Hypothesis Testing Framework

📐 Null & Alternative Hypotheses

Part 1 of 7 — Setting Up the Test

What Is Hypothesis Testing?

Hypothesis testing is a formal procedure for using sample data to decide between two competing claims about a population parameter.

🔑 Key Idea: We assume $H_0$ is true and look for evidence against it. We NEVER prove $H_0$ true — we either reject it or fail to reject it.

Writing Hypotheses

Hypotheses are always about population parameters ($\mu$, $p$), never about sample statistics ($\bar{x}$, $\hat{p}$).

For means:

For proportions:

⚠️ Important: $H_0$ always contains the equals sign ($=$). The alternative contains $\neq$, $>$, or $<$.

Worked Example

Claim: "Students at this school score higher than the national average of 75."

• Parameter: $\mu$ = true mean score of students at this school
• $H_0: \mu = 75$ (no difference from national average)
• $H_a: \mu > 75$ (school average is higher)

This is a right-tailed test because the claim is "higher than."

Significance Level ($\alpha$)

Before testing, we choose a significance level $\alpha$ (usually 0.05):

🔑 AP Tip: Unless told otherwise, assume $\alpha = 0.05$ on the AP exam.

Hypothesis Setup 🎯

Hypothesis Identification 🧮

For each claim, identify the null value ($\mu_0$):

1) Claim: $\mu > 75$. $H_0: \mu =$ ?

2) Claim: $\mu < 50$. $H_0: \mu =$ ?

3) Claim: $\mu \neq 100$. $H_0: \mu =$ ?

Test Direction 🔍

Exit Quiz — Null & Alternative Hypotheses ✅

📊 Test Statistics

Part 2 of 7 — Measuring the Evidence

What Is a Test Statistic?

A test statistic measures how far the sample result falls from the null hypothesis value, expressed in standard-error units.

$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$

🔑 Interpretation: $t$ tells you how many standard errors the sample mean is from the null value. Larger $|t|$ → stronger evidence against $H_0$.

Standard Error (SE)

The standard error measures the typical distance between $\bar{x}$ and $\mu$ due to sampling variability:

$SE = \frac{s}{\sqrt{n}}$

🔑 Key Insight: Quadrupling the sample size halves the standard error (because $\sqrt{4} = 2$).

Degrees of Freedom

For a one-sample $t$-test: $df = n - 1$

The degrees of freedom determine which $t$-distribution to use for finding the P-value. More degrees of freedom → the $t$-distribution looks more like a normal distribution.

Worked Example

A school claims its average SAT math score is 500. A random sample of 36 students gives $\bar{x} = 520$ and $s = 60$.

Step 1 — Standard Error:

$SE = \frac{s}{\sqrt{n}} = \frac{60}{\sqrt{36}} = \frac{60}{6} = 10$

Step 2 — Test Statistic:

$t = \frac{\bar{x} - \mu_0}{SE} = \frac{520 - 500}{10} = \frac{20}{10} = 2.0$

Step 3 — Degrees of Freedom:

$df = n - 1 = 36 - 1 = 35$

Interpretation: The sample mean is 2.0 standard errors above the null value. This is moderate-to-strong evidence against $H_0$.

How Large Is "Large Enough"?

⚠️ Caution: These are rough guidelines. Always compute the P-value for a precise conclusion.

Test Statistic Concepts 🎯

Computing Test Statistics 🧮

1) $s = 14$, $n = 49$. What is the standard error?

2) $\bar{x} = 82$, $\mu_0 = 75$, $SE = 2$. What is the $t$-statistic?

3) $n = 26$. What are the degrees of freedom?

Interpreting Test Statistics 🔍

Exit Quiz — Test Statistics ✅

🔢 P-Values

Part 3 of 7 — How Surprising Is the Evidence?

What Is a P-Value?

The P-value is the probability of obtaining a test statistic as extreme as (or more extreme than) the one observed, assuming $H_0$ is true.

🔑 Plain English: "If nothing special is happening ($H_0$ is true), how likely is it that we'd see data this extreme just by chance?"

Decision Rule

⚠️ Never say "accept $H_0$." We either reject or fail to reject.

Interpreting P-Values

One-Tailed vs Two-Tailed P-Values

🔑 Two-tailed tests double the one-tail probability because evidence in either direction counts.

Worked Example

$\bar{x} = 520$, $\mu_0 = 500$, $SE = 10$, $\alpha = 0.05$, right-tailed test.

