Paired Data

⚖️ Comparing Two Populations

Part 1 of 7 — Two-Sample Z-Test for Proportions

When to Compare Two Proportions

Use when you have two independent groups and want to test whether their population proportions differ.

Example: Is the proportion of smartphone users higher among teens than adults?

Hypotheses

$H_0: p_1 - p_2 = 0 \quad (\text{no difference})$ $H_a: p_1 - p_2 \neq 0 \quad (\text{or } > 0 \text{ or } < 0)$

Test Statistic

$z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$

where $\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}$ is the pooled proportion.

Conditions

1. Random samples from both populations
2. Independent groups (and $n < 10\%$ of population for each)
3. Large counts: $n_1\hat{p}, n_1(1-\hat{p}), n_2\hat{p}, n_2(1-\hat{p}) \geq 10$

🔑 Under $H_0$, we assume $p_1 = p_2$, so we use the pooled proportion $\hat{p}$ for the standard error.

Two-Proportion Test 🎯

Two-Proportion Calculations 🧮

Treatment group: 45 successes out of 150. Control group: 30 successes out of 150.

1) $\hat{p}_1 = 45/150 =$? (Express as a decimal)

2) $\hat{p}_2 = 30/150 =$?

3) Pooled $\hat{p} = (45+30)/(150+150) =$?

📊 Two-Sample T-Test for Means

Part 2 of 7 — Comparing Two Independent Groups

Topics in This Part

🔑 Key Concept: The two-sample $t$-test compares means from two independent groups. The key question: "Is the difference in sample means large enough to conclude the population means differ?"

The Setup

Two independent random samples:

• Group 1: $n_1$ observations, sample mean $\bar{x}_1$, sample SD $s_1$
• Group 2: $n_2$ observations, sample mean $\bar{x}_2$, sample SD $s_2$

Hypotheses

$H_0: \mu_1 - \mu_2 = 0 \quad (\text{or equivalently, } \mu_1 = \mu_2)$ $H_a: \mu_1 - \mu_2 \neq 0 \quad (\text{or } > 0 \text{ or } < 0)$

Test Statistic

$\boxed{t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}}$

Degrees of freedom: Use the technology value (it lies between the smaller of $n_1 - 1$ and $n_2 - 1$ and $n_1 + n_2 - 2$) or the conservative value $\text{df} = \min(n_1 - 1, n_2 - 1)$. The SE is never pooled: each sample keeps its own $s$.

⚠️ AP Tip: The AP formula sheet provides this formula. On the AP exam, use the conservative df or the calculator df — both are acceptable.

Conditions

Worked Example

Study: Does a new teaching method improve test scores?

• Method A (traditional): $n_1 = 35$, $\bar{x}_1 = 74.2$, $s_1 = 8.5$
• Method B (new): $n_2 = 38$, $\bar{x}_2 = 79.6$, $s_2 = 7.9$

Step 1 — Hypotheses: $H_0: \mu_B - \mu_A = 0$ (no difference in mean scores) $H_a: \mu_B - \mu_A > 0$ (new method produces higher mean scores)

Step 2 — Conditions:

• Random: Students randomly assigned to methods ✓
• Independent: Two separate groups ✓
• Normal: $n_1 = 35 \geq 30$ and $n_2 = 38 \geq 30$ ✓
• 10%: Each group is less than 10% of all students ✓

Step 3 — Test statistic: $t = \frac{79.6 - 74.2}{\sqrt{\frac{7.9^2}{38} + \frac{8.5^2}{35}}} = \frac{5.4}{\sqrt{1.642 + 2.064}} = \frac{5.4}{\sqrt{3.706}} = \frac{5.4}{1.925} = 2.805$

df $= \min(34, 37) = 34$ (conservative) or use calculator df

Step 4 — Conclusion: $P \approx 0.004$ (one-sided). Since $P = 0.004 < 0.05 = \alpha$, we reject $H_0$. There is convincing evidence that the new teaching method produces higher mean test scores than the traditional method.

Two-Sample $t$-Test Concepts 🎯

Computing the $t$-Statistic 🧮

Group 1: $\bar{x}_1 = 50$, $s_1 = 10$, $n_1 = 25$ Group 2: $\bar{x}_2 = 44$, $s_2 = 12$, $n_2 = 30$

1) $\bar{x}_1 - \bar{x}_2 =$

2) $s_1^2/n_1 + s_2^2/n_2 =$ (compute to one decimal)

3) Conservative df $= \min(n_1-1, n_2-1) =$

Decisions and Interpretations 🔍

Exit Quiz — Two-Sample $t$-Test ✅

📊 Paired T-Test

Part 3 of 7 — When Data Come in Pairs

Topics in This Part

🔑 Key Concept: A paired $t$-test is a one-sample $t$-test performed on the differences within each pair.

