NUM8ERS · AP Statistics revision

AP Statistics
Formula Sheet

Find the formula. Check the conditions. Explain what your answer means.

A student-friendly reference organised around the five units in our current AP Statistics notes. Each formula comes with its purpose, a reminder about when it applies, and examples you can work through.

2026–27 course5-unit referenceWorked examplesConditions + interpretations

AP Statistics Study Hub Find a Formula

Use this guide alongside the topic lessons. Knowing why a formula fits is just as useful as knowing how to substitute the numbers.

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Jump to the skill you need

This guide follows the course framework effective Fall 2026. Older released questions may use nine-unit numbering; match the statistical skill rather than the old unit label.

Unit 1Data & study design
Unit 2Probability & distributions
Unit 3Proportions & chi-square
Unit 4Means & paired data
Unit 5Regression

Official exam reference: College Board provides formulas and tables for both exam sections. This NUM8ERS guide adds explanations and examples; it is a revision resource. Find the official reference in the Appendix of the current Course and Exam Description.

Choose the method before the formula

Question you are answering	Method to consider	Where to look
How common is a yes/no outcome in one population?	One-proportion $z$ -interval or $z$ -test	Unit 3
Do two independent groups have different proportions?	Two-proportion $z$ -interval or $z$ -test	Unit 3
Do category counts fit a distribution, or are two categorical variables associated?	Chi-square test; identify the study design first	Unit 3
What is one population's mean?	One-sample $t$ -interval or $t$ -test	Unit 4
What is the mean change for matched observations?	Paired $t$ -procedure on the differences	Unit 4
Do two independent groups have different means?	Two-sample $t$ -interval or $t$ -test	Unit 4
How does one quantitative variable help predict another?	Least-squares regression; inspect scatterplot and residuals	Unit 5

Interval or test? Use an interval to estimate a parameter with uncertainty. Use a test to assess evidence about a stated population claim.

Read the notation with confidence

Symbol	Meaning	What to remember
$n, N$	Sample size; population size	A sample is the observed part of the population.
$\bar{x}, μ$	Sample mean; population mean	The sample statistic estimates the population parameter.
$s, σ$	Sample standard deviation; population standard deviation	Both describe variability in individual observations.
$\hat{p}, p, p_{0}$	Sample proportion; population proportion; null-hypothesised proportion	Choose the right proportion for an interval or a test.
$SE, ME$	Standard error; margin of error	Standard error estimates a statistic's sampling variability.
$z^{}, t^{}$	Critical values for confidence intervals	The confidence level determines the critical value; a $t^{*}$ value also depends on degrees of freedom.
$H_{0}, H_{a}, α$	Null hypothesis; alternative hypothesis; significance level	State hypotheses about population parameters.
$\hat{y}, r, r^{2}$	Predicted response; correlation; coefficient of determination	A prediction is different from an observed response.

Unit 1

Exploring One-Variable Data and Collecting Data

Start by identifying the variable and the group being studied. Numerical summaries describe a distribution; the study design determines what conclusions are supported.

Relative frequency and sample proportion

\hat{p} = \frac{x}{n}

Here $x$ is the count in the category of interest. Divide by the total number of observations to get its share.

Remember: A percentage is $100 \hat{p} %$ . Use the correct total for the group being described.

Mean and median

\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}

The mean is the balance point of the data. For the median, order the observations and locate the middle; average the two middle values when the sample size is even.

Remember: The mean is sensitive to extreme values. The median is resistant; choose summaries that fit the shape.

Sample variance and standard deviation

s^{2} = \frac{\sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}}{n - 1}, s = \sqrt{s^{2}}

Standard deviation describes a typical scale of distance from the sample mean. It has the same units as the observations.

Remember: This requires $n > 1$ . The denominator $n - 1$ makes sample variance unbiased under random sampling; sample standard deviation is not exactly unbiased.

Interquartile range and outlier fences

IQR = Q_{3} - Q_{1}

The IQR measures the spread of the middle half of the ordered data. Flag a possible outlier below $Q_{1} - 1.5 IQR$ or above $Q_{3} + 1.5 IQR$ .

