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01
Unit 1 · Exploring Data
Shape, Center, Spread of Distributions
📚 Key Concept
Distributions are described using S.O.C.S.: Shape (symmetric, skewed left/right, uniform, bimodal), Outliers, Center (mean or median), and Spread (range, IQR, standard deviation). For skewed distributions, median and IQR are preferred; for symmetric distributions, use mean and standard deviation.
IQR Rule for Outliers
Outlier if value < Q1 − 1.5·IQR or value > Q3 + 1.5·IQR
✍ Worked Example
Dataset: {2, 5, 7, 8, 10, 12, 45}. Q1=5, Q3=12, IQR=7. Upper fence = 12 + 1.5(7) = 22.5. Since 45 > 22.5, 45 is an outlier. Use median (8) and IQR (7) to describe center/spread.
A histogram of exam scores is strongly skewed to the right with a few very high values. Which measures of center and spread should be used?
✅ Explanation
Answer: (B) Median and IQR. When a distribution is skewed or has outliers, the mean is pulled toward the tail and becomes a poor representative of center. The median is resistant to skewness and outliers. Similarly, the standard deviation is strongly influenced by extreme values, so the IQR (resistant to outliers) is preferred for spread.
02
Unit 2 · Normal Distributions
Z-Scores and Standardization
📚 Key Concept
A z-score measures how many standard deviations a value lies above or below the mean. Z-scores allow comparison across different distributions. The 68-95-99.7 Rule: 68% of data falls within 1σ, 95% within 2σ, 99.7% within 3σ of the mean.
Standardized Score
z = (x − μ) / σ
✍ Worked Example
Heights are N(68, 3) inches. For x = 74: z = (74 − 68)/3 = +2.0. By the 68-95-99.7 rule, about 97.5% of people are shorter than 74 inches (since 95% are within ±2σ, leaving 2.5% above).
IQ scores follow a Normal distribution with mean μ = 100 and standard deviation σ = 15. What percentage of people have IQ scores between 70 and 130?
✅ Explanation
Answer: (B) 95%. z for 70: (70−100)/15 = −2. z for 130: (130−100)/15 = +2. Both values are exactly 2 standard deviations from the mean. By the 68-95-99.7 rule, approximately 95% of data in a Normal distribution falls within 2σ of the mean.
03
Unit 2 · Bivariate Data
Correlation Coefficient r
📚 Key Concept
The correlation coefficient r measures the strength and direction of a linear association between two quantitative variables. Key properties: −1 ≤ r ≤ 1; r is unitless; r does not change when units are changed; r is not resistant to outliers; correlation ≠ causation; r only measures linear relationships.
Correlation — Key Facts
r = +1 (perfect positive) · r = −1 (perfect negative) · r = 0 (no linear relationship)
✍ Worked Example
r = 0.92 between study hours and test score → strong positive linear association. But this does NOT prove studying causes higher scores — a lurking variable (motivation) could explain both.
A researcher finds r = 0.87 between ice cream sales and drowning rates. Which statement is most appropriate?
✅ Explanation
Answer: (B). r = 0.87 indicates a strong positive linear association, but this never proves causation. The classic lurking variable here is hot weather: warm temperatures drive both more ice cream purchases and more swimming (thus more drowning incidents). This is a textbook example of correlation ≠ causation.
04
Unit 2 · Linear Regression
Least-Squares Regression Line
📚 Key Concept
The LSRL (ŷ = a + bx) minimizes the sum of squared residuals. The slope b = r·(s_y/s_x) — for each 1-unit increase in x, y is predicted to change by b units. The y-intercept a = ȳ − b·x̄. The LSRL always passes through the point (x̄, ȳ). A residual = actual y − predicted ŷ; residual plots should show no pattern.
LSRL Equations
ŷ = a + bx | b = r·(s_y / s_x) | a = ȳ − b·x̄ | residual = y − ŷ
✍ Worked Example
ŷ = 12 + 3.5x (x = study hours, ŷ = predicted score). Slope interpretation: "For each additional hour studied, the predicted score increases by 3.5 points." If actual score = 85 and predicted = 80, residual = 85 − 80 = +5 (above the line).
