Overview
Unit Title Exam % Key Topics U1 Exploring Data 15–23% Categorical & quantitative variables, distributions, summary stats, normal model U2 Sampling & Experimentation 12–15% Sampling methods, experimental design, bias, confounding U3 Probability & Distributions 30–40% (U3+U4+U5) Probability rules, geometric & binomial distributions, sampling distributions U4 Statistical Inference 25–35% Confidence intervals, significance tests for means & proportions, chi-square, regression
AP Statistics introduces students to the major concepts and tools for collecting, analysing, and drawing conclusions from data. The course follows four interconnected themes: Exploring Data, Sampling and Experimentation, Probability and Distributions, and Statistical Inference.
Exam format: Section I — 40 MCQ (90 min). Section II — 6 FRQ including one investigative task (90 min).
Exploring Data
📌 AP Statistics Unit 1 — Exploring Data 15–23% of exam · Distributions of categorical & quantitative data · Summary statistics · Normal model · Bivariate data
Exploratory data analysis uses graphical and numerical summaries to describe the distribution of one variable and relationships between two variables. Always describe shape, centre, spread, and outliers (S-C-S-O).
\[\bar{x}=\frac{\sum x_i}{n}\quad(\text{mean}),\qquad\text{Median }M=\text{middle value of ordered data}.\] The mean is non-resistant (pulled toward outliers); the median is resistant. Use median when data are skewed.
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Fixed points: 12, 15, 18, 21, 24 — same set as Worked Example 1.1. Drag the 6th point and watch the mean chase it while the median barely moves.
\[s=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n-1}}\quad(\text{sample SD}),\qquad IQR=Q_3-Q_1.\] The standard deviation \(s\) measures average distance from the mean. The IQR is resistant. Use IQR (and 5-number summary) with the median; use \(s\) (and \(\bar{x}\)) with symmetric data. 1.5×IQR rule for outliers: a value is an outlier if it is below \(Q_1-1.5\cdot IQR\) or above \(Q_3+1.5\cdot IQR.\)
If \(y_i = a + bx_i\), then: Shape does not change; measures of centre shift by \(a+b\cdot(\text{old})\); spread is multiplied by \(|b|\) (unaffected by \(a\)).
Feature What to report Shape Symmetric, skewed left/right, uniform, bimodal Centre Mean or median (state which) Spread \(s\) or IQR (state which) Outliers Individual values; note their direction
For data \((x_i,y_i)\), the least-squares regression line (LSRL) is \(\hat{y}=a+bx\) with The correlation coefficient \(r\) (\(-1\le r\le1\)) measures strength and direction of linear association. \(r^2\) (coefficient of determination) gives the proportion of variation in \(y\) explained by the linear relationship with \(x.\) Residual: \(e=y-\hat{y}.\) A residual plot with no pattern confirms a linear model is appropriate.
Problem. Ages: 12, 15, 18, 21, 24, 100. Find \(\bar{x}\), \(M\), IQR, and flag outliers.
\(\bar{x}=(12+15+18+21+24+100)/6=190/6\approx31.7.\) \(M=(18+21)/2=19.5.\)
Ordered: 12, 15, 18, 21, 24, 100. \(Q_1=15,\;Q_3=24,\;IQR=9.\)
Upper fence: \(24+1.5(9)=37.5.\) The value 100 exceeds 37.5 → outlier. Mean pulled to 31.7; median stays at 19.5 — shows median's resistance.
Problem. Scores in Celsius are shifted: \(F=32+1.8C\). If \(\bar{C}=25\) and \(s_C=5\), find \(\bar{F}\) and \(s_F\).
\(\bar{F}=32+1.8(25)=32+45=77°\text{F}.\quad s_F=|1.8|(5)=9°\text{F}.\)
Problem. A regression of price on size gives \(\hat{y}=15000+80x\) (\(r^2=0.81\)). Interpret the slope, \(y\)-intercept, and \(r^2\).
Slope: For each additional square foot, the predicted price increases by $80.
Intercept: A house with 0 sq ft is predicted to cost $15,000 — not meaningful (extrapolation).
\(r^2=0.81\): 81% of the variation in price is explained by the linear relationship with size.
The following data represents test scores (out of 100) for 12 students:
55, 62, 68, 70, 71, 74, 74, 77, 80, 83, 90, 96.
Find the five-number summary and IQR. Identify any potential outliers using the 1.5×IQR rule.
Min=55, Q1=69, Median=75.5, Q3=81.5, Max=96. IQR=12.5.