Remember: A point exactly on a fence is not outside it. Investigate a flagged value before deciding what it means.

Standardising an observed value

z = \frac{x - \bar{x}}{s}

This sample-based score gives the number of sample standard deviations an observation lies above or below the sample mean.

Remember: This requires $s > 0$ . A positive score is above the mean; a negative score is below. Standardising alone does not make data normal.

Population mean and standard deviation

μ = \frac{\sum_{i = 1}^{N} x_{i}}{N}, σ = \sqrt{\frac{\sum_{i = 1}^{N} (x_{i} - μ)^{2}}{N}}

Use these definitions when the values for the entire population are available.

Remember: The population formula divides by $N$ ; the sample variance formula divides by $n - 1$ .

Worked example · Summarise four travel times

Four students have travel times of $5, 7, 9, 11$ minutes.

\bar{x} = \frac{5 + 7 + 9 + 11}{4} = 8

s = \sqrt{\frac{9 + 1 + 1 + 9}{3}} = \sqrt{\frac{20}{3}} \approx 2.58

The sample mean is $8$ minutes, and the sample standard deviation is about $2.58$ minutes. The median is also $8$ minutes. These summaries describe the students observed; whether they represent the school depends on how the sample was selected.

Design reminder: Random sampling supports generalising to the sampled population. Random assignment supports a cause-and-effect comparison of treatments in a well-designed experiment. A large convenience sample does not remove selection bias.

Study Unit 1 in detail

Unit 2

Probability, Random Variables, and Probability Distributions

Translate words into events before you calculate. Distinguish individual outcomes, counts of successes and statistics that vary across repeated samples.

Complement and addition rules

P (A^{c}) = 1 - P (A)

For “at least one”, it can be easier to subtract the probability of none. For “or”, use $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .

Remember: The “or” includes outcomes in both events. Only omit the overlap term when the events are mutually exclusive.

Conditional probability and multiplication

P (A ∣ B) = \frac{P (A \cap B)}{P (B)}

The condition narrows the denominator to outcomes in $B$ , with $P (B) > 0$ . The general multiplication rule is $P (A \cap B) = P (B) P (A ∣ B)$ .

Remember: For independent events, $P (A \cap B) = P (A) P (B)$ . Mutually exclusive events with positive probabilities are not independent.

Mean and variability of a discrete random variable

μ_{X} = \sum_{i} x_{i} P (X = x_{i})

The mean is a probability-weighted long-run average. Variance is $σ_{X}^{2} = \sum_{i} (x_{i} - μ_{X})^{2} P (X = x_{i})$ , and $σ_{X} = \sqrt{σ_{X}^{2}}$ .

Remember: A valid distribution has $P (X = x_{i}) \geq 0$ and $\sum_{i} P (X = x_{i}) = 1$ . The expected value need not be a possible individual outcome.

Binomial probability

P (X = k) = (\binom{n}{k}) p^{k} (1 - p)^{n - k}

Use for the number of successes in a fixed number of independent trials with two outcomes and a constant success probability. Here $k = 0, 1, \dots, n$ .

Remember: “Exactly” uses a single probability. “At most” sums through the cutoff; “at least” sums from the cutoff, or uses a complement.

Binomial mean and standard deviation

μ_{X} = n p, σ_{X} = \sqrt{n p (1 - p)}

The mean predicts the long-run average count of successes. The standard deviation describes the variability of that count.

Remember: Check the binomial conditions before using either formula. A count and a proportion have different scales.

Normal-model standardisation

z = \frac{x - μ}{σ}, x = μ + z σ

For a normal model, convert a value to a standard normal score to find an area, or convert a percentile score back to an original value.

Remember: For $X ~ N (μ, σ)$ here, the second parameter is the standard deviation. A continuous model gives $P (X = x) = 0$ at any single point.

The empirical rule

P (μ - σ < X < μ + σ) \approx 0.68

For an approximately normal distribution, the intervals $μ \pm 2 σ$ and $μ \pm 3 σ$ contain about $95 %$ and $99.7 %$ of observations.

Remember: The rule describes an approximately normal distribution. It does not apply to every data set.

