AP Statistics Formula Sheet: Examples & Free PDF

A reference for the revised 2026–27 course

This original learning reference follows the revised five-unit AP Statistics framework and the 13-page College Board reference booklet marked ©2026. It explains the main quantitative relationships with variables, conditions and independently written examples. It is not a reproduction of the official booklet, a complete course or an official exam simulation.

The official formula pages are physical pages 4–5 of that booklet. Those pages provide some formulas directly and others as general inference structures with standard errors. We label those distinctions rather than implying that every worked substitution is printed in the official reference. Probability tables and digital-entry notation remain in the official booklet; they are linked, not republished here.

Not included as current-course topics: geometric distributions, combining random variables, chi-square goodness-of-fit and inference for a regression slope. These were explicitly removed in the 2026–27 revision.
Regression description/prediction and chi-square tests of independence or homogeneity remain relevant; removal of slope inference or goodness-of-fit does not remove all regression or all chi-square work.
Sampling design, investigative questions, experiment design and interpretation require reasoning beyond any formula.
Independent qualified educator review is pending; linked catalog questions retain their published foundation scope.

Describe quantitative data and relative position

Distinguish a sample summary from a population parameter. A statistic describes the observations you have; the sampling design determines whether it supports a broader claim.

FORMULA & WORKED EXAMPLE

Sample mean

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

Variables

xi: the ith observed numerical value
n: positive number of observations
x-bar: sample mean

Worked example

Four original delivery times are 2, 5, 8 and 9 minutes. Find the mean.

Sum the times: 2+5+8+9=24 minutes.
Divide by n=4: 24/4=6 minutes.

Answer: 6 minutes

Lesson: summaries Practice 1 linked question

FORMULA & WORKED EXAMPLE

Sample standard deviation

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

Variables

xi: sample values
x-bar: sample mean
n>1: sample size
s: sample standard deviation in the original units

Worked example

Find the sample standard deviation of 1, 3 and 5 cm.

The mean is 3 cm. Squared deviations are 4, 0 and 4 cm².
Divide their sum 8 by n−1=2 to get sample variance 4 cm².
Take the square root: s=2 cm.

Answer: 2 cm

Lesson: summaries Lesson: spread normal Practice 2 linked questions

FORMULA & WORKED EXAMPLE

IQR and the 1.5-IQR outlier fences

\operatorname{IQR}=Q_3-Q_1,\quad L=Q_1-1.5\operatorname{IQR},\quad U=Q_3+1.5\operatorname{IQR}

Variables

Q1,Q3: first and third quartiles using the stated convention
L,U: lower and upper fences

Worked example

For a distribution with Q1=12 and Q3=20, find the fences and decide whether 35 is flagged.

IQR=20−12=8.
L=12−1.5(8)=0; U=20+1.5(8)=32.
35>32, so 35 is an upper flagged value.

Answer: Fences 0 and 32; 35 is flagged.

Lesson: summaries Lesson: distribution views Practice 1 linked question

FORMULA & WORKED EXAMPLE

Standardize a value or recover a normal-model cutoff

z=\frac{x-\mu}{\sigma},\qquad x=\mu+z\sigma

Variables

x: value or cutoff
mu: population/model mean
sigma>0: population/model SD
z: standardized value

Worked example

Under a stated normal model with mean 64 and SD 3, standardize 70 and recover the cutoff with z=−1.5.

For 70, z=(70−64)/3=2.
For z=−1.5, x=64+(−1.5)(3)=59.5.
Interpret 70 as 2 model SDs above the mean, not as a probability.

Answer: z=2; cutoff 59.5.

Lesson: spread normal Lesson: normal quantiles Practice 2 linked questions

Probability and random variables

Specify the chance process and the event before computing. Mutually exclusive events and independent events are different ideas: disjoint nonzero-probability events cannot be independent.