Fences: lower \(=69-18.75=50.25\), upper \(=81.5+18.75=100.25\). No outliers.
SAT scores are approximately \(N(1060,195^2)\). What proportion of students score above 1400?
\[z=\frac{1400-1060}{195}=\frac{340}{195}\approx1.74.\]\(P(Z>1.74)=1-0.9591=0.0409\approx4.1\%\) of students score above 1400.
Sampling & Experimentation
📌 AP Statistics Unit 2 — Sampling & Experimentation 12–15% of exam · Sampling methods & bias · Experimental design · Randomisation & confounding
Good statistical conclusions require well-designed data collection. Observational studies describe associations; experiments can establish cause and effect. Randomisation is the key tool for eliminating bias.
Method Procedure Key point Simple Random (SRS) Every group of \(n\) equally likely Unbiased, gold standard Stratified Random SRS within each stratum Reduces variability; must cover all strata Cluster Randomly select whole clusters Cheaper; cluster must mirror population Systematic Every \(k\)-th from random start Easy; avoid periodic patterns Convenience Whoever is handy Usually biased — avoid Voluntary response People choose to respond Strong-opinion bias — avoid
Randomise — assign subjects to treatments randomly to eliminate confounding. Replicate — use enough subjects so real effects can be detected. Control — use a control group (placebo or standard treatment); hold all other variables constant. Block — group similar subjects into blocks before randomising; reduces unexplained variability (like stratified sampling for experiments).
Placebo effect — improvement due to belief in treatment. Use blinding (subjects unaware of treatment) or double-blinding (neither subjects nor researchers know). Confounding variable — associated with both the explanatory and response variables; may produce a spurious association. Randomisation controls for known and unknown confounders. Observational study vs. experiment — only experiments with random assignment can establish causation.
Random sample? Random assignment? Generalise? Causation? Yes Yes Population ✓ Yes ✓ Yes No Population ✓ No ✗ No Yes Sample only ✗ Yes ✓ No No Sample only ✗ No ✗
Problem. A clinical trial randomly assigns 200 patients to Drug A or placebo. Neither patients nor doctors know who gets which. Classify the design and identify its features.
This is a completely randomised, double-blind experiment.
Blinding: patients don't know → controls placebo effect. Double-blind: doctors don't know → prevents researcher bias in assessment.
Since subjects were randomly assigned, a statistically significant difference in outcomes can be attributed to the drug (causation). However, without a random sample from the population, generalisation to all patients is limited.
Problem. A stratified random sample selects 10 students from each year level (9–12) at a school. Total year-level sizes: 120, 110, 100, 90. Is this sample proportionate?
Proportionate stratified: each stratum contributes proportionally. Total students \(=420\). Sample size \(=40\).
Proportionate sizes: Yr 9: \(10/120\approx8.3\%\); Yr 10: \(10/110\approx9.1\%\); etc. Since each stratum contributes 10 (not proportional), this is disproportionate stratified sampling. Each stratum has equal representation but not in proportion to population size.
Problem. An online poll on a news site asks visitors "Do you support the new tax?" 78% say no. Comment on the reliability.
This is a voluntary response sample — readers choose to respond. People who feel strongly (especially those who oppose the tax) are more likely to participate. The 78% figure cannot be generalised to the general population. Results are biased.
A school wants to know students' opinions on a new homework policy. The principal surveys only students who volunteer to respond. Identify the sampling method and any bias.
This is voluntary response sampling. Students with strong opinions (likely opposed to more homework) are more likely to respond, creating response bias and making results unrepresentative of all students.
Researchers want to test whether a new fertilizer increases crop yield. Describe a well-designed experiment.
Randomised comparative experiment: (1) Use 40 identical plots of land. (2) Randomly assign 20 to treatment (new fertilizer) and 20 to control (standard fertilizer). (3) Measure yield at season end. (4) Compare means. Random assignment controls for confounding variables.