Sampling variability: a mean

μ_{\bar{x}} = μ, σ_{\bar{x}} = \frac{σ}{\sqrt{n}}

For independent observations, these describe the centre and spread of the sampling distribution of the sample mean.

Remember: The central limit theorem concerns the distribution of sample means. Larger samples improve the normal approximation; severe skewness or outliers can require a larger sample.

Sampling variability: a proportion

μ_{\hat{p}} = p, σ_{\hat{p}} = \sqrt{\frac{p (1 - p)}{n}}

For independent Bernoulli observations, these describe repeated-sample proportions, rather than individual binary outcomes.

Remember: A normal approximation requires $n p \geq 10$ and $n (1 - p) \geq 10$ . For sampling without replacement, also check $n \leq 0.10 N$ .

Worked example · Exactly three successes

Suppose $X$ counts successes in $5$ independent trials, each with success probability $0.4$ . The binomial conditions hold.

P (X = 3) = (\binom{5}{3}) (0.4)^{3} (0.6)^{2} = 0.2304

The probability of exactly three successes is $23.04 %$ . This is different from the probability of at least three successes, which also includes $X = 4$ and $X = 5$ .

Standard deviation or standard error? A sampling-distribution standard deviation uses population parameters. A standard error estimates it with sample statistics. For a mean, $SE (\bar{x}) = s / \sqrt{n}$ ; for a proportion, $SE (\hat{p}) = \sqrt{\hat{p} (1 - \hat{p}) / n}$ .

Study Unit 2 in detail

Unit 3

Inference for Categorical Data: Proportions

Use proportions for a yes/no outcome and chi-square procedures for category counts. Confidence intervals estimate; significance tests compare the observed result with a null model.

One-proportion confidence interval

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

Estimate a population proportion. The margin of error is $ME = z^{*} SE (\hat{p})$ .

Remember: Check random sampling or an appropriate randomised design, independence, and observed counts $n \hat{p} \geq 10$ , $n (1 - \hat{p}) \geq 10$ .

One-proportion significance test

z = \frac{\hat{p} - p_{0}}{\sqrt{p_{0} (1 - p_{0}) / n}}

Test $H_{0} : p = p_{0}$ . Choose the tail from the alternative hypothesis before finding the p-value.

Remember: The test uses null counts $n p_{0} \geq 10$ and $n (1 - p_{0}) \geq 10$ , and null-based variability. Do not substitute the sample proportion into this denominator.

Difference of two proportions: interval

({\hat{p}}_{1} - {\hat{p}}_{2}) \pm z^{*} \sqrt{\frac{{\hat{p}}_{1} (1 - {\hat{p}}_{1})}{n_{1}} + \frac{{\hat{p}}_{2} (1 - {\hat{p}}_{2})}{n_{2}}}

Estimate $p_{1} - p_{2}$ for two independent groups. Keep the order of subtraction consistent with your parameter definition.

Remember: Check observed successes and failures in both groups are each at least $10$ . Use an unpooled standard error for the interval.

Difference of two proportions: test

z = \frac{{\hat{p}}_{1} - {\hat{p}}_{2}}{\sqrt{{\hat{p}}_{c} (1 - {\hat{p}}_{c}) (1 / n_{1} + 1 / n_{2})}}

For $H_{0} : p_{1} = p_{2}$ , estimate the common proportion with ${\hat{p}}_{c} = (x_{1} + x_{2}) / (n_{1} + n_{2})$ , where the numerators are success counts.

Remember: Check pooled expected successes and failures: $n_{i} {\hat{p}}_{c} \geq 10$ and $n_{i} (1 - {\hat{p}}_{c}) \geq 10$ for each group. Pool for this equality test, not for the interval.

Sampling distribution of a difference

μ_{{\hat{p}}_{1} - {\hat{p}}_{2}} = p_{1} - p_{2}

For independent samples, $σ_{{\hat{p}}_{1} - {\hat{p}}_{2}} = \sqrt{p_{1} (1 - p_{1}) / n_{1} + p_{2} (1 - p_{2}) / n_{2}}$ . Variances add even though the statistics are subtracted.

