Simple Numbers · Real TI-84 Keystrokes · Self-Check

Don't memorize the formulas — compute them yourself.

Every unit below pairs a deliberately simple dataset with a hand-calculation walkthrough and the exact TI-84 button sequence. Check that your calculator's answer matches the hand calculation, and these procedures will never trip you up on test day.

Z / T distributions Binomial distribution Confidence intervals Hypothesis tests · p-value Type I / II Error Chi-square
01

Normal Distribution & Z-scores

Unit 1.9 / 5.2 — When data follow a normal distribution, standardize a value's position to turn it into a probability.

Core Formula Concept

z = (x − μ) / σ

z tells you "how many standard deviations away from the mean" a value is. Once you know z, the standard normal table (or your calculator) gives you the cumulative probability directly.

Work It With Simple Numbers Example

Scenario
Scores on a math exam are approximately normal with mean μ = 70 and standard deviation σ = 10.
Student A scored 85. What proportion of students scored lower than A?
  1. Compute z: z = (85 − 70) / 10 = 1.5
  2. Look up z = 1.50 in the standard normal table → cumulative probability ≈ 0.9332
  3. Conclusion: about 93.3% of students scored lower than A.
Using normalcdf
2nd → VARS (DISTR) 2 : normalcdf( lower: -1E99 upper: 85 μ: 70 σ: 10 ) ENTER → 0.9331927987

💡 For "negative infinity," type -1E99 on the calculator: (-) → 1 → EE (2nd, comma key) → 99.

Working Backwards: Probability → Score Inverse

Scenario
What score is needed to be in the top 10%? (μ=70, σ=10)
  1. Top 10% = the point below which 90% of the data falls → invNorm(0.90)
  2. From the table, cumulative probability 0.90 corresponds to z ≈ 1.28
  3. x = μ + zσ = 70 + 1.28(10) = 82.8 points
Using invNorm
2nd → VARS (DISTR) 3 : invNorm( area: 0.90 μ: 70 σ: 10 ) ENTER → 82.815... points
02

Binomial Distribution

Unit 4.10–4.11 — When each independent trial has only two outcomes, "success" or "failure," repeated n times.

Conditions & Formulas Concept

Conditions (BINS): Binary · Independent · Number fixed (n) · Same probability p P(X = k) = C(n,k) · p^k · (1−p)^(n−k) Mean μ = np, Standard deviation σ = √(np(1−p))

Simple Example Example

Scenario
A 5-question multiple-choice quiz, with probability p = 0.4 of guessing each question correctly (n=5). What is the probability of getting exactly 2 correct?
  1. C(5,2) = 10
  2. P(X=2) = 10 × (0.4)² × (0.6)³ = 10 × 0.16 × 0.216 = 0.3456
  3. Mean: μ = np = 5(0.4) = 2 questions / σ = √(5×0.4×0.6) = √1.2 ≈ 1.095
binompdf (exactly k)
2nd → VARS (DISTR) A : binompdf( n: 5, p: 0.4, x value: 2 ) ENTER → 0.3456
binomcdf (k or fewer)
2nd → VARS (DISTR) B : binomcdf( n: 5, p: 0.4, x value: 2 ) ENTER → 0.6826 (sum of P(0), P(1), P(2) correct)

💡 For "3 or more correct," use 1 − binomcdf(n,p,2) (the complement rule).

03

Sampling Distributions & the Central Limit Theorem (CLT)

Unit 5.3, 5.5, 5.7 — If you repeatedly draw samples, the sample mean/sample proportion each have their own distribution.

Distribution of Sample Proportion p̂ Concept

Mean: μ(p̂) = p Standard deviation: σ(p̂) = √( p(1−p) / n ) Conditions: np ≥ 10, n(1−p) ≥ 10 (normal approximation valid)
Simple Example
A fair coin has true proportion p = 0.5, flipped n = 100 times.
σ(p̂) = √(0.5×0.5/100) = √0.0025 = 0.05 → p̂ will typically vary around 0.50 by about ±0.05.

