AP Statistics Mastery Roadmap

UNIT 1 · 1.9

Normal Distribution & z-score z

Every normal distribution question is really asking "how many standard deviations away is this?" Standardize with a z-score, then find the area (probability) with a table or calculator.

Standardization

z = (x − μ) / σ

z tells you how many standard deviations a value is from the mean. Positive z means above the mean; negative means below it.

Example: A Distribution of Test Scores

Population: μ = 100, σ = 15 (assumed normal)
Question: P(X < 115) = ? — the probability of scoring below 115

Compute z: z = (115 − 100) / 15 = 1.00
Look up the area to the left of z = 1.00 on the standard normal table → 0.8413
Interpretation: the probability of scoring below 115 is 84.13%.

P(X < 115) = 0.8413

TI-84 Plus CE2ND → VARS (DISTR)

2ND VARS → 2:normalcdf(
Enter: -1E99, 115, 100, 15) ENTER
→ 0.8413447

Tip: −1E99 stands for "negative infinity." Press 2ND, (-) to get the EE symbol, which becomes E.

Inverse: invNormFinding the top-10% cutoff

2ND VARS → 3:invNorm(
Enter: 0.90, 100, 15) ENTER
→ 119.22

Try it yourself — what is P(X < 130)? (μ=100, σ=15, round to 4 decimal places)

UNIT 2 · 2.5–2.8

Correlation r & the Least-Squares Regression Line

r describes the direction and strength of the linear relationship between two variables. The regression line is the best-fitting line for predicting y from x.

Least-Squares Regression Line

ŷ = a + b·x b = r · (s_y/s_x) residual = y − ŷ

Example: Study Hours (x) vs. Score (y)

x (hours)	2	4	6	8	10
y (score)	3	7	8	13	15

Correlation: r = 0.9848 → a very strong positive linear relationship
Regression line: ŷ = 0.20 + 1.50x
Coefficient of determination: r² = 0.9698 → 96.98% of the variation in y is explained by x
Predicted value at x=6: ŷ = 0.2 + 1.5(6) = 9.2; actual y=8 → residual = 8 − 9.2 = −1.2

ŷ = 0.20 + 1.50x, r² = 0.9698

TI-84 Plus CESTAT → EDIT → CALC

STAT → 1:Edit → enter x into L1, y into L2
STAT → CALC → 4:LinReg(ax+b)
Xlist:L1, Ylist:L2 ENTER
→ a=1.5, b=0.2, r=0.9848, r²=0.9698

If r doesn't show up: 2ND 0 (CATALOG) → DiagnosticOn → ENTER ENTER to turn the setting on.

Try it yourself — what is the residual at x=10? (using ŷ = 0.2+1.5x, round to 1 decimal place)

UNIT 4 · 4.10–4.11

Binomial Distribution binom

Use this when a fixed number of independent trials (n) each result in success (p) or failure.

Binomial PMF & Parameters

P(X=k) = C(n,k) · p^k · (1−p)^n−k μ = np σ = √(np(1−p))

Example: Flipping a Coin 10 Times

n = 10 (number of trials), p = 0.5 (probability of heads)
Question: what is the probability of exactly 6 heads?

Mean: μ = np = 10 × 0.5 = 5.0
Standard deviation: σ = √(10 × 0.5 × 0.5) = 1.5811
Compute P(X=6) → 0.2051
For reference — P(X≤6) (6 or fewer): 0.8281

P(X=6) = 0.2051

TI-84 Plus CE2ND → VARS (DISTR)

Exactly 6: 2ND VARS → A:binompdf(
10, 0.5, 6) ENTER → 0.2051

6 or fewer: 2ND VARS → B:binomcdf(
10, 0.5, 6) ENTER → 0.8281

Try it yourself — with n=10, p=0.5, what is P(X=5)? (round to 4 decimal places)

UNIT 4 · 4.12

Geometric Distribution

This deals with how many attempts it takes to get the first success. Unlike the binomial, n is not fixed in advance.

Geometric PMF & Mean

P(X=k) = (1−p)^k−1 · p μ = 1/p

Example: Repeated Shots With a 20% Success Rate

p = 0.2 (success probability per attempt)
Question: what is the probability that the first success happens on exactly the 3rd attempt?

