Interpreting Confidence Intervals

📐 Introduction to Confidence Intervals

Part 1 of 7 — Estimating Population Parameters

Point Estimates vs. Interval Estimates

Confidence Interval Structure

$\text{estimate} \pm \text{margin of error}$

$\text{estimate} \pm z^* \cdot \text{standard error}$

What Does "95% Confident" Mean?

If we repeated the sampling process many times and built a 95% CI each time, about 95% of those intervals would contain the true parameter.

⚠️ It does NOT mean there is a 95% probability that the parameter is in this particular interval. The parameter is fixed — it is either in the interval or it is not.

Common Confidence Levels

🔑 Higher confidence → wider interval → less precise. There is always a tradeoff between confidence and precision.

Confidence Interval Basics 🎯

Confidence Interval Calculations 🧮

A poll finds $\hat{p} = 0.60$ with $n = 400$.

1) Standard error $= \sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{0.24/400}$. Round to 3 decimal places.

2) For a 95% CI, the margin of error $= 1.96 \times SE$. Round to 3 decimal places.

3) The 95% CI upper bound $= 0.60 + ME$. Round to 3 decimal places.

📊 One-Sample Z-Interval for Proportions

Part 2 of 7 — Estimating a Population Proportion

Topics in This Part

🔑 Key Concept: The one-proportion $z$-interval is the most common confidence interval on the AP exam. Master the formula, conditions, and interpretation.

The Formula

$\boxed{\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}}$

Critical Values Reference

✅ Conditions for the One-Proportion $z$-Interval

Before constructing the interval, verify:

1. Random: Data comes from a random sample or randomized experiment.

2. 10% Condition (Independence): $n < 0.10N$ — the sample is less than 10% of the population.

3. Large Counts: $n\hat{p} \geq 10$ and $n(1-\hat{p}) \geq 10$ — enough successes and failures.

⚠️ Warning: Note that for confidence intervals we check $n\hat{p}$ and $n(1-\hat{p})$ (using $\hat{p}$), whereas for hypothesis tests we use $np_0$ and $n(1-p_0)$ (using the null value). This is a subtle but important distinction.

📝 Full Worked Example

Problem: A random sample of 500 U.S. adults finds that 320 support a proposed policy. Construct a 95% confidence interval for the true proportion who support the policy.

Step 1 — Identify: $\hat{p} = 320/500 = 0.64$, $n = 500$, confidence level = 95%, $z^* = 1.960$

Step 2 — Check conditions:

• Random: "A random sample" — stated ✓
• 10%: 500 is less than 10% of all U.S. adults (~260 million) ✓
• Large Counts: $500(0.64) = 320 \geq 10$ ✓ and $500(0.36) = 180 \geq 10$ ✓

Step 3 — Calculate: $SE = \sqrt{\frac{0.64 \times 0.36}{500}} = \sqrt{\frac{0.2304}{500}} = \sqrt{0.0004608} = 0.02147$

$ME = 1.960 \times 0.02147 = 0.04208$

$CI = 0.64 \pm 0.042 = (0.598, 0.682)$

Step 4 — Interpret: We are 95% confident that the true proportion of all U.S. adults who support the proposed policy is between 0.598 and 0.682.

🔑 AP Tip: Always state the interval in the context of the problem. Generic statements like "we are 95% confident the proportion is between 0.598 and 0.682" will lose points if they don't mention WHAT proportion (of WHOM doing WHAT).

One-Proportion Z-Interval Check 🎯

Build a Confidence Interval 🧮

A survey of 800 randomly selected teenagers finds that 480 use social media daily.

1) What is $\hat{p}$?

2) What is the standard error? (Round to 4 decimal places)

3) What is the 95% margin of error? (Round to 3 decimal places)

Interpretation Check 🔍

Exit Quiz — One-Proportion Z-Interval ✅

📊 One-Sample T-Interval for Means

Part 3 of 7 — Estimating a Population Mean

Topics in This Part

🔑 Key Concept: When $\sigma$ is unknown (almost always in practice), we use the $t$-distribution instead of the $z$-distribution. The $t$-interval is wider to account for the extra uncertainty from estimating $\sigma$ with $s$.