Probability & Distributions
📌 AP Statistics Unit 3 — Probability & Distributions 30–40% of exam · Probability rules · Geometric & Binomial · Normal distribution · Sampling distributions
Probability measures the long-run relative frequency of an event. It underpins all statistical inference by describing how variable sample statistics are relative to the true population parameter.
\[0\le P(A)\le1,\quad P(\Omega)=1,\quad P(A^c)=1-P(A),\]\[P(A\cup B)=P(A)+P(B)-P(A\cap B),\quad P(A\mid B)=\frac{P(A\cap B)}{P(B)}.\] Independent events: \(P(A\cap B)=P(A)\cdot P(B)\) (equivalently, \(P(A\mid B)=P(A)\)). Mutually exclusive events: \(P(A\cap B)=0\Rightarrow P(A\cup B)=P(A)+P(B).\)
For a discrete RV \(X\) with probability distribution \(P(X=x_i)=p_i\): Linear combinations: \(\mu_{aX+b}=a\mu_X+b;\quad\sigma_{aX+b}^2=a^2\sigma_X^2.\) For independent \(X,Y\): \(\mu_{X+Y}=\mu_X+\mu_Y;\quad\sigma_{X+Y}^2=\sigma_X^2+\sigma_Y^2.\)
\[P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\quad\mu=np,\quad\sigma=\sqrt{np(1-p)}.\] CONDITIONS (BINS): Binary outcomes only; Independent trials; fixed Number \(n\) of trials; constant Success probability \(p\). Normal approximation: valid when \(np\ge10\) and \(n(1-p)\ge10.\)
\(X\sim N(\mu,\sigma)\). Standardise: \(z=\dfrac{x-\mu}{\sigma}.\)Rule Interval % of data 68% \(\mu\pm\sigma\) 68.27% 95% \(\mu\pm2\sigma\) 95.45% 99.7% \(\mu\pm3\sigma\) 99.73%
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Same standardisation as the worked examples: \(z=(x-\mu)/\sigma\). The shaded area is the proportion below \(z\).
\[\mu_{\bar{x}}=\mu,\quad\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\quad(\text{standard error}).\] CLT: For \(n\ge30\) (or smaller if population is Normal), \(\bar{x}\) is approximately \(N\!\left(\mu,\dfrac{\sigma}{\sqrt{n}}\right)\) regardless of population shape. Sampling distribution of \(\hat{p}\): \(\mu_{\hat{p}}=p,\;\sigma_{\hat{p}}=\sqrt{p(1-p)/n}\); approx Normal when \(np\ge10\) and \(n(1-p)\ge10.\)
Problem. A bag has 3 red, 5 blue balls. Two drawn without replacement. Find \(P(\text{both red})\).
\(P(R_1)=\dfrac{3}{8}.\quad P(R_2\mid R_1)=\dfrac{2}{7}.\quad P(\text{both red})=\dfrac{3}{8}\cdot\dfrac{2}{7}=\dfrac{6}{56}=\dfrac{3}{28}\approx0.107.\)
Problem. \(X\sim B(15,0.4)\). Find \(\mu,\;\sigma,\) and \(P(X\le2)\).
\(\mu=15(0.4)=6.\quad\sigma=\sqrt{15(0.4)(0.6)}=\sqrt{3.6}\approx1.897.\)
\(P(X\le2)=P(X=0)+P(X=1)+P(X=2)\)
\(=\binom{15}{0}(0.4)^0(0.6)^{15}+\binom{15}{1}(0.4)^1(0.6)^{14}+\binom{15}{2}(0.4)^2(0.6)^{13}\)
\(\approx0.00047+0.00470+0.02194\approx0.0271.\)
Problem. Heights are \(N(170, 8)\) cm. (a) Find \(P(X>182)\). (b) Find the 90th percentile.
(a) \(z=(182-170)/8=1.5.\quad P(X>182)=P(Z>1.5)=1-0.9332=0.0668.\)
(b) 90th percentile: \(z_{0.90}\approx1.282.\quad x=170+1.282(8)=170+10.26\approx180.3\text{ cm}.\)
Problem. A population has \(\mu=65,\;\sigma=10.\) Samples of \(n=25\) are taken. Find \(P(\bar{x}<62)\).
\(\sigma_{\bar{x}}=10/\sqrt{25}=2.\quad z=(62-65)/2=-1.5.\quad P(\bar{x}<62)=P(Z<-1.5)=0.0668.\)
A multiple-choice test has 10 questions, each with 4 options. A student guesses randomly. Find the probability of getting (a) exactly 4 correct, (b) at least 3 correct.
\(X\sim B(10,0.25)\).
(a) \(P(X=4)=\binom{10}{4}(0.25)^4(0.75)^6=210\times0.00391\times0.1779\approx0.1460\).
(b) \(P(X\geq3)=1-P(X\leq2)=1-[P(0)+P(1)+P(2)]\approx1-0.5256=0.4744\).
Bags of coffee have weight \(\mu=340\) g and \(\sigma=8\) g. A store buys cases of 25 bags. What is the probability the mean weight \(\bar{x}\) of a case is less than 337 g?