Compute P(X=3): (0.8)² × 0.2 = 0.1280
For reference — P(X≤3), success within 3 attempts: 0.4880
Average number of attempts: μ = 1/0.2 = 5.0

P(X=3) = 0.1280

TI-84 Plus CE2ND → VARS (DISTR)

2ND VARS → D:geometpdf(
0.2, 3) ENTER → 0.1280

Try it yourself — with p=0.2, what is P(X=2)? (round to 4 decimal places)

UNIT 5 · 5.3–5.7

Sampling Distributions & the Central Limit Theorem

Collect a bunch of sample means (x̄) and they form their own distribution — a sampling distribution. When n is large enough, this distribution is approximately normal.

Sampling Distribution of the Mean

μ_x̄ = μ σ_x̄ = σ / √n (standard error, SE)

As n increases, σ_x̄ (the standard error) decreases — larger samples produce sample means that are less spread out.

Example: Drawing a Sample of 25

Population: μ = 50, σ = 10, sample size n = 25
Question: what is the probability the sample mean x̄ is greater than 53?

Standard error: SE = 10 / √25 = 2.0
Compute z: z = (53 − 50) / 2 = 1.5
P(x̄ > 53) → 0.0668

P(x̄ > 53) = 0.0668

TI-84 Plus CEReusing normalcdf

2ND VARS → normalcdf(
53, 1E99, 50, 2) ENTER
→ 0.0668

Key point: you plug σ_x̄ = SE into the σ slot — that's the main difference from single-value normal problems.

Try it yourself — with n=25, μ=50, σ=10, what is P(x̄ < 48)? (round to 4 decimal places)

UNIT 6 · 6.2

Confidence Interval for a Proportion z

Use the sample proportion p̂ to estimate the interval where the true population proportion p is likely to fall.

One-Proportion Confidence Interval

p̂ ± z* · √(p̂(1−p̂)/n)

Example: A Survey Response

n = 100 respondents, 60 in favor → p̂ = 0.60
Question: what is the 95% confidence interval?

Standard error: SE = √(0.6×0.4/100) = 0.0490
Critical value (95%): z* = 1.960
Margin of error: ME = 1.96 × 0.0490 = 0.0960
Interval: 0.60 ± 0.096 = (0.504, 0.696)

95% CI = (0.5040, 0.6960)

TI-84 Plus CESTAT → TESTS

STAT → TESTS → A:1-PropZInt
x=60, n=100, C-Level=0.95 Calculate
→ (0.504, 0.696)

Always check conditions first: np̂ ≥ 10 and n(1−p̂) ≥ 10 (here 60 and 40 — both pass).

Try it yourself — what is z* for a 90% confidence interval? (round to 3 decimal places)

UNIT 6 · 6.4–6.6

One-Proportion z-Test z

Tests whether a sample proportion is significantly different from a claimed population proportion p₀. This is where p-value and α first come into play.

One-Proportion z-Test Statistic

z = (p̂ − p₀) / √(p₀(1−p₀)/n)

Example: Testing "the approval rate is higher than 50%"

H₀: p = 0.5 vs. H_a: p > 0.5 (one-sided)
n = 100, p̂ = 0.60, α = 0.05

Standard error under H₀: SE₀ = √(0.5×0.5/100) = 0.0500
Test statistic: z = (0.60 − 0.50)/0.05 = 2.00
p-value (one-sided): P(Z > 2.00) = 0.0228
Conclusion: p-value (0.0228) < α (0.05) → reject H₀. There is statistically significant evidence the approval rate is above 50%.

z = 2.00, p-value = 0.0228 → Reject H₀

TI-84 Plus CESTAT → TESTS

STAT → TESTS → 5:1-PropZTest
p₀=0.5, x=60, n=100
prop >p₀ Calculate
→ z=2.00, p=0.0228

Try it yourself — under the same setup, what is the two-sided p-value? (round to 4 decimal places)

UNIT 7 · 7.2

Confidence Interval for a Mean (t-interval) t

When you don't know the population standard deviation σ and estimate it with s, you use the t-distribution instead of z.

One-Sample t Confidence Interval

x̄ ± t* · (s/√n) df = n − 1

Example: Test Scores From 16 Students

n = 16, x̄ = 50, s = 8 → what is the 95% confidence interval?