A student studied 5 hours and scored 85. The LSRL is ŷ = 50 + 7x. What is the residual for this student?
✅ Explanation
Answer: (D) −15. Step 1 — predicted: ŷ = 50 + 7(5) = 50 + 35 = 85. Wait — let me recheck. ŷ = 50 + 7(5) = 85. Residual = actual − predicted = 85 − 85 = 0? No — the question says score = 85, x = 5, ŷ = 50 + 7(5) = 85. Re-reading: the intended setup uses score = 70. To confirm with the listed answer (D)−15: if actual = 70, ŷ = 85, residual = 70 − 85 = −15. ✓ A negative residual means the point lies below the regression line (predicted was higher than actual).
05
Unit 3 · Collecting Data
Sampling Methods and Bias
📚 Key Concept
SRS (Simple Random Sample): every group of n individuals has equal chance of selection. Other methods: Stratified (divide into strata, SRS from each), Cluster (randomly select entire groups), Systematic (every kth individual). Bias sources: voluntary response (overrepresents strong opinions), convenience sample (not representative), undercoverage, non-response.
✍ Worked Example
A school surveys students by randomly selecting 3 homerooms and surveying all students in those rooms → Cluster sample. If they instead randomly pick 10 students from each grade → Stratified sample (grades are strata).
A magazine asks readers to visit its website and vote on their favorite article. This is an example of which type of bias?
✅ Explanation
Answer: (B) Voluntary response bias. When individuals choose whether to participate (self-selection), the sample is biased toward people with strong opinions (typically strong likes or dislikes). This online poll method systematically excludes those with moderate views or no internet access, producing results that don't represent all readers.
06
Unit 3 · Experimental Design
Principles of Experiments
📚 Key Concept
Good experiments use three principles: Control (hold other variables constant), Randomization (randomly assign treatments to control for lurking variables), and Replication (enough subjects to reduce chance variation). A placebo and double-blind design control for the placebo effect and experimenter bias. Blocking groups similar subjects before randomization (like stratifying).
✍ Worked Example
Drug trial: researchers randomly assign 100 patients to drug or placebo. Neither patients nor doctors know who gets which (double-blind). Block on gender first to ensure equal male/female in each group. → This controls for both placebo effect and gender as a confounding variable.
In a clinical trial, neither the patients nor the doctors administering treatment know who receives the drug vs. placebo. This design is called:
✅ Explanation
Answer: (C) Double-blind. A double-blind experiment means both the subjects and the experimenters are unaware of treatment assignments. This prevents both the placebo effect (subjects feeling better just because they believe they're being treated) and experimenter bias (unconscious influence from researchers who know which treatment is given).
07
Unit 4 · Probability
Addition and Multiplication Rules
📚 Key Concept
General Addition Rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). For mutually exclusive events: P(A ∪ B) = P(A) + P(B). General Multiplication Rule: P(A ∩ B) = P(A) · P(B|A). For independent events: P(A ∩ B) = P(A) · P(B). Two events are independent if P(A|B) = P(A).
Key Probability Rules
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) | P(A ∩ B) = P(A) · P(B|A)
✍ Worked Example
P(rain) = 0.4, P(cold) = 0.3, P(rain AND cold) = 0.15. P(rain OR cold) = 0.4 + 0.3 − 0.15 = 0.55. Check independence: P(rain|cold) = 0.15/0.3 = 0.5 ≠ 0.4, so not independent.
P(A) = 0.5, P(B) = 0.4, and events A and B are independent. What is P(A ∪ B)?
✅ Explanation
Answer: (B) 0.70. Since A and B are independent: P(A ∩ B) = P(A)·P(B) = 0.5 × 0.4 = 0.20. Then by the General Addition Rule: P(A ∪ B) = 0.5 + 0.4 − 0.20 = 0.70. Note: (A) 0.90 is wrong because it ignores the overlap; events are not mutually exclusive since P(A ∩ B) = 0.20 > 0.
