Sampling Distribution of the Sample Mean

🎯 The Central Limit Theorem

Part 1 of 7 — The Most Important Theorem in Statistics

From Population to Sample

When we take a sample of size $n$ from a population, the sample mean $\bar{x}$ varies from sample to sample. The sampling distribution of $\bar{x}$ describes this variation.

The Central Limit Theorem (CLT)

For a random sample of size $n$ from a population with mean $\mu$ and standard deviation $\sigma$:

$\bar{X} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right) \text{ (approximately, for large } n\text{)}$

🔑 The CLT works regardless of the shape of the population distribution — as long as $n$ is large enough.

Why It Matters

Even if data is skewed, the distribution of sample means will be approximately normal for large $n$. This is why so many inference procedures use the normal distribution.

Seeing a Sampling Distribution by Simulation

A sampling distribution is the distribution of a statistic over all possible samples of a given size. You can approximate it by simulation: assume a value for the parameter, generate many random samples, compute the statistic for each one, and graph the results. The larger the sample size in each simulated sample, the closer the dotplot of $\bar{x}$ values looks to a normal curve.

CLT Concept Check 🎯

CLT Calculations 🧮

A population has $\mu = 50$ and $\sigma = 12$. Samples of size $n = 36$ are drawn.

1) What is the mean of the sampling distribution of $\bar{x}$?

2) What is the standard error of $\bar{x}$?

3) For $n = 144$, what would the standard error be?

📊 Distribution of Sample Means

Part 2 of 7 — Applying the CLT

When Is the CLT Valid?

Finding Probabilities About $\bar{x}$

To find $P(\bar{x} > k)$:

1. Calculate $z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$
2. Use the z-table

Example: Population: $\mu = 100$, $\sigma = 15$. Sample $n = 25$. Find $P(\bar{x} > 106)$.

$z = \frac{106 - 100}{15/\sqrt{25}} = \frac{6}{3} = 2$

$P(\bar{x} > 106) = P(Z > 2) = 1 - 0.9772 = 0.0228$

Sample Means Practice 🎯

Sampling Distribution Calculations 🧮

Exam scores: $\mu = 75$, $\sigma = 10$, $n = 25$.

1) Standard error of $\bar{x}$ = ?

2) $z$-score for $\bar{x} = 79$: $z = (79 - 75) / SE$ = ?

3) $P(\bar{x} > 79) \approx P(Z > 2)$. Using $P(Z \leq 2) = 0.9772$, the probability is?

📊 Distribution of Sample Proportions

Part 3 of 7 — From Counts to Proportions

Topics in This Part

🔑 Key Concept: Parts 1–2 focused on sample means ($\bar{x}$). Now we shift to sample proportions ($\hat{p}$), which are used whenever the variable is categorical (yes/no, success/failure).

What You'll Master in Part 3

• Defining and computing the sample proportion $\hat{p}$
• Describing the sampling distribution of $\hat{p}$ (center, spread, shape)
• Checking conditions for the normal approximation
• Using $z$-scores to find probabilities about $\hat{p}$

📐 The Sample Proportion

When a variable is categorical (e.g., "supports policy" vs. "does not"), we summarize samples with a proportion rather than a mean.

$\boxed{\hat{p} = \frac{\text{number of successes}}{n}}$

Example: In a sample of 200 voters, 120 support a candidate. Then:

$\hat{p} = \frac{120}{200} = 0.60$

Sampling Distribution of $\hat{p}$

If repeated random samples of size $n$ are drawn from a population where the true proportion is $p$:

🔑 Key Concept: The sampling distribution of $\hat{p}$ is centered at the true proportion $p$ — so $\hat{p}$ is an unbiased estimator of $p$.

Normal Approximation Conditions

The sampling distribution of $\hat{p}$ is approximately normal when:

$\boxed{np \geq 10 \quad \text{and} \quad n(1-p) \geq 10}$

This ensures enough successes AND failures for the bell curve to be a good model.

⚠️ Warning: If either condition fails, the normal approximation is not valid. You would need exact binomial probabilities instead.

Finding Probabilities About $\hat{p}$

To find $P(\hat{p} > k)$, standardize:

$z = \frac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Then use the $z$-table.