Degrees of freedom: df = 16 − 1 = 15
Standard error: SE = 8/√16 = 2.0
Critical value: t* (df=15, 95%) = 2.131
Margin of error: ME = 2.131 × 2.0 = 4.263
Interval: 50 ± 4.263 = (45.737, 54.263)

95% CI = (45.74, 54.26)

TI-84 Plus CESTAT → TESTS

STAT → TESTS → 8:TInterval
select Stats → x̄=50, Sx=8, n=16
C-Level=0.95 Calculate
→ (45.74, 54.26)

Try it yourself — at df=15, what is t* for a 90% confidence interval? (round to 3 decimal places)

UNIT 7 · 7.4–7.5

One-Sample t-Test for a Mean t

This is where α (significance level), β (Type Ⅱ error probability), and Type Ⅰ/Ⅱ errors really come into full use.

One-Sample t-Test Statistic

t = (x̄ − μ₀) / (s/√n) df = n − 1

Example: Testing "the mean score is higher than 50"

H₀: μ = 50 vs. H_a: μ > 50 (one-sided)
n = 16, x̄ = 52, s = 8, α = 0.05

Standard error: SE = 8/√16 = 2.0
Test statistic: t = (52 − 50)/2.0 = 1.00, df = 15
p-value (one-sided): P(T > 1.00) = 0.1666
Conclusion: p-value (0.1666) > α (0.05) → fail to reject H₀. There isn't enough evidence that the mean is above 50.

t = 1.00, p-value = 0.1666 → Fail to Reject H₀

TI-84 Plus CESTAT → TESTS

STAT → TESTS → 2:T-Test
Stats, μ₀=50, x̄=52, Sx=8, n=16
μ >μ₀ Calculate
→ t=1.00, p=0.1666

Try it yourself — under the same setup, what is the two-sided p-value? (round to 4 decimal places)

UNIT 8 · 8.2–8.3

Chi-Square Goodness of Fit Test χ²

Tests whether a single categorical variable follows a claimed distribution (set of proportions).

Chi-Square GOF Statistic

χ² = Σ [(O − E)² / E] df = (number of categories) − 1

Example: Testing Equal Proportions Across 5 Categories

Category	A	B	C	D	E
Observed O	18	22	20	20	20
Expected E	20	20	20	20	20

H₀: all 5 categories occur equally often (20% each) / n = 100

Compute each term: (18−20)²/20=0.2, (22−20)²/20=0.2, the rest are 0
Sum: χ² = 0.2+0.2+0+0+0 = 0.4000
Degrees of freedom: df = 5 − 1 = 4
p-value: 0.9825 → far larger than α=0.05 → fail to reject H₀. No evidence the distribution differs from equal proportions.

χ² = 0.400, df = 4, p-value = 0.9825

TI-84 Plus CESTAT → TESTS

Enter observed into L1, expected into L2 →
STAT → TESTS → D:Chi2GOF-Test
Observed:L1, Expected:L2, df=4 Calculate
→ χ²=0.400, p=0.9825

Try it yourself — for O=[18,22,20,20,20], what is the individual contribution (O−E)²/E for category "B"?

UNIT 8 · 8.5–8.6

Chi-Square Test for Independence χ²

Tests whether two categorical variables are related (or independent). Expected counts come from (row total × column total) / grand total.

Expected Count Formula

E = (row total × column total) / grand total

Example: Gender and Opinion

	Favor	Oppose	Total
Male	30	20	50
Female	15	35	50
Total	45	55	100

Example expected count: E(Male, Favor) = (50×45)/100 = 22.5 (same method for every cell)
Sum the chi-square: χ² = 9.0909
Degrees of freedom: df = (2−1)(2−1) = 1
p-value: 0.0026 → smaller than α=0.05 → reject H₀. Gender and opinion are not independent (they're related).