Remember: These are population-based quantities; estimated standard errors are used when the population proportions are unknown.

Chi-square statistic and goodness of fit

χ^{2} = \sum_{i} \frac{(O_{i} - E_{i})^{2}}{E_{i}}

For goodness of fit, expected counts are $E_{i} = n p_{i}$ , where $p_{i}$ is the null proportion for category $i$ . With $k$ specified categories, $df = k - 1$ .

Remember: Use counts, not percentages. Check a random sample, independence and every expected count greater than $5$ , following the current course framework. The p-value is a right-tail area.

Chi-square: two-way table expected counts

E_{i j} = \frac{(row total) (column total)}{grand total}

For a table with $R$ rows and $C$ columns, $df = (R - 1) (C - 1)$ . Sum chi-square contributions over all cells.

Remember: Independence studies association in one population. Homogeneity compares a categorical distribution across populations or treatments. Check design, independence and every expected count greater than $5$ .

Confidence critical values and sample-size planning

z^{*} \approx 1.645 (90 %), 1.960 (95 %), 2.576 (99 %)

For a target margin of error $m$ , plan a one-proportion sample with $n \geq (z^{*} / m)^{2} p^{*} (1 - p^{*})$ . Here $p^{*}$ is a planning estimate.

Remember: Use $p^{*} = 0.5$ if no estimate is available and round the required sample size up. Express the margin of error as a proportion, not a whole-number percentage.

Worked example · Estimate the proportion who prefer a study app

In a simple random sample of $200$ students from a school with at least $2000$ students, $120$ prefer a study app. The independence approximation is reasonable, and the observed success and failure counts, $120$ and $80$ , meet the large-count condition.

\hat{p} = \frac{120}{200} = 0.60

0.60 \pm 1.96 \sqrt{\frac{0.60 (0.40)}{200}} \approx (0.532, 0.668)

We are $95 %$ confident that between about $53.2 %$ and $66.8 %$ of students at this school prefer the app. The interval estimates a population proportion, not the range of individual responses.

Design and independence: State the random sample or random assignment. For samples taken without replacement, check $n_{i} \leq 0.10 N_{i}$ for each relevant population. This sampling fraction check is unnecessary for a randomised experiment. Two-group formulas also require independent groups.

Study Unit 3 in detail

Unit 4

Inference for Quantitative Data: Means

Use a mean procedure for a quantitative response. First decide whether the data form one sample, matched pairs, or two independent samples.

One-sample mean interval

\bar{x} \pm t^{*} \frac{s}{\sqrt{n}}, df = n - 1

Estimate a population mean when the population standard deviation is unknown. The standard error is $s / \sqrt{n}$ .

Remember: Use a random sample or appropriate randomised design. Check independence and the distribution or sample-size condition; use $t^{*}$ , not a fixed normal critical value.

One-sample mean test

t = \frac{\bar{x} - μ_{0}}{s / \sqrt{n}}, df = n - 1

Test $H_{0} : μ = μ_{0}$ against the specified alternative. The statistic measures how far the sample mean is from the null value in standard-error units.

Remember: Use a $t$ -distribution with the appropriate degrees of freedom to obtain the p-value. Choose one or two tails to match the alternative.

Paired differences: interval and test

\bar{d} \pm t^{*} \frac{s_{d}}{\sqrt{n}}

Define each difference consistently, then treat the differences as one sample. The test statistic is $t = (\bar{d} - μ_{d, 0}) / (s_{d} / \sqrt{n})$ , with $df = n - 1$ .

Remember: Here $n$ counts pairs. Check the distribution of differences, not the two original distributions separately.

Two independent means: interval

({\bar{x}}_{1} - {\bar{x}}_{2}) \pm t^{*} \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}

Estimate $μ_{1} - μ_{2}$ . The square root term is the standard error of the difference between independent sample means.

Remember: Check conditions for both groups. Use unpooled two-sample inference; do not assume equal population variances.

Two independent means: test

t = \frac{({\bar{x}}_{1} - {\bar{x}}_{2}) - Δ_{0}}{\sqrt{s_{1}^{2} / n_{1} + s_{2}^{2} / n_{2}}}

Here $Δ_{0}$ is the hypothesised difference $μ_{1} - μ_{2}$ , often $0$ . State that difference and the subtraction order before calculating.

