Use a paired t-test when:
Each observation in one group is naturally matched to exactly one observation in the other (e.g., the same person measured twice, or siblings, or matched experimental pairs).
Worked example (math scores):
A teacher randomly selects 10 girls and 10 boys from a class and assigns numbers alphabetically within each group.
The correct procedure is a two-sample t-test (answer: C).
The alphabetical numbering creates an arbitrary pairing, not a natural one. Being in the same class does not make the samples dependent. Having equal sample sizes does not justify pairing either.
One-sample t-interval:
x̄ ± t* × (s / √n), where df = n − 1
One-sample t-test statistic:
t = (x̄ − μ₀) / (s / √n), where df = n − 1
Key t values to recognise (df = 9 example from the t-table):*
90% confidence (tail = 0.05): t* = 1.833
95% confidence (tail = 0.025): t* = 2.262
99% confidence (tail = 0.005): t* = 3.250
Confidence intervals for means are used constantly in quality control (e.g., checking that a manufacturing process produces items within specification), clinical trials (estimating treatment effects), and market research (estimating average customer spend). Any time you see a poll or study report a result "plus or minus" some number, that is a confidence interval in action.
Students often say "there is a 95% probability the true mean is in this interval." That is incorrect. The true mean is fixed; the interval either contains it or does not. The 95% refers to the long-run success rate of the method.
Students confuse when to use z* versus t*. If the population standard deviation σ is unknown (which is nearly always the case in practice), you must use t*.
Students assume that equal sample sizes between two groups means the data is paired. Equal n does not create pairing; a natural, meaningful link between specific observations does.
Students forget to check conditions before running a t-test. The AP exam awards points specifically for stating and verifying Random, 10%, and Normal/Large Sample conditions.
⚠️ Free-response questions almost always require you to state hypotheses, check conditions, compute the test statistic, and interpret the result in context. Skipping any step costs marks.
⚠️ Know how to read the t-table: match df on the row, tail probability on the column. This is tested directly.
⚠️ The distinction between paired and two-sample t-tests appears regularly as a multiple-choice trap. Look for a natural, meaningful pairing, not just equal group sizes.
⚠️ When interpreting a confidence interval for a difference in means, check whether the interval contains zero. If it does not, there is evidence of a difference.
⚠️ Always state conclusions in context. "Reject H₀" on its own is incomplete. Say what it means for the specific scenario.
1. True or false: A 95% confidence interval means there is a 95% chance the true population mean lies within the interval.
Answer: False. The 95% describes the method's long-run capture rate, not the probability for any single interval.
2. Fill in the blank: For a one-sample t-procedure with n = 15, the degrees of freedom are ___.
Answer: 14.
3. True or false: If a 98% confidence interval for μ₁ − μ₂ is (3, 12), there is evidence that μ₁ is greater than μ₂.
Answer: True. The entire interval is positive.
4. Fill in the blank: You should use a paired t-test instead of a two-sample t-test when there is a natural ___ between observations in the two groups.
Answer: Pairing (or one-to-one match).
5. True or false: When the population standard deviation is known, you should use the t-distribution.
Answer: False. When σ is known, you use the z-distribution (though this is rare in practice).
Q: A sample of 10 cups of coffee has a mean weight of 12.5 oz and a standard deviation of 0.5 oz. Write the formula for a 95% confidence interval for the population mean weight.
A: 12.5 ± 2.262 × (0.5 / √10). The t* value of 2.262 comes from the t-table at df = 9, tail probability = 0.025.
Q: A 98% confidence interval for the difference in means (treatment minus control) is (−27, −4). Does this support the claim that the treatment lowers the outcome?
A: Yes. The interval is entirely negative, meaning every plausible value of the difference indicates the treatment group's mean is lower than the control group's mean.
Q: A father collects 20 TV-watching times to test whether his son's peers watch less than 2 hours per day. He gets t = 1.81 and p = 0.0431 at α = 0.05. What is the conclusion?
A: Since 0.0431 < 0.05, reject H₀. There is sufficient evidence at the 5% level that the mean TV watching time of the son's peers is less than 2 hours per day.
Q: A teacher randomly selects 10 girls and 10 boys from a class to compare their mean maths scores. She numbers them alphabetically within each group. Should she use a paired or two-sample t-test?
A: Two-sample t-test. The boys and girls are independently and randomly selected. The alphabetical numbering is arbitrary and does not create a natural pairing. Equal sample sizes alone do not justify a paired test.
Q: What three conditions must be checked before performing a one-sample t-test?
A: Random (the data comes from a random sample or random assignment), 10% condition (the sample is less than 10% of the population), and Normal/Large Sample (the data shows no strong skew or outliers, or n is large enough for the CLT to apply).
This material connects directly to Unit 6 (inference for proportions), which uses z-procedures instead of t-procedures. The logic of confidence intervals and hypothesis testing is identical; only the formula and distribution change. Understanding both lets you recognise which tool to reach for based on whether the parameter of interest is a mean or a proportion.
The conditions-checking framework here (Random, 10%, Normal/Large Sample) carries forward into every inference procedure for the rest of the course, including chi-square tests and regression inference.
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