Why $t$ Instead of $z$?

For a population mean, $\sigma$ is not known, so the standard error uses the sample SD $s$, and critical values come from a $t$-distribution instead of the standard normal.

The $t$-distribution:

• Is bell-shaped and symmetric, like the normal
• Has heavier tails (more spread) than the normal
• Depends on degrees of freedom: $df = n - 1$
• Approaches the normal distribution as $df \to \infty$

The Formula

$\boxed{\bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}}$

where $t^*$ is the critical value from the $t$-distribution with $df = n - 1$.

Selected $t^*$ Values (95% Confidence)

⚠️ Warning: Notice that $t^*$ is always larger than $z^* = 1.960$ for finite $df$. This makes $t$-intervals wider than $z$-intervals — by design.

✅ Conditions for the One-Sample $t$-Interval

1. Random: Data comes from a random sample or randomized experiment.

2. 10% Condition: $n < 0.10N$ (if sampling without replacement).

3. Normal/Large Sample:

🔑 AP Tip: For the Normal condition with means, you should reference the sample data (boxplot, dotplot, or histogram). Saying "no strong skewness or outliers in the sample" is the expected language.

📝 Full Worked Example

Problem: A random sample of 35 commuters has a mean commute of $\bar{x} = 28.4$ minutes with $s = 8.6$ minutes. Construct a 95% confidence interval for the true mean commute time.

Step 1 — Identify: $\bar{x} = 28.4$, $s = 8.6$, $n = 35$, $df = 34$, $t^* \approx 2.032$ (from table, $df = 34$)

Step 2 — Check conditions:

• Random: Stated — random sample ✓
• 10%: $35$ is less than 10% of all commuters ✓
• Normal: $n = 35 \geq 30$, so by the CLT, the sampling distribution of $\bar{x}$ is approximately normal ✓

Step 3 — Calculate: $SE = \frac{s}{\sqrt{n}} = \frac{8.6}{\sqrt{35}} = \frac{8.6}{5.916} = 1.454$

$ME = 2.032 \times 1.454 = 2.954$

$CI = 28.4 \pm 2.954 = (25.45, 31.35)$

Step 4 — Interpret: We are 95% confident that the true mean commute time for all commuters is between 25.45 and 31.35 minutes.

$t$-Interval Concepts 🎯

$t$-Interval Calculations 🧮

A random sample of 25 test scores: $\bar{x} = 82$, $s = 10$. Build a 95% CI. Use $t^* = 2.064$ ($df = 24$).

1) What is the standard error?

2) What is the margin of error?

3) What is the lower bound of the 95% CI? (Round to 1 decimal)

$z$ vs. $t$ Decision 🔍

Exit Quiz — One-Sample $t$-Interval ✅

📊 Choosing Sample Size

Part 4 of 7 — Planning Your Study

Topics in This Part

🔑 Key Concept: Before collecting data, researchers choose a sample size that will produce a margin of error small enough to be useful. This is called "planning for a desired margin of error."

Margin of Error Review

Recall the margin of error for each type of interval:

The idea: set ME equal to your desired margin of error, then solve for $n$.

Sample Size for Proportions

Starting from $ME = z^* \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$, solve for $n$:

$\boxed{n = \left(\frac{z^*}{ME}\right)^2 \hat{p}(1 - \hat{p})}$

What if you don't know $\hat{p}$? Use $\hat{p} = 0.5$ — this maximizes $\hat{p}(1-\hat{p}) = 0.25$ and gives the most conservative (largest) sample size.

Worked Example — Proportions

Problem: A pollster wants a 95% CI for the proportion of voters who support a candidate, with a margin of error of no more than 3%. What sample size is needed?

$n = \left(\frac{1.96}{0.03}\right)^2 \times 0.5(0.5) = (65.33)^2 \times 0.25 = 4268.44 \times 0.25 = 1067.1$

$\boxed{n = 1068}$

⚠️ Always round UP to the next whole number. Rounding down gives a margin of error that exceeds your target.