By CLT: \(\bar{X}\sim N\!\left(340,\dfrac{64}{25}\right)=N(340,2.56^2)\).
\[P(\bar{X}<337)=P\!\left(Z<\frac{337-340}{1.6}\right)=P(Z<-1.875)\approx0.030.\]Statistical Inference
📌 AP Statistics Unit 4 — Statistical Inference 25–35% of exam · Confidence intervals · Significance tests · Chi-square tests · Inference for regression
Inference uses sample data to draw conclusions about populations. The two main tools are confidence intervals (estimating a parameter) and hypothesis tests (assessing evidence against a claim).
\[\text{estimate} \pm \underbrace{(\text{critical value})\times(\text{standard error})}_{\text{margin of error}}\] Interpretation: "We are C% confident the true [parameter] lies between [L, U]." Do NOT say "probability C% the parameter is in this interval" — the parameter is fixed.Parameter Interval Conditions One mean \(\mu\) \(\bar{x}\pm t^*\dfrac{s}{\sqrt{n}}\), \(df=n-1\) Random; Normal/large \(n\) One proportion \(p\) \(\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\) Random; \(n\hat{p}\ge10,\;n(1-\hat{p})\ge10\) Two means \(\mu_1-\mu_2\) \((\bar{x}_1-\bar{x}_2)\pm t^*\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}\) Random; Normal/large \(n\) Two proportions \(p_1-p_2\) \((\hat{p}_1-\hat{p}_2)\pm z^*\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}\) Random; all four counts \(\ge10\)
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Same \(\bar{x}=74,\,s=12\) as Worked Example 4.1, using \(z^*\approx1.96\) as a large-\(n\) stand-in for \(t^*\). Drag \(n\) up and watch the interval shrink.
1. State: \(H_0\) (null, equality) and \(H_a\) (alternative: \(\neq,\;>,\;<\)), significance level \(\alpha.\) 2. Plan: Choose test; verify conditions. 3. Do: Compute test statistic and \(p\)-value. 4. Conclude: If \(p<\alpha\): reject \(H_0\), "there is significant evidence that \(H_a\) is true." If \(p\ge\alpha\): fail to reject \(H_0\).Test Statistic \(df\) One-sample \(t\) (mean) \(t=\dfrac{\bar{x}-\mu_0}{s/\sqrt{n}}\) \(n-1\) One-proportion \(z\) \(z=\dfrac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}\) — Two-sample \(t\) \(t=\dfrac{(\bar{x}_1-\bar{x}_2)-0}{\sqrt{s_1^2/n_1+s_2^2/n_2}}\) use calculator \(df\) Two-proportion \(z\) \(z=\dfrac{(\hat{p}_1-\hat{p}_2)-0}{\sqrt{\hat{p}_c(1-\hat{p}_c)(1/n_1+1/n_2)}}\) —
Power increases when: \(\alpha\) increases, sample size \(n\) increases, or the true parameter is farther from \(H_0.\)Row label \(H_0\) true \(H_0\) false Reject \(H_0\) Type I error (prob \(=\alpha\)) Correct ✓ (Power \(=1-\beta\)) Fail to reject Correct ✓ Type II error (prob \(=\beta\))
\[\chi^2=\sum\frac{(O-E)^2}{E}.\] Goodness-of-fit: \(df=k-1\) (k categories); tests observed vs. claimed distribution. Independence/Homogeneity: \(df=(r-1)(c-1)\); \(E_{ij}=\dfrac{(\text{row }i\text{ total})(\text{col }j\text{ total})}{\text{grand total}}.\) Conditions: random; expected counts \(\ge5\) in each cell.
\[H_0:\beta_1=0\quad vs.\quad H_a:\beta_1\neq0,\qquad t=\frac{b_1}{SE_{b_1}},\quad df=n-2.\] Conditions: Linear (scatter plot + residual plot); Independent errors; Normal residuals; Equal variance (constant spread in residuals); Random sample.
Problem. A sample of \(n=25\) has \(\bar{x}=74,\;s=12.\) Construct a 95% CI for \(\mu\). (Use \(t^*_{24}=2.064.\))
Check conditions: random sample; assume Normal population or \(n\) large enough. ✓
\(ME=2.064\cdot\dfrac{12}{\sqrt{25}}=2.064\cdot2.4=4.95.\)
\(CI=(74-4.95,\;74+4.95)=(69.05,\;78.95).\)
Interpret: We are 95% confident the true population mean lies between 69.05 and 78.95.