Remember: Use technology for the Welch degrees of freedom. A conservative table-based choice is $min (n_{1} - 1, n_{2} - 1)$ ; do not use a paired procedure for unrelated groups.

Sampling distribution of a difference in means

μ_{{\bar{x}}_{1} - {\bar{x}}_{2}} = μ_{1} - μ_{2}

For independent samples, $σ_{{\bar{x}}_{1} - {\bar{x}}_{2}} = \sqrt{σ_{1}^{2} / n_{1} + σ_{2}^{2} / n_{2}}$ . Estimating the population standard deviations gives the two-sample standard error.

Remember: The population formula describes repeated samples. It is different from the spread of individual observations in either group.

Worked example · Estimate a population mean travel time

A simple random sample of $25$ students from a school with at least $250$ students has $\bar{x} = 18$ minutes and $s = 5$ minutes. Suppose the travel-time distribution is reasonably normal with no concerning outliers.

SE = \frac{5}{\sqrt{25}} = 1, df = 24

For a $95 %$ interval, $t^{*} \approx 2.064$ .

18 \pm 2.064 (1) = (15.936, 20.064)

We are $95 %$ confident that the school population mean travel time is between about $15.94$ and $20.06$ minutes. The sample standard deviation describes individual travel times; the standard error describes uncertainty in estimating the mean.

Conditions for mean procedures: Explain the random selection or assignment and independence. For sampling without replacement, check $n \leq 0.10 N$ . For a small sample, inspect the relevant distribution for skewness and outliers; an approximately normal population supports the procedure. A sample of at least $30$ observations is a common course guideline, but severe skewness or outliers still deserve attention. For independent groups, check each; for paired data, check differences.

Study Unit 4 in detail

Unit 5

Regression Analysis

Describe the scatterplot before using a line. A useful linear prediction needs a suitable relationship, sensible input values and residuals that do not show a clear curved pattern.

Predicted response and residual

\hat{y} = a + b x, e = y - \hat{y}

Use the explanatory variable $x$ to predict the response. The residual compares the observed response with its prediction.

Remember: A positive residual means the observation is above the line; a negative residual means it is below. Avoid extrapolating beyond the observed range without justification.

Slope and intercept relationships

b = r \frac{s_{y}}{s_{x}}, a = \bar{y} - b \bar{x}

These identities link the least-squares line to the sample summaries. Technology supplies the fitted coefficients. The line passes through $(\bar{x}, \bar{y})$ .

Remember: Interpret slope as the predicted change in the response for a one-unit increase in the explanatory variable. The intercept describes the prediction at $x = 0$ , which may not be meaningful.

Correlation and explained variation

- 1 \leq r \leq 1, 0 \leq r^{2} \leq 1

Correlation measures direction and strength of a linear association. In simple linear regression, $r^{2}$ is the proportion of variability in the response explained by its linear relationship with the explanatory variable.

Remember: A value near $0$ does not rule out a curved association. Neither correlation nor a large $r^{2}$ establishes causation.

What least squares minimises

SSE = \sum_{i = 1}^{n} (y_{i} - {\hat{y}}_{i})^{2}

The least-squares line minimises the sum of squared vertical residuals. Use a residual plot to check whether a straight-line description is reasonable.

Remember: Check unusual points, curvature and changing spread. A small residual for one observation does not validate the whole model.

Worked example · Predict and interpret a residual

A fitted model predicts a quiz score from weekly study hours:

\hat{y} = 60 + 4 x

For a student who studies $5$ hours and scores $77$ :

\hat{y} = 60 + 4 (5) = 80, e = 77 - 80 = - 3

The model predicts $80$ points. The student scored $3$ points below the prediction. The slope describes an increase of $4$ predicted score points per additional weekly study hour. An observational association alone does not show that extra study caused that increase.

Interpret explained variation carefully: If $r^{2} = 0.64$ , say that $64 %$ of the variability in quiz scores is explained by the linear relationship with study hours in this model. It does not mean $64 %$ of students are predicted correctly.