Distribution of Sample Mean x̄ Concept

Mean: μ(x̄) = μ Standard deviation: σ(x̄) = σ / √n (standard error, SE)
Simple Example
Population σ = 20, sample size n = 25 → SE = 20/√25 = 20/5 = 4. As n increases, SE decreases (inversely proportional to √n).
04

Confidence Interval for a Proportion (1-PropZInt)

Unit 6.2 — Use a sample proportion to estimate a range for the population proportion p.

Formula Concept

p̂ ± z* · √( p̂(1−p̂) / n ) Conditions: random sample, np̂≥10, n(1−p̂)≥10, population at least 10× the sample size (10% condition)

Simple Example Example

Scenario
Of n = 50 students surveyed, 30 said they prefer online classes. Construct a 95% confidence interval.
  1. p̂ = 30/50 = 0.6
  2. SE = √(0.6×0.4/50) = √0.0048 ≈ 0.0693
  3. z* (95%) = 1.96 → ME = 1.96 × 0.0693 ≈ 0.1358
  4. Confidence interval: 0.6 ± 0.136 → (0.464, 0.736)
Using 1-PropZInt
STAT → TESTS → A : 1-PropZInt x: 30 n: 50 C-Level: 0.95 Calculate → (.46425, .73575) p̂ = 0.6

Interpretation sentence (memorize!): "We are 95% confident that the true population proportion is between 0.464 and 0.736."

05

Hypothesis Test for a Proportion (1-PropZTest)

Unit 6.4–6.7 — Where p-value, significance level α, and Type I/II errors first appear.

Procedure & Formula Concept

H0: p = p0 vs Ha: p ≠ p0 (or <, >) Test statistic: z = (p̂ − p0) / √( p0(1−p0)/n ) p-value: the probability of getting a result at least this extreme, computed using z Decision rule: p-value < α → reject H0 (statistically significant)

Simple Example Example

Scenario
Testing whether a coin is fair (p0 = 0.5). Out of n = 100 flips, 60 landed heads. α = 0.05.
H0: p = 0.5, Ha: p ≠ 0.5
  1. p̂ = 60/100 = 0.6
  2. z = (0.6 − 0.5) / √(0.5×0.5/100) = 0.1 / 0.05 = 2.0
  3. Two-sided p-value = 2 × P(Z > 2.0) = 2 × 0.0228 = 0.0456
  4. 0.0456 < 0.05(α) → reject H0. There is significant evidence the coin is not fair.
Using 1-PropZTest
STAT → TESTS → 5 : 1-PropZTest p0: 0.5 x: 60 n: 100 prop ≠ p0 (two-sided) Calculate → z = 2.0 p = 0.0455...

What Are Type I and Type II Errors? Must-Know

Type I Error: rejecting H0 when H0 is actually true → probability = α Type II Error: failing to reject H0 when H0 is actually false → probability = β Power = 1 − β = the probability of correctly rejecting H0 when it is false

Memory tip: "α is the number I set in advance (usually 0.05)", "β shrinks as sample size n grows", "a larger n leaves α unchanged but reduces β → power increases."

06

Confidence Interval for a Mean (TInterval)

Unit 7.2 — When σ (the population standard deviation) is unknown, use the t-distribution instead of z.

Formula Concept

x̄ ± t* · (s / √n), df = n − 1 Conditions: random sample, approximate normality (check a graph if n<30), 10% condition

Simple Example Example

Scenario — 6 Data Points
Prep time (minutes) for drinks at a café: 4, 5, 5, 6, 7, 9 (n = 6)
  1. x̄ = (4+5+5+6+7+9)/6 = 36/6 = 6
  2. Sum of squared deviations = (−2)²+(−1)²+(−1)²+0²+1²+3² = 4+1+1+0+1+9 = 16
  3. s = √(16/(6−1)) = √3.2 ≈ 1.789
  4. df = 5, 90% confidence level → t* ≈ 2.015
  5. ME = 2.015 × (1.789/√6) ≈ 2.015 × 0.730 ≈ 1.471
  6. Confidence interval: 6 ± 1.47 → (4.53, 7.47)
1️⃣ Enter the Data
STAT → 1:Edit Enter 4,5,5,6,7,9 into L1
2️⃣ Using TInterval
STAT → TESTS → 8 : TInterval Inpt: Data List: L1 Freq: 1 C-Level: 0.90 Calculate → (4.5286, 7.4714) x̄=6, Sx≈1.789, n=6
07

Hypothesis Test for a Mean (T-Test)

Unit 7.4–7.5 — Use a sample to test whether a population mean differs from a claimed value.