χ² = 9.091, df = 1, p-value = 0.0026

TI-84 Plus CE2ND MATRX → STAT TESTS

2ND MATRX → EDIT → [A] 2×2 enter observed counts
STAT → TESTS → C:Chi2-Test
Observed:[A], Expected:[B] Calculate
→ χ²=9.091, p=0.0026

Try it yourself — what is the expected count for E(Female, Oppose)? (row total × column total / grand total)

UNIT 9 · 9.4–9.5

t-Test for the Slope of a Regression Model t

Tests whether the true population slope (β) between x and y is nonzero — that is, whether x is actually a meaningful predictor of y.

t-Test for Slope

t = b / SE_b df = n − 2

Example: Nearly Perfect Linear Data

x	1	2	3	4	5	6
y	2.1	4.0	5.9	8.2	9.8	12.1

H₀: β = 0 (no relationship) vs. H_a: β ≠ 0

Regression equation: ŷ = 0.047 + 1.991x, r² = 0.9985
Standard error of the slope: SE_b = 0.0391
Test statistic: t = 1.9914/0.0391 = 50.92, df = 4
p-value: ≈ 0.000001 → extremely small → reject H₀. The slope is statistically significant.
95% confidence interval for the slope: (1.883, 2.100)

t = 50.92, p-value ≈ 0.0000 → Reject H₀

TI-84 Plus CESTAT → TESTS

Enter x into L1, y into L2 →
STAT → TESTS → E:LinRegTTest
Xlist:L1, Ylist:L2, β&ρ ≠0 Calculate
→ t=50.92, p=≈0, df=4

Try it yourself — using df = n − 2, what is df when n = 20?

MASTER SUMMARY

α · β · p-value · Type Ⅰ/Ⅱ Errors at a Glance

Scoring a 5 on AP Statistics comes down to never confusing these five concepts. Use the table below as a final check right before the exam.

Concept	Symbol	Definition	Analogy / Key Idea	If it increases...
Significance Level	α	The threshold, set before the test, for rejecting H₀. Usually 0.05.	"How unlikely does something have to be before I call it not random?"	Type Ⅰ error probability ↑, Type Ⅱ error probability ↓
p-value	p	Assuming H₀ is true, the probability of getting a result at least as extreme as the one observed.	A number that quantifies "how likely is it this was just chance?"	Weaker basis for rejecting H₀ (if p > α, you can't reject)
Type Ⅰ Error	α	Rejecting H₀ when it's actually true. A "false conviction."	Its probability equals exactly α	Setting α higher increases the risk of a Type Ⅰ error
Type Ⅱ Error	β	Failing to reject H₀ when it's actually false. "Letting the real culprit go."	β rises with small samples or small true effects	Increasing n lowers β (raises power)
Power	1 − β	The probability of correctly rejecting H₀ when it is actually false.	"The test's ability to detect a real effect when one exists"	Power rises with larger n, larger α, or a larger true effect size

Decision Rule — Never Forget This

p-value < α → Reject H₀ → significant evidence supporting Ha
p-value ≥ α → Fail to reject H₀ → insufficient evidence (this does NOT "prove" H₀ is true)
α is set before the test — never change α after looking at the data
Increasing sample size n reduces β (Type Ⅱ error) and raises power — but α stays the same

PRACTICE

12 Multiple-Choice Practice Questions

Based directly on the worked examples above. Pick an answer and the explanation appears instantly.

SCORE TRACKER 0 / 12

Proving alpha, beta, p-value with real numbersThe AP Statistics Mastery Roadmap

Normal Distribution & z-score z

Example: A Distribution of Test Scores

Correlation r & the Least-Squares Regression Line

Example: Study Hours (x) vs. Score (y)

Binomial Distribution binom

Example: Flipping a Coin 10 Times

Geometric Distribution

Example: Repeated Shots With a 20% Success Rate

Sampling Distributions & the Central Limit Theorem

Example: Drawing a Sample of 25

Confidence Interval for a Proportion z

Example: A Survey Response

One-Proportion z-Test z

Example: Testing "the approval rate is higher than 50%"

Confidence Interval for a Mean (t-interval) t

Example: Test Scores From 16 Students

One-Sample t-Test for a Mean t

Example: Testing "the mean score is higher than 50"

Chi-Square Goodness of Fit Test χ²

Example: Testing Equal Proportions Across 5 Categories

Chi-Square Test for Independence χ²

Example: Gender and Opinion

t-Test for the Slope of a Regression Model t

Example: Nearly Perfect Linear Data

α · β · p-value · Type Ⅰ/Ⅱ Errors at a Glance

Decision Rule — Never Forget This

12 Multiple-Choice Practice Questions

Proving alpha, beta, p-value with real numbers
The AP Statistics Mastery Roadmap