Problem. A factory claims mean filling \(\mu=500\)ml. A sample of \(n=36\) gives \(\bar{x}=497,\;s=9\). Test at \(\alpha=0.05.\)
\(H_0:\mu=500;\quad H_a:\mu<500.\)
One-sample \(t\)-test, \(df=35.\) \(t=\dfrac{497-500}{9/\sqrt{36}}=\dfrac{-3}{1.5}=-2.0.\)
\(p\text{-value}=P(T_{35}<-2.0)\approx0.027.<\alpha=0.05.\)
Conclude: Reject \(H_0.\) There is statistically significant evidence that the mean filling is less than 500 ml.
Problem. A die is rolled 120 times. Test goodness-of-fit at \(\alpha=0.05\) given observed: \(O=[15,18,25,22,20,20]\).
Expected each face: \(E=120/6=20.\)
\(\chi^2=\dfrac{(15-20)^2}{20}+\dfrac{(18-20)^2}{20}+\dfrac{(25-20)^2}{20}+\dfrac{(22-20)^2}{20}+\dfrac{(20-20)^2}{20}+\dfrac{(20-20)^2}{20}\)
\(=\dfrac{25+4+25+4+0+0}{20}=\dfrac{58}{20}=2.9.\quad df=5.\quad p>0.25.\)
Conclude: Fail to reject \(H_0.\) No significant evidence the die is unfair.
A company claims its batteries last 100 hours. A sample of 16 batteries has \(\bar{x}=97\) hours and \(s=6\) hours. Test at \(\alpha=0.05\) whether there is evidence the batteries last less than claimed.
Conditions: random sample, \(n=16<30\) so assume population is approximately normal.
\(H_0:\mu=100\), \(H_1:\mu<100\). Test stat: \(t=\dfrac{97-100}{6/\sqrt{16}}=\dfrac{-3}{1.5}=-2\), df=15.
\(p\)-value \(=P(t_{15}<-2)\approx0.032<0.05\). Reject \(H_0\). There is sufficient evidence the mean battery life is less than 100 hours.
In a poll of 500 voters, 265 support a ballot measure. Construct a 95% CI for the true proportion.
\(\hat p=265/500=0.53\). SE \(=\sqrt{\dfrac{0.53\times0.47}{500}}=\sqrt{0.000498}\approx0.0223\).
\[0.53\pm1.96\times0.0223=(0.486,\;0.574).\]We are 95% confident the true proportion in favour is between 48.6% and 57.4%.
Practice Set
A data set has mean \(\bar{x}=72\) and standard deviation \(s=8\). What \(z\)-score corresponds to \(x=88\)?
\(z=(88-72)/8=2.0.\) The value 88 is exactly 2 standard deviations above the mean.
Interpret \(r=0.85\) in context and state \(r^2\).
Strong positive linear association. \(r^2=0.7225.\) About 72.25% of the variability in \(y\) is explained by the linear relationship with \(x.\)
\(X\sim B(20,0.3).\) Find \(\mu,\sigma,\) and \(P(X=6).\)
\(\mu=6,\;\sigma\approx2.05,\;P(X=6)\approx0.1916.\)
A 95% CI is \((47.2,\,52.8)\). Interpret it and state the margin of error.
We are 95% confident the true mean lies between 47.2 and 52.8. MOE \(=(52.8-47.2)/2=2.8.\) Note: the mean is fixed — the interval is the random quantity.
\(H_0:\mu=100\) vs \(H_a:\mu>100.\) \(p=0.03,\;\alpha=0.05.\) Conclude.
Since \(p<\alpha\), reject \(H_0.\) Statistically significant evidence that \(\mu>100.\)
Design a completely randomised experiment to test whether a new tutoring programme improves exam scores, using 60 volunteer students. Identify one potential confounding variable and explain how randomisation addresses it.
Number the 60 students 1–60. Use a random number generator to assign 30 to treatment (tutoring) and 30 to control (no tutoring). Compare mean exam scores after the programme.
Confounder: prior academic ability. Randomisation distributes prior ability roughly equally between groups on average, so the groups are comparable at baseline. The scope of inference is limited to volunteer students — results may not generalise to all students.
Events \(A\) and \(B\) are independent with \(P(A)=0.4\) and \(P(B)=0.5\). Find \(P(A\cup B)\).
\(P(A\cap B)=0.4\times0.5=0.2.\quad P(A\cup B)=0.4+0.5-0.2=0.70.\)
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AP Statistics full exam — Section I: 40 MCQ (90 min) + Section II: 6 FRQ incl. investigative task (90 min), matching College Board format.
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