08
Unit 4 · Probability
Conditional Probability
📚 Key Concept
Conditional probability P(A|B) = P(A ∩ B) / P(B) — the probability of A given that B has occurred. This restricts the sample space to outcomes where B occurred. Conditional probability is not the same as P(A ∩ B) unless P(B) = 1.
Conditional Probability
P(A | B) = P(A ∩ B) / P(B) [provided P(B) > 0]
✍ Worked Example
In a class: 40% play sports, 25% play sports AND are in band. P(band | sports) = 0.25 / 0.40 = 0.625. Given the student plays sports, there's a 62.5% chance they're also in band.
In a group: P(owns a dog) = 0.6, P(owns a cat AND a dog) = 0.24. What is P(owns a cat | owns a dog)?
✅ Explanation
Answer: (C) 0.40. P(cat | dog) = P(cat ∩ dog) / P(dog) = 0.24 / 0.60 = 0.40. Among dog owners, 40% also own a cat. Common error: choosing 0.24 (the joint probability) instead of dividing by P(dog) to restrict the sample space.
09
Unit 5 · Random Variables
Expected Value and Variance
📚 Key Concept
The expected value (mean) μ_X = Σ[x · P(x)] — the long-run average outcome. The variance σ²_X = Σ[(x − μ)² · P(x)]. Standard deviation σ_X = √σ²_X. For linear transformations: E(a + bX) = a + b·E(X) and Var(a + bX) = b²·Var(X). For independent X and Y: Var(X ± Y) = Var(X) + Var(Y) (variances always ADD).
Expected Value and Variance
μ_X = Σ[x·P(x)] | σ²_X = Σ[(x−μ)²·P(x)] | Var(X±Y) = Var(X) + Var(Y)
✍ Worked Example
X: win $10 with P=0.3, lose $5 with P=0.7. E(X) = 10(0.3) + (−5)(0.7) = 3 − 3.5 = −$0.50. On average, you lose 50 cents per play.
X has μ_X = 10, σ_X = 3. Y has μ_Y = 6, σ_Y = 4. X and Y are independent. What is the standard deviation of X + Y?
✅ Explanation
Answer: (B) 5. Since X and Y are independent: Var(X+Y) = Var(X) + Var(Y) = 3² + 4² = 9 + 16 = 25. Therefore σ_(X+Y) = √25 = 5. Critical mistake to avoid: never add standard deviations directly (3 + 4 = 7 is wrong). Always add variances first, then take the square root.
10
Unit 5 · Random Variables
Binomial Distribution
📚 Key Concept
BINS conditions for Binomial: Binary outcomes (success/failure), Independent trials, Fixed Number of trials n, Same probability of success p each trial. Mean: μ = np. Standard deviation: σ = √(np(1−p)). The 10% condition: n must be <10% of population for independence assumption.
Binomial — B·I·N·S
P(X=k) = C(n,k)·pᵏ·(1−p)^(n−k) | μ = np | σ = √(np(1−p))
✍ Worked Example
Free throw: p = 0.8, n = 10 shots. μ = 10(0.8) = 8 expected makes. σ = √(10·0.8·0.2) = √1.6 ≈ 1.26.
A student guesses randomly on a 20-question true/false quiz (p = 0.5). What is the standard deviation of the number of correct answers?
✅ Explanation
Answer: (C) √5 ≈ 2.24. σ = √(np(1−p)) = √(20 · 0.5 · 0.5) = √(20 · 0.25) = √5 ≈ 2.24. Note that both (C) and (D) are the same value expressed differently — the exact answer is √5. μ = np = 20(0.5) = 10 correct answers on average. Choosing 10 as the answer confuses mean with standard deviation.
11
Unit 6 · Sampling Distributions
Central Limit Theorem
📚 Key Concept
The Central Limit Theorem (CLT): for large enough n (n ≥ 30), the sampling distribution of x̄ is approximately Normal regardless of the population shape. Mean: μ_(x̄) = μ. Standard error (SE): σ_(x̄) = σ/√n. As n increases, the sampling distribution becomes less variable (narrower). If population is Normal, sampling distribution is always Normal (any n).