Formula Concept

Test statistic: t = (x̄ − μ0) / (s/√n), df = n−1

Simple Example Example

Scenario — 5 Data Points
A manufacturer claims battery life averages 10 hours (μ0=10). A sample of 5 batteries: 9, 9, 10, 11, 11
H0: μ = 10, Ha: μ ≠ 10, α = 0.05
  1. x̄ = (9+9+10+11+11)/5 = 50/5 = 10
  2. Sum of squared deviations = 1+1+0+1+1 = 4 → s = √(4/4) = 1
  3. t = (10 − 10)/(1/√5) = 0
  4. t = 0 → p-value = 1 (nothing could be less extreme) → fail to reject H0. No evidence battery life differs from 10 hours.
Using T-Test
STAT → TESTS → 2 : T-Test Inpt: Data μ0: 10 List: L1 (enter 9,9,10,11,11) μ ≠ μ0 Calculate → t = 0 p = 1 x̄=10, Sx=1, n=5
08

Chi-Square Tests

Unit 8.2–8.6 — Test the difference between "observed" and "expected" counts for categorical data.

Formula Concept

χ² = Σ ( (observed − expected)² / expected ) Goodness of Fit (GOF): df = number of categories − 1 Independence / Homogeneity: df = (rows−1)(columns−1)

Simple Goodness-of-Fit Example Example A

Scenario
Testing whether a die is fair by rolling it 60 times. (Expected count per face = 60/6 = 10)
Face123456
Observed8912111010
Expected101010101010
  1. χ² = (8−10)²/10 + (9−10)²/10 + (12−10)²/10 + (11−10)²/10 + (10−10)²/10 + (10−10)²/10
  2. = 0.4 + 0.1 + 0.4 + 0.1 + 0 + 0 = χ² = 1.0
  3. df = 6 − 1 = 5 → p-value(χ²=1.0, df=5) ≈ 0.963 (very large)
  4. p-value > α → fail to reject H0. No evidence the die is unfair.
Using χ²GOF-Test
STAT → EDIT: L1=observed, L2=expected STAT → TESTS → D : χ²GOF-Test Observed: L1 Expected: L2 df: 5 Calculate → χ²=1.0, p=0.9626

Simple Independence Test Example Example B

Scenario — 2×2 Table
Testing whether "gender" and "club membership" are independent (n=40)
MemberNot a member
Male128
Female812
  1. Row/column totals: Male 20, Female 20 / Member 20, Not a member 20 → total 40
  2. Expected count (same for all cells) = (row total × column total)/total = (20×20)/40 = 10 (all 4 cells)
  3. χ² = (12−10)²/10 × 4 cells = 4×(4/10) = 1.6
  4. df = (2−1)(2−1) = 1 → p-value(χ²=1.6, df=1) ≈ 0.206
  5. p > α(0.05) → not enough evidence of an association (gender and club membership can be treated as independent)
Using the 2-way χ²-Test
2nd → x⁻¹ (MATRX) → EDIT → [A] Enter 2×2 matrix: 12,8 / 8,12 STAT → TESTS → C : χ²-Test Observed: [A] Expected: [B] Calculate → χ²=1.6, p=0.2059

α · β · p-value · Type I/II Error — Side by Side

The five most commonly confused terms on the exam, laid out in one table.