FORMULA & WORKED EXAMPLE

Complement and addition rules

P(A^c)=1-P(A),\qquad P(A\cup B)=P(A)+P(B)-P(A\cap B)

Variables

A,B: events
A^c: complement of A
union means at least one event occurs

Worked example

Suppose P(A)=0.35, P(B)=0.40 and P(A and B)=0.10. Find P(A or B) and P(not A).

P(A or B)=0.35+0.40−0.10=0.65.
P(not A)=1−0.35=0.65.

Answer: Both requested probabilities are 0.65.

Lesson: probability Practice 1 linked question

FORMULA & WORKED EXAMPLE

Conditional probability and the general multiplication rule

P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(A\cap B)=P(B)P(A\mid B)

Variables

A,B: events
P(B)>0: probability of the conditioning event

Worked example

If P(B)=0.30 and P(A and B)=0.18, find P(A given B) and check the product rule.

P(A given B)=0.18/0.30=0.60.
The product 0.30×0.60 returns P(A and B)=0.18.

Answer: P(A given B)=0.60.

Lesson: probability Lesson: two way association Practice 2 linked questions

FORMULA & WORKED EXAMPLE

Mean and SD of a discrete random variable

\mu_X=\sum_i x_i p_i,\qquad \sigma_X=\sqrt{\sum_i(x_i-\mu_X)^2p_i}

Variables

xi: possible numerical outcomes
pi=P(X=xi): nonnegative outcome probabilities summing to 1
mu_X: expected value
sigma_X: long-run SD

Worked example

An original random variable takes 0, 2 and 4 with probabilities 0.25, 0.50 and 0.25. Find its mean and SD.

mu=0(0.25)+2(0.50)+4(0.25)=2.
Variance=(0−2)^2(0.25)+(2−2)^2(0.50)+(4−2)^2(0.25)=2.
SD=sqrt(2), approximately 1.414.

Answer: Mean 2; SD sqrt(2), approximately 1.414.

Lesson: random variables Practice 2 linked questions

FORMULA & WORKED EXAMPLE

Binomial count probability

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

Variables

n: fixed number of trials
p: common success probability
k: integer success count from 0 through n
C(n,k): number of ways to choose the k successful trials

Worked example

Five independent trials each succeed with probability 0.4. Find the probability of exactly two successes.

The stated process satisfies the binomial conditions with n=5, p=0.4 and k=2.
C(5,2)=10, so P(X=2)=10(0.4)^2(0.6)^3.
Evaluate 10×0.16×0.216=0.3456.

Answer: 0.3456

Lesson: binomial conditions Lesson: random variables Practice 2 linked questions

FORMULA & WORKED EXAMPLE

Mean and SD of a binomial count

\mu_X=np,\qquad \sigma_X=\sqrt{np(1-p)}

Variables

n: number of independent trials
p: fixed success probability
X: number of successes

Worked example

For 50 independent trials with p=0.2, find the mean and SD of the success count.

Mean=50×0.2=10.
SD=sqrt(50×0.2×0.8)=sqrt(8), approximately 2.828.

Answer: Mean 10 successes; SD approximately 2.828 successes.

Lesson: binomial conditions Lesson: sampling distributions Practice 2 linked questions

Sampling distributions: distinguish SD from estimated SE

A population SD describes the repeated-sampling spread under specified population parameters. A standard error substitutes sample information to estimate that spread. Neither quantity is the spread of all individual measurements in one sample.

FORMULA & WORKED EXAMPLE

Spread of a sample proportion

\mu_{\hat p}=p,\quad \sigma_{\hat p}=\sqrt{\frac{p(1-p)}n},\quad \operatorname{SE}_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}n}

Variables

p: population success proportion
p-hat: sample success proportion
n: sample size

Worked example

For independent samples of size 100 with true p=0.4, find SD(p-hat). In one sample p-hat=0.45; find its estimated SE.

True-model SD=sqrt(0.4×0.6/100)=sqrt(0.0024), approximately 0.04899.
Estimated SE=sqrt(0.45×0.55/100)=sqrt(0.002475), approximately 0.04975.
The SD uses true p; the SE estimates it from p-hat.