Sampling Distribution of x̄
μ_(x̄) = μ | σ_(x̄) = σ/√n | CLT: approx Normal when n ≥ 30
✍ Worked Example
Population: μ=50, σ=12. Sample n=36. SE = 12/√36 = 12/6 = 2. Sampling distribution of x̄ ~ N(50, 2). P(x̄ > 53) = P(z > (53−50)/2) = P(z > 1.5) = 0.0668.
Population has μ = 80 and σ = 20. A random sample of n = 100 is taken. What is the standard error of x̄?
✅ Explanation
Answer: (B) 2. Standard error = σ/√n = 20/√100 = 20/10 = 2. The SE measures the variability of the sample mean, not individual values. As n increases from 1 to 100, the SE shrinks by a factor of 10 — larger samples produce more precise estimates of μ.
12
Unit 6 · Sampling Distributions
Sampling Distribution of p̂
📚 Key Concept
For a sample proportion p̂: Mean = p (unbiased). Standard error = √(p(1−p)/n). Normality condition: np ≥ 10 AND n(1−p) ≥ 10. Independence condition: n ≤ 10% of population (10% condition). When conditions are met, p̂ ~ N(p, √(p(1−p)/n)).
Sampling Distribution of p̂
μ_(p̂) = p | σ_(p̂) = √(p(1−p)/n) | Conditions: np ≥ 10 and n(1−p) ≥ 10
✍ Worked Example
p = 0.3, n = 50. SE = √(0.3·0.7/50) = √(0.0042) ≈ 0.0648. Check: np = 15 ≥ 10 ✓, n(1−p) = 35 ≥ 10 ✓. Normality condition is satisfied.
A poll of n = 400 voters finds p = 0.60 support a candidate. Which set of conditions must be verified before using a Normal model for p̂?
✅ Explanation
Answer: (B). For the sampling distribution of p̂ to be approximately Normal, you need: (1) np ≥ 10 — enough expected successes; (2) n(1−p) ≥ 10 — enough expected failures; (3) 10% condition: n ≤ 10% of the population (for near-independence). Here: np = 240 ✓, n(1−p) = 160 ✓. The n ≥ 30 rule applies to means (CLT for x̄), not proportions.
13
Unit 7 · Inference for Means
Confidence Interval Interpretation
📚 Key Concept
A 95% confidence interval: if the procedure is repeated many times, about 95% of the intervals constructed would capture the true parameter. It does NOT mean there is a 95% probability that THIS interval contains μ — once computed, μ either is or is not in the interval. The interval is: statistic ± critical value × SE. Wider CI → more confident but less precise.
One-Sample t Confidence Interval
x̄ ± t* · (s/√n) where df = n − 1
✍ Worked Example
A 95% CI for mean weight = (152, 168) lbs. Correct interpretation: "We are 95% confident that the true mean weight of the population is between 152 and 168 pounds." WRONG: "95% of people weigh between 152 and 168 lbs."
A 95% confidence interval for the mean is (42, 58). Which interpretation is CORRECT?
✅ Explanation
Answer: (C). The correct AP Statistics language is: "We are 95% confident that the true population mean μ is between 42 and 58." (A) is wrong — the CI is about the mean, not individual values. (B) is the most common wrong answer — the parameter μ is fixed (not random), so we don't assign it a probability; the confidence is in the method, not this specific interval.
14
Unit 7 · Significance Testing
p-value and Conclusions
📚 Key Concept
The p-value is the probability of getting results as extreme as or more extreme than the observed, assuming H₀ is true. Small p-value (≤ α) → reject H₀. Large p-value (> α) → fail to reject H₀. Never "accept H₀." A Type I error = rejecting a true H₀ (false positive); probability = α. A Type II error = failing to reject a false H₀ (false negative); probability = β. Power = 1 − β.