TermDefinitionWho/When It's SetEffect of Sample Size nMemory Sentence
α (significance level) The maximum acceptable probability of rejecting H0 when it's actually true. Chosen before the test (commonly 0.05, 0.01, 0.10) Set by the researcher before running the test Independent of n — it's a number you choose, so it doesn't change "The biggest mistake I'm willing to risk"
p-value Assuming H0 is true, the probability of observing a result at least as extreme as the one obtained Calculated from the sample data (a result of the test) All else equal, larger n tends to make the p-value smaller for the same effect size "How surprising is this result, if it were really just chance?"
Type I Error Rejecting H0 when it is actually true (a false alarm) — probability = α Lowering α lowers the probability of a Type I error Independent of n "Convicting an innocent person"
Type II Error (β) Failing to reject H0 when it is actually false (a miss) — probability = β Determined indirectly by α, effect size, and n (not chosen directly) As n increases, β decreases (power increases) "Letting a guilty person go free (missed it)"
Power (= 1−β) The probability of correctly rejecting H0 when it is false Increases with larger α, larger effect size, or larger n Power increases as n increases "The test's ability to catch a real difference"

The One-Line Decision Rule Must Memorize

p-value < α → Reject H0 (the result is statistically significant) p-value ≥ α → Fail to reject H0 (insufficient evidence)
Q

10 Practice Multiple-Choice Questions

Click "Show answer" under each question to check your work. Try solving without a calculator first, then verify.

Q1 · Unit 1

Exam scores follow a normal distribution with μ=70, σ=10. What is the z-score for a score of 60?

A −0.5
B −1.0
C 1.0
D −10
Show answer
Answer: B. z=(60−70)/10 = −1.0
Q2 · Unit 2

For a binomial distribution with n=5, p=0.4, which is the correct calculator command for P(X=2)?

A binomcdf(5,0.4,2)
B binompdf(5,0.4,2)
C normalcdf(5,0.4,2)
D invNorm(5,0.4,2)
Show answer
Answer: B. "Exactly k" uses the pdf (probability mass function); "k or fewer" (cumulative) uses the cdf.
Q3 · Unit 3

For a population with σ=20, if the sample size increases from n=25 to n=100, how does the standard error (SE) change?

A It becomes 4 times larger
B It stays the same
C It is cut in half (1/2)
D It becomes 4 times smaller
Show answer
Answer: C. SE=σ/√n. If n becomes 4 times larger, √n doubles, so SE is cut in half. (20/5=4 → 20/10=2)
Q4 · Unit 4

For a sample with n=50, x=30, what value of p̂ does the calculator automatically compute when running 1-PropZInt?

A 0.5
B 0.6
C 30
D 50
Show answer
Answer: B. p̂ = x/n = 30/50 = 0.6
Q5 · Unit 5

A 1-PropZTest gives p-value = 0.0456 with α = 0.05. What is the correct conclusion?

A Reject H0 (significant)
B Fail to reject H0
C Reject Ha
D The sample size is too small
Show answer
Answer: A. p-value (0.0456) < α (0.05), so we reject H0.
Q6 · Mixed

A new drug actually works (H0: "no effect" is false), but the test concluded "no effect." What is this error called?

A Type I Error
B Type II Error
C Sampling bias
D Chi-square error
Show answer
Answer: B. H0 ("no effect") is false but was not rejected, so this is a Type II Error (β).
Q7 · Unit 6

If sample size n increases (all else equal), which of the following is true?

A α automatically increases
B β decreases and power increases
C The p-value always increases
D Nothing changes
Show answer
Answer: B. n↑ → SE↓ → the test can detect the same effect size more reliably → β↓, power (=1−β)↑. α is fixed by the researcher and doesn't change.
Q8 · Unit 7

When the population standard deviation σ is unknown and you're estimating a population mean from sample data, which distribution is used?

A Standard normal (z) distribution
B Binomial distribution
C t-distribution (df = n−1)
D Chi-square distribution
Show answer
Answer: C. When σ is unknown and estimated by s, use the t-distribution, with degrees of freedom n−1.
Q9 · Unit 8

For a 2×3 contingency table, what is the degrees of freedom (df) for a chi-square independence test?

A 2
B 3
C 5
D 6
Show answer
Answer: A. df=(rows−1)(columns−1)=(2−1)(3−1)=1×2=2
Q10 · Mixed

Which of the following is the correct interpretation of a "p-value"?

A The probability that H0 is true
B Assuming H0 is true, the probability of getting the observed result (or something more extreme)
C Always equal to the probability of a mistake (α)
D The same as the sample size
Show answer
Answer: B. The p-value is a conditional probability calculated under the assumption that H0 is true — it is NOT the probability that H0 is true (the most common misconception in AP Stats!).