Answer: SD approximately 0.04899; estimated SE approximately 0.04975.

Lesson: sampling distributions Practice 1 linked question

FORMULA & WORKED EXAMPLE

Spread of a sample mean

\mu_{\bar x}=\mu,\quad \sigma_{\bar x}=\frac{\sigma}{\sqrt n},\quad \operatorname{SE}_{\bar x}=\frac{s}{\sqrt n}

Variables

mu,sigma: population mean and SD
s: sample SD
n: sample size

Worked example

A population has SD 12 and independent samples of size 36 are drawn. In one such sample s=15. Find SD(x-bar) and its estimated SE.

The model SD is 12/sqrt(36)=12/6=2.
The sample-based SE is 15/sqrt(36)=15/6=2.5.

Answer: SD 2; estimated SE 2.5, in the measurement units.

Lesson: sampling distributions Lesson: inference conditions Practice 1 linked question

Confidence intervals and tests for proportions

Write the population parameter and alternative direction before using a calculator. An interval uses estimated uncertainty around the observed estimate; a test uses uncertainty under its null model. Pool proportions only when the null hypothesis warrants it.

FORMULA & WORKED EXAMPLE

Confidence-interval structure

\widehat\theta\pm c^*\operatorname{SE}(\widehat\theta)

Variables

theta-hat: estimate of the target parameter
c*: critical multiplier for the chosen method and confidence level
SE: appropriate standard-error estimate

Worked example

For an appropriate stated method, an estimate is 72, its SE is 3 and the supplied critical multiplier is 2. Find the interval.

Margin of error=2×3=6.
Endpoints are 72−6=66 and 72+6=78.
No confidence level is inferred from the supplied multiplier alone.

Answer: (66,78)

Lesson: intervals Practice 1 linked question

FORMULA & WORKED EXAMPLE

One-proportion z interval

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

Variables

p-hat=x/n: observed success proportion
n: sample size
z*: standard-normal critical value

Worked example

A random sample of 400 from a much larger population contains 240 successes. Build a 95% interval using z*=1.96.

p-hat=240/400=0.60; observed counts 240 and 160 exceed 10.
SE=sqrt(0.6×0.4/400)=0.0244949.
Margin=1.96×0.0244949≈0.0480100. Endpoints are approximately 0.55199 and 0.64801.

Answer: Approximately (0.5520,0.6480) for the population proportion.

Lesson: intervals Lesson: inference conditions Practice 2 linked questions

CHECK BEFORE YOU USE IT

Conditions & boundaries

Use data from a random sample or a suitably randomized experiment; the design limits the population/causal conclusions. For sampling without replacement, check n<=10% of the population to approximate independence.
For the usual normal approximation, check observed successes n*p-hat>=10 and observed failures n*(1-p-hat)>=10.
Use the unpooled sample proportion in the interval SE, not a hypothesized null p0.

Exam reference note

The reference booklet supplies the general confidence-interval/test-statistic structure and relevant standard errors on physical pages 4–5. The method-specific substitution below is explained here rather than printed as a standalone formula.

Reference: AP Statistics Exam Reference Information; AP Statistics Course and Exam Description

FORMULA & WORKED EXAMPLE

One-proportion z test statistic

z=\frac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}

Variables

p0: proportion specified by H0
p-hat: observed sample proportion
n: sample size

Worked example

In a random sample of 200 from a much larger population, 120 are successes. For H0:p=0.5, calculate z.

p-hat=120/200=0.60. Null expected successes and failures are both 100.
Null SE=sqrt(0.5×0.5/200)=sqrt(0.00125)≈0.0353553.
z=(0.60−0.50)/0.0353553≈2.8284.

Answer: z≈2.8284; select a p-value tail from the prespecified alternative.

Lesson: tests Lesson: one proportion conclusions Practice 2 linked questions

CHECK BEFORE YOU USE IT

Conditions & boundaries

Use data from a random sample or a suitably randomized experiment; the design limits the population/causal conclusions. For sampling without replacement, check n<=10% of the population to approximate independence.
Require 0<p0<1 and adequate null expected counts n*p0>=10 and n*(1-p0)>=10 for the usual normal approximation.
The null SE uses p0; the alternative determines the appropriate p-value tail.