Step 1 — Test statistic: $t = \frac{520 - 500}{10} = 2.0$

Step 2 — P-value (using calculator): $P = \text{tcdf}(2.0, 10^{99}, 35) \approx 0.027$

Step 3 — Decision: $0.027 < 0.05 \Rightarrow \text{Reject } H_0$

Step 4 — Conclusion in context: "There is convincing evidence ($t = 2.0$, $P = 0.027$) that the true mean SAT math score at this school is greater than 500."

Writing AP Conclusions

Always include four elements:

1. Decision — Reject or fail to reject $H_0$
2. Evidence — Cite $t$-statistic and P-value
3. Context — Refer to the specific problem
4. Direction — "Greater than," "less than," or "different from"

🔑 AP Tip: "Fail to reject" does NOT mean the null is proven true — only that we lack sufficient evidence.

P-Value Concepts 🎯

P-Value Decisions 🧮

1) The most common significance level $\alpha$ is:

2) $P = 0.03$ vs $P = 0.08$: which is more significant? (enter the P-value)

3) $P = 0.12$, $\alpha = 0.05$. Do we reject? (enter "yes" or "no")

P-Value Interpretation 🔍

Exit Quiz — P-Values ✅

📈 Type I & Type II Errors

Part 4 of 7 — Making the Wrong Decision

Decision Table

Every hypothesis test has four possible outcomes:

Type I Error (False Positive)

Definition: Rejecting $H_0$ when it is actually true.

• Probability = $\alpha$ (the significance level)
• You conclude there IS an effect when there really isn't one

Real-world example: A medical test says the patient has a disease, but they are actually healthy.

🔑 Key Connection: Choosing $\alpha = 0.05$ means you accept a 5% chance of a Type I error.

Type II Error (False Negative)

Definition: Failing to reject $H_0$ when it is actually false.

• Probability = $\beta$
• You conclude there is NO effect when there really IS one

Real-world example: A medical test says the patient is healthy, but they actually have the disease.

Power

Definition: The probability of correctly rejecting a false $H_0$.

$\text{Power} = 1 - \beta$

🔑 AP Tip: Power is typically considered adequate when it is at least 0.80 (80%).

The $\alpha$–$\beta$ Tradeoff

🔑 The only way to reduce BOTH errors is to increase the sample size.

Worked Example

A jury trial: $H_0$: The defendant is innocent.

The justice system sets a very low $\alpha$ ("beyond reasonable doubt") because Type I errors have severe consequences.

Error Identification 🎯

Error Probabilities 🧮

1) Rejecting a true $H_0$ is a Type ___ error.

2) If $\alpha = 0.05$, the probability of a Type I error is:

3) If $\beta = 0.20$, the power is:

Error Concepts 🔍

Exit Quiz — Type I & Type II Errors ✅

🧮 One-Sample t-Test

Part 5 of 7 — The Complete Procedure

When to Use a One-Sample t-Test

Use a one-sample $t$-test when:

• You have one quantitative variable
• You want to test a claim about the population mean $\mu$
• The population standard deviation $\sigma$ is unknown (use $s$ instead)

Conditions (CHECK EVERY TIME)

⚠️ AP Tip: You MUST state and verify all three conditions to earn full credit on the free response.

The Four-Step Process

Step 1 — STATE:

• Define the parameter: "Let $\mu$ = the true mean ..."
• Write hypotheses: $H_0: \mu = \mu_0$ vs $H_a$

Step 2 — PLAN:

• Name the test: "One-sample $t$-test"
• Check all three conditions

Step 3 — DO:

$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}, \quad df = n - 1$

• Find P-value using $t$-distribution with $df = n-1$

Step 4 — CONCLUDE:

• Compare P-value to $\alpha$
• State conclusion in context

Worked Example

A manufacturer claims their batteries last 500 hours. A random sample of 25 batteries gives $\bar{x} = 490$, $s = 20$. Test at $\alpha = 0.05$.

STATE:

• $\mu$ = true mean battery life (hours)
• $H_0: \mu = 500$, $H_a: \mu < 500$ (left-tailed — suspect batteries last LESS)

PLAN:

• One-sample $t$-test
• ✅ Random: stated "random sample"
• ✅ Independent: 25 < 10% of all batteries produced
• ✅ Normal: $n = 25 < 30$, so the sample data must be free from strong skewness and outliers (a dotplot of the 25 lifetimes shows no strong skew or outliers)

DO: $SE = \frac{s}{\sqrt{n}} = \frac{20}{\sqrt{25}} = \frac{20}{5} = 4$ $t = \frac{490 - 500}{4} = \frac{-10}{4} = -2.5$ $df = 25 - 1 = 24$ $P = \text{tcdf}(-10^{99}, -2.5, 24) \approx 0.0098$

CONCLUDE: Since $P = 0.0098 < \alpha = 0.05$, we reject $H_0$. There is convincing evidence that the true mean battery life is less than 500 hours.

z Procedures vs. t Procedures on the AP Exam

t-Test Concepts 🎯

Computing a t-Test 🧮

$n = 36$, $\bar{x} = 52$, $s = 6$, $\mu_0 = 50$:

1) $SE = s/\sqrt{n} =$ ?