Study Unit 5 in detail

Finish with a statistical explanation

Confidence intervals

Name the population parameter, give the interval with units, and state the confidence level. The confidence level describes the long-run success rate of the interval-building method.

Avoid saying that the confidence interval contains a stated percentage of individual observations.

Significance tests

A p-value is the probability, assuming $H_{0}$ is true, of obtaining a test statistic as extreme or more extreme in the direction of $H_{a}$ .

If $p-value \leq α$ , reject $H_{0}$ . Otherwise, fail to reject it. State the evidence for the alternative in the context of the population claim.

Errors and power

A Type I error rejects a true $H_{0}$ . A Type II error fails to reject a false $H_{0}$ . Power is the chance of correctly rejecting $H_{0}$ for a particular true alternative.

power = 1 - β

Here $β$ is the Type II error probability for that alternative. Which error matters more depends on the consequences in context.

Precision and sample size

With other factors held fixed, more data usually reduce standard error. A higher confidence level uses a larger critical value and makes an interval wider.

SE \propto \frac{1}{\sqrt{n}}

For a fixed population variability, multiplying the sample size by $4$ halves the sampling standard deviation. More data do not fix systematic bias.

Common mistakes to catch before you submit

Wrong denominator: for conditional probability, divide by the total in the given group.
Wrong proportion: a one-proportion test uses $p_{0}$ in its standard deviation; an interval estimates variability with $\hat{p}$ .
Pooling everywhere: use the pooled proportion for a test of equal population proportions, and separate sample proportions for an interval.
Ignoring pairs: matched observations require a procedure on the differences.
Adding standard deviations: independent variances add; take the square root afterwards.
Using percentages in chi-square: use observed and expected counts.
Equating evidence with certainty: failing to reject $H_{0}$ does not prove it is true.
Reporting numbers without meaning: identify the population, response, subtraction order and units.

A quick check of your understanding

Try these before opening the answers.

Why does a one-proportion test use the null proportion?

The test asks how unusual the sample is under the null model. Its sampling variability is therefore calculated assuming $p = p_{0}$ , rather than assuming the observed proportion is the population value.

Two classes measure improvement before and after teaching. Which mean procedure fits?

If each student's before and after scores are matched, define a difference for each student and use paired inference on those differences. If the question compares two unrelated groups' improvement scores, use an independent two-sample procedure on those scores.

A residual is positive. Is the observed response above or below the prediction?

It is above the prediction because $e = y - \hat{y} > 0$ . The model underpredicted that observation.

Questions students ask

Do I need to memorise every formula?

College Board supplies formulas and tables for the exam. Practise locating the reference you need, identifying each symbol, checking conditions and explaining the result. Our guide adds teaching reminders and examples; consult the official Appendix to see the provided exam reference.

Why are there five units here instead of nine?

The course framework effective Fall 2026 groups content into five units. Our AP Statistics Study Hub follows that sequence. Older released questions and textbooks may have different numbering and additional content, so match resources by skill and course year.

What is the difference between standard deviation and standard error?

Standard deviation describes spread in a distribution. A standard error estimates the standard deviation of a statistic across repeated samples. For example, $s$ describes individual observations, while $s / \sqrt{n}$ estimates the sampling variability of their mean.

Can I use a normal critical value for a population mean?

In the mean-inference procedures in our current course notes, the population standard deviation is unknown. Estimate it with the sample standard deviation and use a $t$ -procedure with the appropriate degrees of freedom. A large sample makes the $t$ -distribution closer to the standard normal distribution; it does not change which procedure you should name.

Is a p-value the probability that the null hypothesis is true?

No. A p-value is calculated assuming $H_{0}$ is true. It describes how unusual the observed result or a more extreme result would be under that assumption, in the direction of the alternative.

Can I print this guide?

Yes. Use your browser's Print option. Expand any worked examples or answers you want included before printing; the print layout simplifies the formula cards.

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Course organisation and scope checked against the College Board AP Statistics Course and Exam Description, effective Fall 2026. Examples and explanations are original NUM8ERS teaching material. Updated 8 October 2026.

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