Exam reference note

The reference booklet supplies the general confidence-interval/test-statistic structure and relevant standard errors on physical pages 4–5. The method-specific substitution below is explained here rather than printed as a standalone formula.

Reference: AP Statistics Exam Reference Information; AP Statistics Course and Exam Description

FORMULA & WORKED EXAMPLE

Plan a proportion interval sample size

n\geq\frac{(z^*)^2\widetilde p(1-\widetilde p)}{m^2}

Variables

m>0: desired margin of error
p-tilde: planning estimate of the population proportion
z*: critical value for the confidence level

Worked example

Plan a 95% proportion interval with margin at most 0.05 using z*=1.96 and the conservative planning value 0.5.

Compute n>=1.96²×0.5×0.5/0.05²=384.16.
Round upward to the next whole observation.

Answer: At least 385 observations under the planning approximation.

Lesson: intervals Lesson: inference conditions Practice 2 linked questions

FORMULA & WORKED EXAMPLE

Two independent 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}}

Variables

p-hat1,p-hat2: sample success proportions in the stated order
n1,n2: sample sizes
z*: normal critical value

Worked example

Two independent random samples from much larger populations each have size 200. Their success proportions are 0.60 and 0.45. Find a 95% interval for p1−p2 using z*=1.96.

Estimate=0.60−0.45=0.15. Success/failure counts are 120/80 and 90/110.
SE=sqrt(0.24/200+0.2475/200)=sqrt(0.0024375)≈0.0493710.
Margin≈1.96×0.0493710=0.0967672. Add and subtract from 0.15.

Answer: Approximately (0.0532,0.2468) for p1−p2.

Lesson: two group inference Lesson: group inference Practice 2 linked questions

FORMULA & WORKED EXAMPLE

Two independent proportions: pooled null test

\hat p_c=\frac{x_1+x_2}{n_1+n_2},\qquad z=\frac{\hat p_1-\hat p_2}{\sqrt{\hat p_c(1-\hat p_c)(1/n_1+1/n_2)}}

Variables

xi: success count in group i
ni: sample size in group i
p-hat_i=xi/ni
p-hat_c: estimated common null proportion

Worked example

Two independent random samples from large populations have 60 successes out of 100 and 40 out of 100. Test H0:p1=p2 by calculating the pooled proportion and z.

The pooled proportion is (60+40)/(100+100)=0.50; each null expected count is 50.
SE0=sqrt(0.5×0.5×(1/100+1/100))≈0.0707107.
z=(0.60−0.40)/0.0707107≈2.8284.

Answer: Pooled proportion 0.50; z≈2.8284.

Lesson: two group inference Lesson: tests Practice 1 linked question

Means: one sample, paired differences or independent groups

The observational unit chooses the method. Before-and-after measurements on the same person are paired, so analyze one difference per person; two unrelated samples use an independent two-sample method.

FORMULA & WORKED EXAMPLE

One-mean t interval and test statistic

\bar x\pm t^*_{n-1}\frac{s}{\sqrt n},\qquad t=\frac{\bar x-\mu_0}{s/\sqrt n}

Variables

x-bar,s: sample mean and sample SD
n>1: sample size
mu0: null mean
t*: critical value with n−1 degrees of freedom

Worked example

A random sample of 25 from a much larger, approximately normal population has mean 52 and sample SD 10. Use t*=2.064 for a 95% interval, and compute t for H0:mu=50.

SE=10/sqrt(25)=2; df=24.
Margin=2.064×2=4.128, so the interval is (47.872,56.128).
The test statistic is t=(52−50)/2=1.

Answer: 95% interval (47.872,56.128); test statistic t=1 with df=24.