Worked Example: A factory produces items with 5% defect rate ($p = 0.05$). In a sample of $n = 500$:

$\sigma_{\hat{p}} = \sqrt{\frac{0.05 \times 0.95}{500}} = \sqrt{\frac{0.0475}{500}} = \sqrt{0.000095} = 0.00975$

Check: $np = 25 \geq 10$ ✓ and $n(1-p) = 475 \geq 10$ ✓

$P(\hat{p} > 0.07) = P\left(z > \frac{0.07 - 0.05}{0.00975}\right) = P(z > 2.05) = 1 - 0.9798 = 0.0202$

Sample Proportions Concept Check 🎯

Standard Deviation of $\hat{p}$ Practice 🧮

A political poll samples $n = 900$ voters from a state where 40% support a ballot measure ($p = 0.40$).

1) What is the mean of the sampling distribution of $\hat{p}$?

2) Calculate $\sigma_{\hat{p}} = \sqrt{p(1-p)/n}$. Round to 4 decimal places.

3) Is the normal approximation valid? How many expected successes are there? ($np = ?$)

Proportions Concepts Check 🔍

📝 AP-Style Worked Example

Problem: A large university reports that 70% of students live off campus. A random sample of 150 students is selected.

(a) Describe the sampling distribution of $\hat{p}$.

Solution:

• Shape: Check: $np = 150(0.70) = 105 \geq 10$ ✓ and $n(1-p) = 150(0.30) = 45 \geq 10$ ✓. Approximately normal.
• Center: $\mu_{\hat{p}} = 0.70$
• Spread: $\sigma_{\hat{p}} = \sqrt{\frac{0.70 \times 0.30}{150}} = \sqrt{\frac{0.21}{150}} = \sqrt{0.0014} \approx 0.0374$

(b) Find the probability that $\hat{p} < 0.65$.

$z = \frac{0.65 - 0.70}{0.0374} = \frac{-0.05}{0.0374} = -1.34$

$P(\hat{p} < 0.65) = P(z < -1.34) = 0.0901$

🔑 AP Tip: Always structure your answer as Shape–Center–Spread and show the condition checks. The AP rubric awards separate points for each.

Exit Quiz — Distribution of Sample Proportions ✅

� Standard Error

Part 4 of 7 — Measuring the Precision of Estimates

Topics in This Part

🔑 Key Concept: Standard error (SE) measures how much a sample statistic typically varies from sample to sample. Smaller SE = more precise estimate.

What You'll Master in Part 4

• Distinguishing standard deviation from standard error
• Computing SE for both $\bar{x}$ and $\hat{p}$
• Understanding how sample size affects precision
• Applying the 10% condition for independence

📐 Standard Deviation vs. Standard Error

These two concepts are frequently confused on the AP exam:

⚠️ Warning: The AP exam specifically tests whether you know the difference. Standard deviation describes the data; standard error describes the statistic.

Standard Error Formulas

🔑 Key Concept: In practice, we rarely know the population parameters $\sigma$ and $p$. So we estimate the SE by plugging in sample statistics ($s$ and $\hat{p}$).

The 10% Condition

When sampling without replacement, observations are not truly independent. However, if the sample is less than 10% of the population, the dependence is negligible:

$\boxed{n < 0.10 \times N}$

where $N$ is the population size.

Why it matters: The SE formulas assume independent observations. The 10% condition ensures this assumption is approximately met.

Example: Polling 1,000 voters from a city of 50,000 registered voters.

• Check: $1000 < 0.10 \times 50{,}000 = 5{,}000$ ✓ → Independence is reasonable.

How Sample Size Affects SE

$SE \propto \frac{1}{\sqrt{n}}$

🔑 AP Tip: To halve the margin of error, you must quadruple the sample size. This is a frequently tested relationship.

Standard Error Concepts 🎯

SE Calculations 🧮

1) A sample of 100 students has $s = 15$. Find $SE_{\bar{x}}$.

2) In a sample of 400, $\hat{p} = 0.30$. Find $SE_{\hat{p}}$ (round to 4 decimal places).

3) A researcher surveys 500 people from a town of 4,000. Is the 10% condition satisfied? Enter "yes" or "no".

Concepts Check 🔍

📝 AP-Style Worked Example

Problem: A state health department reports that the average wait time at emergency rooms is 45 minutes. A hospital takes a random sample of 50 patients and finds $\bar{x} = 52$ minutes with $s = 20$ minutes.

(a) Calculate the standard error of $\bar{x}$.

$SE_{\bar{x}} = \frac{s}{\sqrt{n}} = \frac{20}{\sqrt{50}} = \frac{20}{7.071} \approx 2.828 \text{ minutes}$

(b) Interpret the SE in context.