Sample Size for Means

Starting from $ME = z^* \cdot \dfrac{\sigma}{\sqrt{n}}$ (using $z^*$ as an approximation when planning):

$\boxed{n = \left(\frac{z^* \cdot \sigma}{ME}\right)^2}$

You need a preliminary estimate of $\sigma$ — use a pilot study, similar past data, or range/4 as a rough estimate.

Worked Example — Means

Problem: Estimate the mean waiting time at a clinic to within 2 minutes (95% confidence). A pilot study found $s \approx 8$ minutes.

$n = \left(\frac{1.96 \times 8}{2}\right)^2 = \left(\frac{15.68}{2}\right)^2 = (7.84)^2 = 61.47$

$\boxed{n = 62}$

Sample Size Concepts 🎯

Sample Size Calculations 🧮

1) A researcher wants a 95% CI for a proportion with ME $\leq$ 0.04. Use $\hat{p} = 0.5$, $z^* = 1.96$. What minimum $n$ is needed?

2) A scientist wants a 99% CI for the mean weight of a species to within 5 grams. Past data: $\sigma \approx 20$ g, $z^* = 2.576$. What minimum $n$ is needed?

3) If a survey originally required $n = 400$ for ME = 5%, what $n$ is needed for ME = 2.5%?

Sample Size Decisions 🔍

Exit Quiz — Choosing Sample Size ✅

📊 Interpreting Confidence Intervals

Part 5 of 7 — What a CI Really Means

Topics in This Part

🔑 Key Concept: A confidence interval is about the process, not the specific interval. "95% confident" means the method produces intervals that capture the true parameter 95% of the time in repeated sampling.

The Correct Interpretation

The AP exam requires this precise language:

$\text{"We are [C]\% confident that the true [parameter in context] is between [lower] and [upper]."}$

Breaking it down:

Full-Credit Example

Context: 95% CI for mean commute time: $(25.4, 31.4)$

✅ "We are 95% confident that the true mean commute time for all workers in this city is between 25.4 and 31.4 minutes."

Common Misinterpretations (All Are WRONG)

⚠️ AP Scoring Note: Using "probability" instead of "confidence" when interpreting a CI loses points on the AP exam.

What "95% Confident" Really Means

If we took many samples and built a CI from each one:

• About 95% of those intervals would contain the true $\mu$
• About 5% would miss it
• Any single interval either contains $\mu$ or does not — we just don't know which

This is the frequentist interpretation: confidence describes the long-run success rate of the method.

Connecting CIs to Hypothesis Tests

A confidence interval can be used as a two-sided test:

Relationship: A 95% CI corresponds to $\alpha = 0.05$ two-sided test. A 99% CI corresponds to $\alpha = 0.01$ two-sided test.

Example: If a 95% CI for $\mu$ is $(25.4, 31.4)$, then:

• $H_0: \mu = 28$ → fail to reject (28 is in the interval)
• $H_0: \mu = 35$ → reject (35 is outside the interval)

Interpretation Concepts 🎯

Right or Wrong? 🔍

Classify each interpretation of a 95% CI $(50, 70)$ for the mean test score.

CI and Hypothesis Test Connection 🧮

A 95% CI for $\mu$ is $(18.2, 24.8)$.

1) What is the point estimate $\bar{x}$? (Hint: center of the interval)

2) What is the margin of error?

3) A test of $H_0: \mu = 25$ at $\alpha = 0.05$ (two-sided) would ___? Enter "reject" or "fail to reject".

Exit Quiz — Interpreting CIs ✅

📊 Problem-Solving Workshop

Part 6 of 7 — Full AP Free-Response Practice

Topics in This Part

🔑 Key Concept: On AP free-response questions, you must clearly show all four steps: Identify, Conditions, Calculate, and Interpret. Missing any step costs points.

The 4-Step CI Framework

Worked Example 1: One-Proportion Z-Interval

Problem: In a random sample of 500 adults, 320 support a new policy. Construct a 95% CI for the proportion who support the policy.