Lesson: t procedures Lesson: intervals Practice 2 linked questions

CHECK BEFORE YOU USE IT

Conditions & boundaries

Use data from a random sample or a suitably randomized experiment; the design limits the population/causal conclusions. For sampling without replacement, check n<=10% of the population to approximate independence.
Population SD is unknown and s estimates it. Assess the normality/large-sample condition; small samples need especially careful attention to nonnormality and outliers.
Use the chosen confidence level and df to obtain t*; use the prespecified alternative for a test’s p-value.
The test statistic requires s>0. A zero estimated SE does not produce a defined t statistic; do not divide by zero.

Exam reference note

The reference booklet supplies the general confidence-interval/test-statistic structure and relevant standard errors on physical pages 4–5. The method-specific substitution below is explained here rather than printed as a standalone formula.

Reference: AP Statistics Exam Reference Information; AP Statistics Course and Exam Description

FORMULA & WORKED EXAMPLE

Paired observations: one t procedure on differences

d_i=\text{after}_i-\text{before}_i,\qquad t=\frac{\bar d-\mu_{d,0}}{s_d/\sqrt n}

Variables

di: within-pair difference in a stated direction
d-bar,sd: mean and sample SD of differences
n: independent pairs, not number of measurements
mu_d,0: null population mean difference

Worked example

For four independently sampled pairs, after-minus-before differences are 2, 4, 6 and 8. Assume the population of differences is approximately normal. Calculate t for H0:mu_d=0.

Mean difference=5. Squared deviations from 5 sum to 9+1+1+9=20.
Sample SD=sqrt(20/3)≈2.58199; SE=sqrt(20/3)/sqrt(4)≈1.29099.
t=(5−0)/1.29099≈3.87298, with df=3.

Answer: t≈3.873 with df=3.

Lesson: t procedures Practice 1 linked question

CHECK BEFORE YOU USE IT

Conditions & boundaries

Preserve each pair and use a consistent subtraction direction.
The pairs should be independently and suitably collected; assess normality/large-sample conditions for the distribution of differences, not the two raw measurement distributions separately.
Use the sample SD of differences; the two marginal SDs cannot simply be substituted.
The test statistic requires a positive sample SD of differences; it is undefined when that estimated SE is zero.

Exam reference note

The reference booklet supplies the general confidence-interval/test-statistic structure and relevant standard errors on physical pages 4–5. The method-specific substitution below is explained here rather than printed as a standalone formula.

Reference: AP Statistics Exam Reference Information; AP Statistics Course and Exam Description

FORMULA & WORKED EXAMPLE

Independent means: unpooled standard error

\operatorname{SE}=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}},\quad t=\frac{(\bar x_1-\bar x_2)-\Delta_0}{\operatorname{SE}},\quad (\bar x_1-\bar x_2)\pm t^*\operatorname{SE}

Variables

x-bar1,x-bar2: independent sample means
s1,s2: sample SDs
n1,n2: sample sizes
Delta0: hypothesized population mean difference

Worked example

Two independent random samples from much larger populations have (n1,mean1,s1)=(40,24,6) and (n2,mean2,s2)=(50,20,8), with suitable distribution conditions. Calculate SE and t for a null difference of 0.

Estimate=24−20=4.
SE=sqrt(6²/40+8²/50)=sqrt(0.9+1.28)=sqrt(2.18)≈1.47648.
t=4/1.47648≈2.70914; use the unequal-variance procedure’s degrees of freedom for a p-value.

Answer: SE≈1.4765; t≈2.7091.

Lesson: t procedures Lesson: group inference Practice 1 linked question

CHECK BEFORE YOU USE IT

Conditions & boundaries

Groups and observations must support the independent-samples model and suitable random collection/assignment.
For sampling without replacement check the 10% condition separately; assess shape/large-sample requirements in each group.
This is the unequal-variance (unpooled) method. Use technology’s appropriate degrees of freedom for its p-value or critical t; do not use n1+n2−2 without equal-variance justification.
The test statistic requires a positive combined SE; do not divide by zero if both sample variances are zero.