When to Use a Paired Test

Use a paired $t$-test when:

• The same subjects are measured twice (before/after)
• Subjects are matched in pairs (e.g., twins, partners)
• Each observation in one group has a natural partner in the other

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

• The two groups are independent (different subjects, no pairing)

The Procedure

Step 1: Compute the differences: $d_i = x_{1i} - x_{2i}$ for each pair

Step 2: Treat the differences as a single sample and perform a one-sample $t$-test:

$\boxed{t = \frac{\bar{d} - 0}{s_d / \sqrt{n}}}$

where:

• $\bar{d}$ = mean of the differences
• $s_d$ = standard deviation of the differences
• $n$ = number of pairs
• df $= n - 1$

Hypotheses

$H_0: \mu_d = 0 \quad (\text{no difference, on average})$ $H_a: \mu_d \neq 0 \quad (\text{or } > 0 \text{ or } < 0)$

⚠️ AP Tip: Define $d$ clearly: "$d$ = After $-$ Before" or "$d$ = Treatment $-$ Control." The direction matters for one-sided tests.

Conditions

Worked Example

Study: Does a tutoring program improve SAT scores? 12 students take the SAT, receive tutoring, then retake it.

Step 1 — Hypotheses: $H_0: \mu_d = 0$ (tutoring has no effect on mean SAT scores) $H_a: \mu_d > 0$ (tutoring increases mean SAT scores) where $d$ = After $-$ Before

Step 2 — Conditions:

• Random: Students randomly selected ✓
• Independent: 12 students < 10% of all SAT takers ✓
• Normal: Histogram of differences shows no strong skew ✓

Step 3 — Test statistic: $t = \frac{28.5 - 0}{22.3/\sqrt{12}} = \frac{28.5}{6.436} = 4.43$ df $= 12 - 1 = 11$

Step 4 — Conclusion: $P < 0.001$ (one-sided). Since $P < 0.001 < 0.05$, we reject $H_0$. There is convincing evidence that the tutoring program increases mean SAT scores.

Paired vs. Two-Sample: Why It Matters

🔑 Key Insight: Pairing reduces variability by eliminating subject-to-subject differences, making it easier to detect a treatment effect.

Paired $t$-Test Concepts 🎯

Computing the Paired $t$ 🧮

$n = 16$ pairs, $\bar{d} = 5.0$, $s_d = 8.0$

1) SE of $\bar{d}$ $= s_d / \sqrt{n} =$

2) $t = \bar{d} / \text{SE} =$

3) df $=$

Paired or Two-Sample? 🔍

Exit Quiz — Paired $t$-Test ✅

📊 Confidence Intervals for Differences

Part 4 of 7 — Estimating How Much Two Groups Differ

Topics in This Part

🔑 Key Concept: A CI for the difference gives a range of plausible values for how much two population means (or the mean difference) differ.

Two-Sample CI for $\mu_1 - \mu_2$

$\boxed{(\bar{x}_1 - \bar{x}_2) \pm t^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

• Same conditions as the two-sample $t$-test (Random, Independent, Normal/Large, 10%)
• df: use calculator (Welch) or conservative $\min(n_1 - 1, n_2 - 1)$

Paired CI for $\mu_d$

$\boxed{\bar{d} \pm t^* \cdot \frac{s_d}{\sqrt{n}}}$

• Same conditions as the paired $t$-test (Random, Independent pairs, Normal differences)
• df $= n - 1$ (where $n$ = number of pairs)

Interpretation Templates

Two-Sample: "We are [C]% confident that the true difference in mean [context] between [group 1] and [group 2] is between [lower] and [upper] [units]."

Paired: "We are [C]% confident that the true mean difference in [context] is between [lower] and [upper] [units]."

CI ↔ Test Connection

Worked Example — Two-Sample CI

Group A (old drug): $n_1 = 40$, $\bar{x}_1 = 82$, $s_1 = 12$ Group B (new drug): $n_2 = 45$, $\bar{x}_2 = 88$, $s_2 = 10$

95% CI for $\mu_B - \mu_A$:

• $\bar{x}_B - \bar{x}_A = 88 - 82 = 6$
• SE $= \sqrt{10^2/45 + 12^2/40} = \sqrt{2.222 + 3.6} = \sqrt{5.822} = 2.413$
• Conservative df $= \min(39, 44) = 39$, $t^* \approx 2.023$
• $6 \pm 2.023(2.413) = 6 \pm 4.881 = (1.119, 10.881)$

Interpretation: "We are 95% confident that the true difference in mean recovery scores (new $-$ old) is between 1.1 and 10.9 points. Since 0 is not in the interval, there is evidence the new drug produces higher mean scores."