Step 1 — Identify: We will construct a one-proportion $z$-interval for $p$, the true proportion of all adults who support the policy. $C = 95\%$.

Step 2 — Conditions:

• Random: The problem states a random sample ✓
• 10%: $500 < 10\%$ of all adults ✓
• Large Counts: $n\hat{p} = 500(0.64) = 320 \geq 10$ ✓ and $n(1-\hat{p}) = 500(0.36) = 180 \geq 10$ ✓

Step 3 — Calculate: $\hat{p} = \frac{320}{500} = 0.64$

$SE = \sqrt{\frac{0.64(0.36)}{500}} = \sqrt{\frac{0.2304}{500}} = \sqrt{0.000461} = 0.02147$

$ME = 1.96 \times 0.02147 = 0.04209$

$CI = 0.64 \pm 0.042 = (0.598, 0.682)$

Step 4 — Interpret: We are 95% confident that the true proportion of all adults who support the new policy is between 0.598 and 0.682.

Worked Example 2: One-Sample T-Interval

Problem: A nutritionist measures the sodium content (mg) of 40 randomly selected frozen dinners. Results: $\bar{x} = 894$, $s = 124$. Construct a 90% CI for the mean sodium content. ($t^* = 1.685$ for $df = 39$.)

Step 1 — Identify: We will construct a one-sample $t$-interval for $\mu$, the true mean sodium content of all frozen dinners of this brand. $C = 90\%$.

Step 2 — Conditions:

• Random: "randomly selected" ✓
• 10%: $40 < 10\%$ of all frozen dinners produced ✓
• Normal: $n = 40 \geq 30$, so by CLT the sampling distribution of $\bar{x}$ is approximately normal ✓

Step 3 — Calculate: $SE = \frac{124}{\sqrt{40}} = \frac{124}{6.325} = 19.61$

$ME = 1.685 \times 19.61 = 33.04$

$CI = 894 \pm 33.04 = (860.96, 927.04)$

Step 4 — Interpret: We are 90% confident that the true mean sodium content of all frozen dinners of this brand is between 860.96 mg and 927.04 mg.

⚠️ Common AP Mistakes

🔑 AP Advice: On FRQs, write more rather than less. Lost points from missing steps are harder to recover than time spent writing clearly.

Framework Check 🎯

Practice Calculations 🧮

Problem: In a random sample of 250 households, 85 have a pet. Build a 99% CI for the true proportion. ($z^* = 2.576$)

1) What is $\hat{p}$?

2) What is the standard error? (Round to 4 decimal places)

3) What is the margin of error? (Round to 4 decimal places)

Which Procedure? 🔍

Select the correct procedure for each scenario.

Exit Quiz — Problem-Solving Workshop ✅

📊 Review & Applications

Part 7 of 7 — Comprehensive Review

Topics in This Part

🔑 Key Concept: This part brings together everything from the Confidence Intervals unit. Use it as your final review before the exam.

CI Procedure Decision Chart

Formula Reference

What Affects CI Width?

The Complete 4-Step Process

1. Identify: Name the procedure, state the parameter in context, give the confidence level.

2. Conditions: Check Random, 10%, and Normal/Large Counts (proportions) or Normal/Large Sample (means).

3. Calculate: Show the formula, plug in values, give both the ME and the interval.

4. Interpret: "We are C% confident that the true [parameter in context] is between [lower] and [upper]."

⚠️ Never say "probability" in the interpretation. Say "confident."

Interpretation vs. Meaning

Comprehensive Concept Check 🎯

Mixed Practice 🧮

Problem 1: $n = 400$, $\hat{p} = 0.55$, 95% CI. Calculate the margin of error. (Use $z^* = 1.96$, round to 3 decimal places)

Problem 2: $n = 50$, $\bar{x} = 72$, $s = 10$, $t^* = 2.010$ ($df = 49$). What is the upper bound of the CI?

Problem 3: You want ME $\leq 0.02$ for a 95% CI for a proportion (use $\hat{p} = 0.5$). What minimum $n$ is needed?

Quick Decisions 🔍

Final Exam — Confidence Intervals Unit ✅