Exam reference note

The reference booklet supplies the general confidence-interval/test-statistic structure and relevant standard errors on physical pages 4–5. The method-specific substitution below is explained here rather than printed as a standalone formula.

Reference: AP Statistics Exam Reference Information; AP Statistics Course and Exam Description

Chi-square for independence or homogeneity

Use a two-way count table, not percentages as if they were counts. Independence investigates two categorical variables within a sample; homogeneity compares a categorical outcome across populations or treatment groups. The data collection, not just the table shape, distinguishes them.

FORMULA & WORKED EXAMPLE

Expected count under no association

E_{ij}=\frac{(\text{row }i\text{ total})(\text{column }j\text{ total})}{N}

Variables

Eij: expected count in row i and column j
N>0: grand total count

Worked example

In a 100-observation two-way table, a cell’s row total is 20 and column total is 30. Find its expected count under independence.

Multiply the marginal totals: 20×30=600.
Divide by the grand total: E=600/100=6.

Answer: Expected count 6.

Lesson: chi square investigation Lesson: two way association Practice 1 linked question

FORMULA & WORKED EXAMPLE

Chi-square statistic and table degrees of freedom

\chi^2=\sum_{i,j}\frac{(O_{ij}-E_{ij})^2}{E_{ij}},\qquad df=(R-1)(C-1)

Variables

Oij,Eij: observed and null-expected cell counts
R,C: table dimensions
df: degrees of freedom

Worked example

A suitable random sample of 100 gives a 2×2 table with rows [30,20] and [20,30]. Find the independence-test chi-square statistic and df.

Every row and column total is 50, so each expected count is 50×50/100=25. All are at least 5.
Each cell contributes (5²)/25=1; the four contributions sum to 4.
df=(2−1)(2−1)=1. A test decision additionally needs the right-tail p-value and stated significance level.

Answer: chi-square=4 with df=1.

Lesson: chi square investigation Lesson: tests Practice 2 linked questions

CHECK BEFORE YOU USE IT

Conditions & boundaries

Use suitably random, independent observations; each observation contributes to one cell. Apply the finite-population sampling fraction condition where appropriate.
For the usual chi-square approximation, every expected cell count should be at least 5; checking observed counts is not the same condition.
The df formula assumes an ordinary contingency table without structural zeros or extra fitted constraints. This resource does not teach the removed goodness-of-fit procedure.

Exam reference note

The chi-square statistic is printed on physical page 4. The contingency-table df relationship is a supplemental course relationship, not separately printed on formula pages 4–5.

Reference: AP Statistics Exam Reference Information; AP Statistics Course and Exam Description

Regression prediction, residuals and explained variation

The revised course retains regression analysis. A line summarizes association and supports predictions within an appropriate domain; it does not establish causation. The removed topic is inference for a population regression slope.

FORMULA & WORKED EXAMPLE

Fitted line and residual

\widehat y=a+bx,\qquad e=y-\widehat y

Variables

a: fitted y-intercept
b: fitted slope
x: explanatory value
y: observed response; y-hat: predicted response

Worked example

An original fitted model is y-hat=12+1.5x. At x=8, the observed response is 22. Find the prediction and residual.

Prediction=12+1.5(8)=24.
Residual=22−24=−2.
The observation is 2 response units below the model.

Answer: Prediction 24; residual −2.

Lesson: regression Lesson: regression diagnostics Practice 2 linked questions

FORMULA & WORKED EXAMPLE

Coefficient of determination in simple linear regression

R^2=r^2

Variables

r: sample correlation, between −1 and 1
R²: proportion of response variation explained by the fitted linear relationship

Worked example

For a suitable simple linear regression fit with r=−0.8, calculate r² and interpret it.

r²=(−0.8)^2=0.64.
About 64% of the variation in the response is explained by its fitted linear relationship with the explanatory variable.
The negative r indicates a decreasing association; 0.64 itself has no direction.

Answer: r²=0.64, or 64% explained variation in this model.