Worked Example — Paired CI

15 patients measured before and after treatment. $\bar{d} = 8.3$ (After $-$ Before), $s_d = 6.2$

95% CI for $\mu_d$: df $= 14$, $t^* = 2.145$ $8.3 \pm 2.145 \cdot \frac{6.2}{\sqrt{15}} = 8.3 \pm 2.145(1.601) = 8.3 \pm 3.434$ $(4.866, 11.734)$

Interpretation: "We are 95% confident that the true mean change in [outcome] after treatment is between 4.9 and 11.7 units."

⚠️ AP Tip: Always define $d$ (e.g., After $-$ Before) and include context and units.

CI for Differences Concepts 🎯

Building CIs 🧮

Two-Sample: $\bar{x}_1 - \bar{x}_2 = 10$, SE $= 4$, $t^* = 2.0$

1) Margin of error $=$

2) Lower bound $=$

Paired: $\bar{d} = -3$, $s_d = 6$, $n = 36$, $t^* = 2.030$

3) SE of $\bar{d}$ $=$

Interpretation Decisions 🔍

Exit Quiz — CIs for Differences ✅

📊 Power and Sample Size

Part 5 of 7 — Detecting Real Differences

Topics in This Part

🔑 Key Concept: Power is the probability of correctly rejecting $H_0$ when $H_0$ is actually false. Higher power = better ability to detect a real effect.

Error Types

$\text{Power} = 1 - \beta = P(\text{reject } H_0 \mid H_0 \text{ is false})$

Type I Error ($\alpha$)

• Rejecting $H_0$ when it is true (false positive)
• Probability $= \alpha$ (the significance level)
• Example: Concluding a drug works when it actually does not

Type II Error ($\beta$)

• Failing to reject $H_0$ when it is false (false negative)
• Probability = $\beta$
• Example: Concluding a drug does not work when it actually does

Factors That Increase Power

⚠️ AP Tip: You will NOT be asked to calculate power on the AP exam, but you MUST understand conceptually how each factor affects power.

Intuition for Each Factor

Larger $n$: More data → smaller SE → easier to detect a difference $\text{SE} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \quad \text{→ larger } n \text{ → smaller SE → larger } |t|$

Larger $\alpha$: Easier rejection threshold → more likely to reject (but more risk of Type I error)

Larger effect: A bigger real difference is easier to detect than a tiny one

Smaller $\sigma$: Less noise → the signal (difference) stands out more clearly

The Power-$\alpha$ Tradeoff

$\alpha \downarrow \Rightarrow \text{Power} \downarrow \Rightarrow \beta \uparrow$

Decreasing $\alpha$ (e.g., from 0.05 to 0.01) reduces Type I error but increases Type II error (reduces power). The only way to reduce BOTH errors simultaneously is to increase sample size.

Sample Size Planning

Before collecting data, researchers choose $n$ to achieve desired power (typically 80% or higher):

1. Specify the smallest meaningful effect size
2. Estimate population variability ($\sigma$)
3. Choose $\alpha$ (usually 0.05)
4. Use a power table or software to find the required $n$

🔑 Key Insight: Larger samples are always better for power, but they cost more. Sample size planning balances statistical needs with practical constraints.

Power Concepts 🎯

Error and Power Calculations 🧮

1) If $\beta = 0.15$, what is the power? (give as decimal)

2) If $\alpha = 0.01$, what is the probability of a Type I error?

3) A test has power $= 0.80$. What is $\beta$?

Power Factors 🔍

Exit Quiz — Power & Sample Size ✅

📊 Problem-Solving Workshop

Part 6 of 7 — Complete Worked Examples

Worked Example 1: Two-Sample T-Test

Problem

A fitness company wants to compare two training programs. 45 volunteers are randomly assigned: 22 to Program A, 23 to Program B. After 8 weeks, weight loss (in pounds) is recorded:

Test whether Program A produces greater average weight loss at $\alpha = 0.05$.

Step 1: State Hypotheses

$H_0: \mu_A - \mu_B = 0$ $H_a: \mu_A - \mu_B > 0$

Where $\mu_A$ = true mean weight loss for Program A, $\mu_B$ = true mean weight loss for Program B.

Step 2: Check Conditions

✅ Random: Volunteers randomly assigned to groups (experiment) ✅ Normal: Both samples are smaller than 30, so we need the sample data in each group to be free from strong skewness and outliers (dotplots of both groups show no strong skew or outliers) ✅ Independent: Groups are independent; each person in only one program. Both $< 10\%$ of all potential participants.