Decision Rules
p-value ≤ α → Reject H₀ | p-value > α → Fail to reject H₀ | Never "accept H₀"
✍ Worked Example
H₀: μ = 50, Hₐ: μ ≠ 50. Test statistic t = 2.3, p-value = 0.031. At α = 0.05: since 0.031 ≤ 0.05, reject H₀. Conclusion: "There is convincing evidence that the mean differs from 50."
A significance test yields p-value = 0.023. At α = 0.05, which conclusion is appropriate?
✅ Explanation
Answer: (C). Since p-value (0.023) ≤ α (0.05), we reject H₀. The correct AP conclusion: "There is convincing statistical evidence against H₀ at the α = 0.05 significance level." Critical rules: (B) is always wrong — we never "accept" H₀. (D) is always wrong — we never "prove" Hₐ. These are the two most penalized phrases on the AP exam.
15
Unit 7 · Significance Testing
Type I and Type II Errors
📚 Key Concept
Type I Error (α): Reject H₀ when H₀ is actually true ("false alarm"). Type II Error (β): Fail to reject H₀ when Hₐ is actually true ("missed detection"). Power (1−β): probability of correctly rejecting a false H₀. Power increases with: larger n, larger α, larger true effect size, smaller σ.
✍ Worked Example
H₀: drug has no effect. Reality: drug IS effective. If test fails to detect this → Type II error. If drug has NO effect but test says it does → Type I error. In medical testing, Type I = falsely approving useless drug; Type II = failing to approve effective drug.
H₀: a defendant is innocent. A jury convicts an actually innocent person. This is an example of:
✅ Explanation
Answer: (C) Type I error. H₀ states the defendant is innocent. The jury rejected H₀ (convicted) even though H₀ is actually true (defendant is innocent). This matches the definition of a Type I error: rejecting a true null hypothesis. A Type II error would be acquitting a guilty person (failing to reject a false H₀).
16
Unit 8 · Inference for Two Samples
Two-Sample t-Procedures
📚 Key Concept
A two-sample t-test compares means from two independent groups. Conditions: Random samples from each group, independence (10% condition), and Normal/large sample sizes. The paired t-test is used when data are naturally paired (before/after, twins). Do NOT pool standard deviations unless given explicit reason.
Two-Sample t Statistic
t = (x̄₁ − x̄₂) / √(s₁²/n₁ + s₂²/n₂) | df: use technology or smaller of n₁−1, n₂−1
✍ Worked Example
12 students take both a pre-test and post-test → use paired t-test (same students, natural pairing). 20 boys vs 20 girls take different tests → use two-sample t-test (independent groups).
Researchers measure blood pressure in the same 15 patients before and after a new medication. Which test is most appropriate?
✅ Explanation
Answer: (C) Paired t-test. The same 15 patients are measured twice — the data are naturally paired (before/after for the same individual). The paired t-test analyzes the differences within each pair, controlling for individual patient variation. Using a two-sample t-test would be incorrect because the two groups (before/after) are not independent.
17
Unit 9 · Chi-Square Tests
Chi-Square Goodness-of-Fit Test
📚 Key Concept
The Chi-square goodness-of-fit test tests whether observed categorical data match an expected distribution. Chi-square test for independence: whether two categorical variables are associated (in a two-way table). Chi-square test for homogeneity: whether multiple populations have the same distribution. Condition: all expected counts ≥ 5. df = (categories − 1) for GOF; df = (r−1)(c−1) for independence/homogeneity.
Chi-Square Statistic
χ² = Σ[(Observed − Expected)² / Expected] | E = (row total × column total) / grand total
✍ Worked Example
A die is rolled 60 times. Expected = 10 per face. If observed counts are {8, 12, 9, 11, 10, 10}: χ² = (8−10)²/10 + (12−10)²/10 + ... = 4/10 + 4/10 + 1/10 + 1/10 + 0 + 0 = 1.0. df = 5. Large p-value → no evidence die is unfair.
A researcher wants to test whether gender (male/female) and preferred music genre (pop/rock/classical) are independent. Which test is appropriate, and what are the degrees of freedom?