2) $t = (\bar{x} - \mu_0)/SE =$ ?

3) $df =$ ?

t-Test Procedure 🔍

Exit Quiz — One-Sample t-Test ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Putting It All Together

Worked Example 1: Cereal Box Weights

A cereal company advertises 16 oz boxes. A consumer group suspects the boxes are underfilled. They weigh a random sample of 40 boxes and find $\bar{x} = 15.7$ oz, $s = 0.8$ oz. Test at $\alpha = 0.05$.

STATE:

• $\mu$ = true mean weight of cereal boxes (oz)
• $H_0: \mu = 16$ vs $H_a: \mu < 16$ (left-tailed — suspect underfilling)

PLAN:

• One-sample $t$-test
• ✅ Random: "random sample" stated
• ✅ Independent: 40 boxes $< 10\%$ of all boxes produced
• ✅ Normal/Large: $n = 40 \geq 30$ → CLT applies

DO: $SE = \frac{0.8}{\sqrt{40}} = \frac{0.8}{6.325} \approx 0.1265$

$t = \frac{15.7 - 16}{0.1265} = \frac{-0.3}{0.1265} \approx -2.372$

$df = 40 - 1 = 39$

$P = \text{tcdf}(-10^{99}, -2.372, 39) \approx 0.0114$

CONCLUDE: Since $P = 0.0114 < \alpha = 0.05$, we reject $H_0$. There is convincing evidence that the true mean weight of cereal boxes is less than 16 oz. The consumer group's suspicion is supported.

Worked Example 2: Study Hours

A college claims students study an average of 15 hours per week. A professor surveys a random sample of 50 students and finds $\bar{x} = 13.2$ hours, $s = 5.1$ hours. Test at $\alpha = 0.05$ whether the true mean differs from 15.

STATE:

• $\mu$ = true mean weekly study hours for students at this college
• $H_0: \mu = 15$ vs $H_a: \mu \neq 15$ (two-tailed — "differs from")

PLAN:

• One-sample $t$-test
• ✅ Random: "random sample" stated
• ✅ Independent: $50 < 10\%$ of all college students
• ✅ Normal/Large: $n = 50 \geq 30$ → CLT applies

DO: $SE = \frac{5.1}{\sqrt{50}} = \frac{5.1}{7.071} \approx 0.7212$

$t = \frac{13.2 - 15}{0.7212} = \frac{-1.8}{0.7212} \approx -2.495$

$df = 49$

$P = 2 \times \text{tcdf}(-10^{99}, -2.495, 49) \approx 2 \times 0.0080 = 0.0160$

CONCLUDE: Since $P = 0.016 < \alpha = 0.05$, we reject $H_0$. There is convincing evidence that the true mean weekly study hours for students at this college differs from 15 hours.

Common AP Mistakes to Avoid

Workshop Practice 🎯

Workshop Calculations 🧮

$n = 25$, $\bar{x} = 84$, $s = 10$, $\mu_0 = 80$:

1) $SE = s/\sqrt{n} =$ ?

2) $t = (\bar{x} - \mu_0)/SE =$ ?

3) $df =$ ?

AP Process Steps 🔍

Exit Quiz — Problem-Solving Workshop ✅

🏆 Review & Applications

Part 7 of 7 — Complete Reference Guide

Formula Reference

Hypothesis Test Decision Guide

Decision Rule

$\text{If } P < \alpha \Rightarrow \text{Reject } H_0 \qquad \text{If } P \geq \alpha \Rightarrow \text{Fail to reject } H_0$

Error Summary

Conditions Checklist

AP Four-Step Process

1. STATE — Define parameter; write $H_0$ and $H_a$
2. PLAN — Name the test; check all three conditions
3. DO — Compute SE, $t$, $df$, P-value
4. CONCLUDE — Decision + evidence + context

Power Factors

Common Mistakes on the AP Exam

Comprehensive Review 🎯

Quick Calculations 🧮

1) $n = 30$. $df =$ ?

2) $s = 12$, $n = 36$. $SE =$ ?

3) $\bar{x} = 48$, $\mu_0 = 50$, $SE = 2$. $t =$ ?

Key Concepts 🔍

Exit Quiz — Hypothesis Testing Review ✅