Step 3: Calculate

$SE = \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}} = \sqrt{\frac{3.8^2}{22} + \frac{4.1^2}{23}} = \sqrt{\frac{14.44}{22} + \frac{16.81}{23}}$

$= \sqrt{0.6564 + 0.7309} = \sqrt{1.3873} \approx 1.178$

$t = \frac{(\bar{x}_A - \bar{x}_B) - 0}{SE} = \frac{12.4 - 9.7}{1.178} = \frac{2.7}{1.178} \approx 2.291$

Using technology with $df \approx 42$: $P$-value $\approx 0.0136$

Step 4: Conclude

Since $P = 0.0136 < 0.05 = \alpha$, we reject $H_0$.

AP-Style Conclusion: There is convincing evidence that the true mean weight loss for Program A is greater than the true mean weight loss for Program B.

Worked Example 2: Paired T-Test

Problem

A researcher tests whether a meditation app reduces stress. 30 participants rate their stress (1–100) before and after 4 weeks of daily use:

Test whether the app reduces stress at $\alpha = 0.05$.

Step 1: State Hypotheses

$H_0: \mu_d = 0$ $H_a: \mu_d > 0$

Where $\mu_d$ = true mean difference in stress scores (Before − After) for all users of this app.

Step 2: Check Conditions

✅ Random: 30 participants randomly selected (or assume representative) ✅ Normal: $n = 30 \geq 30$ (CLT applies for differences) ✅ Independent (10%): The 30 participants' differences are independent of each other; $30 < 10\%$ of all potential users

⚠️ Key: We check conditions on the DIFFERENCES, not the individual scores.

Step 3: Calculate

$SE = \frac{s_d}{\sqrt{n}} = \frac{11.3}{\sqrt{30}} = \frac{11.3}{5.477} \approx 2.063$

$t = \frac{\bar{x}_d - 0}{SE} = \frac{8.7}{2.063} \approx 4.217$

$df = n - 1 = 29$, $P$-value $< 0.001$

Step 4: Conclude

Since $P < 0.001 < 0.05 = \alpha$, we reject $H_0$.

AP-Style Conclusion: There is convincing evidence that the meditation app reduces mean stress scores.

Common AP Mistakes

Workshop Practice 🎯

Computation Practice 🧮

1) Two groups: $\bar{x}_1 = 45$, $s_1 = 6$, $n_1 = 36$; $\bar{x}_2 = 41$, $s_2 = 8$, $n_2 = 36$. Calculate SE (round to 2 decimal places).

2) Using the values above, calculate the t-statistic (round to 2 decimal places).

3) Paired data: $\bar{x}_d = 3.2$, $s_d = 5.0$, $n = 25$. Calculate the t-statistic (round to 2 decimal places).

Procedure Selection 🔍

Exit Quiz — Problem-Solving Workshop ✅

📊 Review & Applications

Part 7 of 7 — Comprehensive Review

Complete Formula Reference

Two-Sample T-Test for Means

Paired T-Test

Decision Guide: Paired vs. Two-Sample

Conditions Summary

Two-Sample Tests/CIs

1. Random: Both samples from random processes (or random assignment)
2. Normal: $n_1 \geq 30$ and $n_2 \geq 30$, or no strong skewness/outliers
3. Independent: Samples independent of each other; each $< 10\%$ of its population

Paired Tests/CIs

1. Random: Random sample of pairs (or randomly determine order)
2. Normal: $n \geq 30$ differences, or differences show no strong skewness/outliers
3. Independent: Individual pairs are independent; $n < 10\%$ of all pairs in population

Interpretation Templates

Hypothesis Test Conclusion

"Since $P$-value = ___ is [less/greater] than $\alpha =$ ___, we [reject/fail to reject] $H_0$. There [is/is not] convincing evidence that [context: what the alternative hypothesis claims]."

Confidence Interval Interpretation

"We are ___% confident that the true difference in means [context: $\mu_1 - \mu_2$ in words] is between ___ and ___."

Confidence Interval and Significance

• If 0 is NOT in the CI → Reject $H_0$ (at the corresponding $\alpha$)
• If 0 IS in the CI → Fail to reject $H_0$

Key Concepts from Every Part

Power Quick Review

$\text{Power} = 1 - \beta = P(\text{reject } H_0 \mid H_0 \text{ is false})$

Power increases when: $n \uparrow$, $\alpha \uparrow$, effect size $\uparrow$, $\sigma \downarrow$

🔑 Final Tip: The AP exam tests your ability to (1) choose the right test, (2) check conditions, (3) calculate correctly, and (4) interpret in context. Practice the full 4-step process until it becomes automatic.

Comprehensive Review 🎯

Formula Application 🧮

1) Two-sample data: $\bar{x}_1 = 78$, $\bar{x}_2 = 72$, $SE = 2.5$. Calculate $t$ (round to 1 decimal place).

2) Paired data: $\bar{x}_d = -4.5$, $s_d = 6$, $n = 36$. Calculate $SE$ (round to 1 decimal place).

3) Using the values from #2, what is $df$?

Concept Connections 🔍

Final Exam — Comparing Populations ✅