In repeated random samples of 50 patients from this hospital, the sample mean wait time would typically differ from the true mean by about 2.83 minutes.

(c) How large a sample would be needed to achieve $SE \leq 1$ minute?

$1 \geq \frac{20}{\sqrt{n}} \implies \sqrt{n} \geq 20 \implies n \geq 400$

🔑 AP Tip: When interpreting SE, always connect it to the context: "The sample [statistic] would typically vary by about [SE] [units] from the true [parameter] across repeated samples."

Exit Quiz — Standard Error ✅

✅ Conditions for Inference

Part 5 of 7 — The Three Conditions You Must Always Check

Topics in This Part

🔑 Key Concept: Every inference procedure on the AP exam requires you to check conditions before performing the test or building a confidence interval. Missing conditions = lost points.

The Three Conditions Framework

Every inference procedure requires these three conditions:

⚠️ Warning: Simply writing "conditions met" earns zero credit. You must name the condition, check it with numbers, and state whether it is satisfied.

🎲 Condition 1: Random

What to check: Was the data collected using a random process?

• Random sample from a population → inference about population parameters
• Random assignment in an experiment → inference about causation

How to state it on the AP exam:

• ✅ "The problem states that a random sample of 200 adults was selected."
• ❌ "Random ✓" (too vague — no credit)

📏 Condition 2: Independence (10% Condition)

What to check: Are individual observations approximately independent?

$\boxed{n \leq 0.10 \times N}$

How to state it:

• ✅ "The sample of 150 is less than 10% of all registered voters in the state (over 5 million), so observations are approximately independent."
• ❌ "10% condition met" (no numbers shown)

🔑 Key Concept: This condition is automatically satisfied when sampling from a functionally infinite population (all U.S. adults, all manufactured items, etc.).

📊 Condition 3: Normal/Large Sample

This condition differs depending on the parameter:

For Means ($\bar{x}$):

For Proportions ($\hat{p}$):

$\boxed{np \geq 10 \quad \text{and} \quad n(1-p) \geq 10}$

When checking with sample data (no known $p$): $n\hat{p} \geq 10 \quad \text{and} \quad n(1-\hat{p}) \geq 10$

How to state it:

• ✅ "The sample size is $n = 50 \geq 30$, so by the CLT, the sampling distribution of $\bar{x}$ is approximately normal."
• ✅ "$n\hat{p} = 200(0.65) = 130 \geq 10$ and $n(1-\hat{p}) = 200(0.35) = 70 \geq 10$."

⚠️ Warning: For proportions, you must check BOTH $np$ and $n(1-p)$. Checking only one is incomplete.

Conditions Concept Check 🎯

Checking Conditions Practice 🧮

A random sample of 80 students from a university of 12,000 finds that 52 prefer online classes.

1) What fraction of the population is the sample? Express as a percentage (round to 1 decimal).

2) How many "successes" (prefer online) are in the sample?

3) How many "failures" (do not prefer online) are in the sample?

Condition Identification 🔍

📝 AP-Style Condition Check (Full Credit Response)

Problem: A quality engineer takes a random sample of 150 batteries from a day's production of 2,000 batteries and finds that 12 are defective. She wants to construct a 95% confidence interval for the true proportion of defective batteries.

Check all conditions for inference.

Model Response:

1. Random: The problem states that the engineer selected a random sample of 150 batteries. ✓

2. 10% Condition: $n = 150$ is less than 10% of the day's production of $N = 2{,}000$ batteries ($150 < 200$). Observations are approximately independent. ✓

3. Normal (Large Counts):

• Successes: $n\hat{p} = 12 \geq 10$ ✓
• Failures: $n(1-\hat{p}) = 150 - 12 = 138 \geq 10$ ✓

The sampling distribution of $\hat{p}$ is approximately normal.

All conditions are met. We may proceed with the one-proportion $z$-interval.

🔑 AP Tip: This three-part format (name → check → conclude) is exactly what AP readers look for. Practice writing it out every time.

Exit Quiz — Conditions for Inference ✅

� Problem-Solving Workshop

Part 6 of 7 — Putting It All Together

Topics in This Part

🔑 Key Concept: AP free-response questions on sampling distributions typically require you to (1) describe the sampling distribution, (2) check conditions, and (3) calculate a probability. Practice doing all three in sequence.