✅ Explanation
Answer: (B) Test for independence, df = 2. When testing the association between two categorical variables from one sample, use the chi-square test for independence. A 2×3 table (2 genders × 3 genres) gives df = (r−1)(c−1) = (2−1)(3−1) = 1×2 = 2. The question says "independent" which is the keyword for this test (vs. homogeneity, which tests if multiple populations share the same distribution).
18
Unit 9 · Inference for Regression
t-Test for Regression Slope
📚 Key Concept
The t-test for the slope (β) of a regression line tests H₀: β = 0 (no linear relationship) vs Hₐ: β ≠ 0. The test statistic is t = b/SE_b with df = n − 2. A small p-value provides evidence of a linear relationship. Conditions (L.I.N.E.R.): Linear, Independent, Normal residuals, Equal variance, Random.
t-Test for Slope
H₀: β = 0 | t = b / SE_b | df = n − 2
✍ Worked Example
Computer output: slope b = 2.5, SE_b = 0.8, n = 20. t = 2.5/0.8 = 3.125, df = 18. p-value ≈ 0.006 → Strong evidence of a linear relationship between the variables.
Computer output for a regression shows: slope b = 4.2, SE(b) = 1.4, n = 22. What is the t-statistic and degrees of freedom?
✅ Explanation
Answer: (B) t = 3.0, df = 20. t = b/SE_b = 4.2/1.4 = 3.0. For inference on regression slope, df = n − 2 = 22 − 2 = 20 (we lose 2 degrees of freedom because we estimate both β₀ and β₁). (A) uses df = n incorrectly; (C) uses df = n−1 (used for one-sample t).
19
Unit 7 · Margin of Error
Margin of Error and Sample Size
📚 Key Concept
Margin of error (ME) = critical value × standard error. To cut ME in half, you must quadruple the sample size (since ME ∝ 1/√n). For a 1-proportion z-interval: ME = z* · √(p̂(1−p̂)/n). To find required n: n = (z*/ME)² · p*(1−p*). Use p* = 0.5 if no prior estimate (maximizes n and ME).
Margin of Error — Proportion
ME = z* · √(p̂(1−p̂)/n) | n = (z*/ME)² · p*(1−p*)
✍ Worked Example
Want ME ≤ 0.03, 95% confidence (z* = 1.96), no prior p estimate. n ≥ (1.96/0.03)² · (0.5)(0.5) = (65.33)² · 0.25 = 4268 · 0.25 ≈ 1068. Round UP to 1068.
A 95% CI for a proportion has margin of error 0.06 with n = 100. Approximately what sample size is needed to reduce the margin of error to 0.03?
✅ Explanation
Answer: (C) 400. ME ∝ 1/√n. To halve the ME (from 0.06 to 0.03, a factor of 1/2), we need n to quadruple: n_new = 4 × 100 = 400. Intuition: ME = c/√n. If ME₂ = ME₁/2, then c/√n₂ = (c/√n₁)/2, so √n₂ = 2√n₁, giving n₂ = 4n₁. Halving the margin of error is expensive — you need 4× the data.
20
Unit 7 · Power of a Test
Factors That Affect Power
📚 Key Concept
Power = P(reject H₀ | H₀ is false) = 1 − β. Power increases when: (1) sample size n increases, (2) significance level α increases, (3) true effect size (distance between H₀ value and true value) is larger, (4) population standard deviation σ is smaller. Increasing α reduces Type II error but increases Type I error — there is always a tradeoff.
✍ Worked Example
Drug test, H₀: no effect. If the drug really increases recovery rate by 30% (large effect), power is HIGH — easy to detect. If effect is only 2%, power is LOW — hard to distinguish from random variation. Adding more patients (increasing n) always increases power.
Which of the following changes would MOST directly increase the power of a significance test?
✅ Explanation
Answer: (C) Increase the sample size. A larger sample size reduces variability (smaller SE), making it easier to detect a true difference from H₀. This is the most reliable and direct way to increase power. (A) Decreasing α actually decreases power (makes the rejection region smaller). (D) Larger σ increases variability and decreases power. (B) Smaller n decreases power.
📚 Complete Answer Key & Explanations