Lesson: regression Lesson: regression diagnostics Practice 1 linked question

Choose the method before the numbers

First state the parameter: one proportion, a difference of proportions, one mean, a mean paired difference, a difference of independent means, or a categorical association. Then check how data were collected and what independence/shape conditions the selected method needs.

For an interval, report endpoints and interpret the population parameter in context. For a test, state the null and alternative, use the correct tail, compare a p-value with the prespecified significance level, and describe the evidence without saying that a nonsignificant result proves the null. A small p-value does not measure the size or practical importance of an effect.

Question structure	Method family	Uncertainty to use
One binary outcome, one population	One-proportion z interval/test	Interval uses p-hat; test uses null p0
Two independent binary-outcome groups	Two-proportion z interval/test	Unpooled interval SE; pooled null SE only for equal-proportion test
One quantitative group	One-sample t	Sample SD divided by sqrt(n)
Before/after or matched units	One-sample t on paired differences	SD of differences, and n counts pairs
Two independent quantitative groups	Unequal-variance two-sample t	sqrt(s1²/n1+s2²/n2)
Two categorical variables or comparison of category distributions	Chi-square independence/homogeneity	Null expected cell counts, not observed counts

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Sources and review notes

The references establish the stated framework and source basis. Original examples and study tasks are independently authored; reference links do not imply endorsement or permission to reproduce official test items.

AP Statistics Exam Reference Information
Version / date: ©2026 College Board, 13-page reference booklet retrieved 2026-10-28
Locator: Physical page 4: descriptive statistics, probability/distributions, generic inference, chi-square statistic. Physical page 5: sampling-distribution SD/SE for proportions and means. Both formula pages visually checked.
AP Statistics Course and Exam Description
Version / date: Revised 2026–27 course framework; retrieved 2026-10-28
Locator: Units 1–5; physical pp45–46 summary statistics, pp93–112 proportion inference, pp115–118 categorical association, pp129–145 mean inference, pp158–159 regression.
AP Statistics Revisions
Version / date: Revisions effective 2026–27; current page retrieved 2026-10-28
Locator: Course Revisions and Exam Revisions; removed geometric distributions, combining random variables, goodness-of-fit and inference for slopes; five-unit framework.

Lesson: summaries Lesson: spread normal Lesson: distribution views Lesson: normal quantiles Lesson: probability Lesson: two way association Lesson: random variables Lesson: binomial conditions Lesson: sampling distributions Lesson: inference conditions Lesson: intervals Lesson: tests Lesson: one proportion conclusions Lesson: two group inference Lesson: group inference Lesson: t procedures Lesson: chi square investigation Lesson: regression Lesson: regression diagnostics

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A reference for the revised 2026–27 course

Describe quantitative data and relative position

Sample mean

Worked example

Sample standard deviation

Worked example

IQR and the 1.5-IQR outlier fences

Worked example

Standardize a value or recover a normal-model cutoff

Worked example

Probability and random variables

Complement and addition rules

Worked example

Conditional probability and the general multiplication rule

Worked example

Mean and SD of a discrete random variable

Worked example

Binomial count probability

Worked example

Mean and SD of a binomial count

Worked example

Sampling distributions: distinguish SD from estimated SE

Spread of a sample proportion

Worked example

Spread of a sample mean

Worked example

Confidence intervals and tests for proportions

Confidence-interval structure

Worked example

One-proportion z interval

Worked example

One-proportion z test statistic

Worked example

Plan a proportion interval sample size

Worked example

Two independent proportions: interval

Worked example

Two independent proportions: pooled null test

Worked example

Means: one sample, paired differences or independent groups

One-mean t interval and test statistic

Worked example

Paired observations: one t procedure on differences

Worked example

Independent means: unpooled standard error

Worked example

Chi-square for independence or homogeneity

Expected count under no association

Worked example

Chi-square statistic and table degrees of freedom

Worked example

Regression prediction, residuals and explained variation

Fitted line and residual

Worked example

Coefficient of determination in simple linear regression

Worked example

Choose the method before the numbers

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Sources and review notes