Problem-Solving Framework

For any sampling distribution problem, follow these steps:

1. Identify the parameter and statistic ($\mu$ and $\bar{x}$, or $p$ and $\hat{p}$)
2. Check conditions (Random, 10%, Normal/Large Sample)
3. Describe the sampling distribution (Shape, Center, Spread)
4. Calculate using the $z$-score formula
5. Interpret in context

📝 Worked Example 1: Sample Means

Problem: The weights of apples at a farm are normally distributed with $\mu = 150$ g and $\sigma = 20$ g. A random sample of 25 apples is selected.

(a) Describe the sampling distribution of $\bar{x}$.

Solution:

• Shape: Population is normal, so the sampling distribution of $\bar{x}$ is exactly normal for any $n$.
• Center: $\mu_{\bar{x}} = 150$ g
• Spread: $\sigma_{\bar{x}} = 20/\sqrt{25} = 20/5 = 4$ g

$\bar{X} \sim N(150, 4)$

(b) Find $P(\bar{x} > 158)$.

$z = \frac{158 - 150}{4} = \frac{8}{4} = 2.00$

$P(\bar{x} > 158) = P(Z > 2.00) = 1 - 0.9772 = 0.0228$

(c) Find $P(145 < \bar{x} < 155)$.

$z_1 = \frac{145 - 150}{4} = -1.25 \qquad z_2 = \frac{155 - 150}{4} = 1.25$

$P(145 < \bar{x} < 155) = P(-1.25 < Z < 1.25) = 0.8944 - 0.1056 = 0.7888$

📝 Worked Example 2: Sample Proportions

Problem: A website reports a 25% click-through rate ($p = 0.25$). A marketer samples 200 visitors.

(a) Check whether the normal approximation is valid.

• $np = 200(0.25) = 50 \geq 10$ ✓
• $n(1-p) = 200(0.75) = 150 \geq 10$ ✓

(b) Find $P(\hat{p} < 0.20)$.

$\sigma_{\hat{p}} = \sqrt{\frac{0.25 \times 0.75}{200}} = \sqrt{\frac{0.1875}{200}} = \sqrt{0.0009375} = 0.03062$

$z = \frac{0.20 - 0.25}{0.03062} = \frac{-0.05}{0.03062} = -1.63$

$P(\hat{p} < 0.20) = P(Z < -1.63) = 0.0516$

🔑 AP Tip: About 5% of the time, a sample of 200 would show less than 20% click-through — rare enough to be noteworthy, but not extreme.

Problem-Solving Check 🎯

Multi-Step Practice 🧮

The mean household income in a city is $\mu =$ $65,000 with $\sigma =$ $15,000. A random sample of 225 households is drawn.

1) What is the standard error of $\bar{x}$?

2) What is the $z$-score for $\bar{x} =$ $67,000?

3) What is $P(\bar{x} > 67{,}000)$ for $\bar{x}$ in dollars (i.e. $67,000)? (Use $P(Z > 2) = 0.0228$)

⚠️ Common Mistakes on the AP Exam

🔑 AP Tip: The most common error is using $\sigma$ where you need $\sigma/\sqrt{n}$. The sampling distribution is ALWAYS narrower than the population distribution.

Error Identification 🔍

Exit Quiz — Problem-Solving Workshop ✅

🎓 Review & Applications

Part 7 of 7 — Comprehensive Review

Complete Summary

Three Conditions for Inference

🔑 AP Tip: This entire topic is foundational — confidence intervals and hypothesis tests (the core of AP Stats) all rely on sampling distributions. Master this, and the rest follows.

📊 Means vs. Proportions: Side-by-Side

⚠️ Warning: A common AP exam trap is using the wrong formula — applying the proportions SE when the problem involves means, or vice versa. Always identify the data type first.

Key Relationships to Remember

1. Larger $n$ → smaller SE → narrower sampling distribution → more precision
2. To halve SE: quadruple $n$
3. SE measures precision, not accuracy — a biased method has small SE but wrong center
4. Unbiased: $E(\bar{x}) = \mu$ and $E(\hat{p}) = p$ — both estimators are unbiased
5. The 10% condition matters most for small populations — for large populations (states, countries), it is automatically satisfied

Comprehensive Review 🎯

Mixed Practice 🧮

1) Population: $\mu = 80$, $\sigma = 12$, $n = 144$. Find $\sigma_{\bar{x}}$.

2) Population: $p = 0.35$, $n = 500$. Find $\sigma_{\hat{p}}$ (round to 4 decimal places).

3) To reduce $\sigma_{\bar{x}}$ from $4$ to $1$, you must multiply $n$ by what factor?

Quick Identification 🔍

Final Exam — Sampling Distributions ✅