Chapter 9: Sampling Distributions
Learning Objectives
- Distinguish between parameters and statistics using the PP-SS mnemonic
- Describe and construct a sampling distribution from all possible samples
- Apply the Central Limit Theorem to describe the sampling distribution of $\bar{x}$
- Calculate probabilities using the sampling distribution of $\bar{x}$
- Verify the Large Counts condition and find the sampling distribution of $\hat{p}$
- Distinguish bias from variability in sampling distributions
9.1 Parameters vs. Statistics
Before we can make sense of statistical inference, we need to clearly separate two types of numerical summaries: those that describe populations and those that describe samples.
Definition: Parameter vs. Statistic — PP-SS
A parameter is a numerical summary of a Population. It is fixed (though usually unknown). Common notation: $\mu$ (population mean), $\sigma$ (population standard deviation), $p$ (population proportion).
A statistic is a numerical summary of a Sample. It varies from sample to sample. Common notation: $\bar{x}$ (sample mean), $s$ (sample standard deviation), $\hat{p}$ (sample proportion).
Mnemonic — PP-SS: Population → Parameter, Sample → Statistic.
Sampling variability refers to the natural fact that different random samples from the same population will produce different values of a statistic. This is expected and not a flaw — it is the very thing that sampling distributions describe.
The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples of size $n$ from the same population. Understanding this distribution is the foundation of all statistical inference.
Example 9.1 — Building a Sampling Distribution of $\bar{x}$
Consider a small population of 5 values: $\{2, 4, 6, 8, 10\}$. The population mean is $\mu = (2+4+6+8+10)/5 = 6$.
Take all possible samples of size $n = 2$ (without replacement). There are $\binom{5}{2} = 10$ such samples:
| Sample | $\bar{x}$ | Sample | $\bar{x}$ |
|---|---|---|---|
| (2, 4) | 3 | (4, 6) | 5 |
| (2, 6) | 4 | (4, 8) | 6 |
| (2, 8) | 5 | (4, 10) | 7 |
| (2, 10) | 6 | (6, 8) | 7 |
| (4, 8) — see (4,8) | 6 | (6, 10) | 8 |
Wait — let's list them cleanly. The 10 samples and their $\bar{x}$ values: $(2,4)\to3$, $(2,6)\to4$, $(2,8)\to5$, $(2,10)\to6$, $(4,6)\to5$, $(4,8)\to6$, $(4,10)\to7$, $(6,8)\to7$, $(6,10)\to8$, $(8,10)\to9$.
The $\bar{x}$ values are: 3, 4, 5, 5, 6, 6, 7, 7, 8, 9.
The mean of all 10 sample means: $(3+4+5+5+6+6+7+7+8+9)/10 = 60/10 = 6$.
Key result: The sampling distribution of $\bar{x}$ is centered exactly at the population mean $\mu = 6$. This is not a coincidence — $\bar{x}$ is an unbiased estimator of $\mu$.
Using the sampling distribution from Example 9.1 (values: 3, 4, 5, 5, 6, 6, 7, 7, 8, 9), find $P(\bar{x} \geq 7)$.
Show Answer
$P(\bar{x} \geq 7) = 4/10 = \mathbf{0.40}$.
9.2 Sampling Distribution of $\bar{x}$ and the Central Limit Theorem
For most real applications, it is impractical to list every possible sample. Instead, we rely on two powerful theoretical results about the sampling distribution of $\bar{x}$.
When the population is Normal: If the population distribution is $N(\mu, \sigma)$, then the sampling distribution of $\bar{x}$ is exactly $N\!\left(\mu,\, \dfrac{\sigma}{\sqrt{n}}\right)$ for any sample size $n$.
Theorem: The Central Limit Theorem (CLT)
For a random sample of size $n$ from any population with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of $\bar{x}$ is approximately Normal when $n$ is large enough (rule of thumb: $n \geq 30$):
- Mean: $\mu_{\bar{x}} = \mu$
- Standard deviation (standard error): $\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}}$
The approximation improves as $n$ increases. The shape of the original population does not matter — the sampling distribution of $\bar{x}$ becomes approximately Normal for large $n$.
The quantity $\sigma_{\bar{x}} = \sigma/\sqrt{n}$ is called the standard error of $\bar{x}$. It measures how much $\bar{x}$ typically varies from sample to sample. Note carefully: it is $\sigma/\sqrt{n}$, not $\sigma$ itself.
Example 9.2 — Normal Population, Exact Distribution
Heights of adult males follow a Normal distribution with $\mu = 70$ in and $\sigma = 3$ in. Random samples of $n = 25$ are taken.
Since the population is Normal, $\bar{x} \sim N\!\left(70,\, \dfrac{3}{\sqrt{25}}\right) = N(70,\, 0.6)$ exactly.
Find $P(\bar{x} > 71)$:
$$z = \frac{71 - 70}{0.6} = \frac{1}{0.6} \approx 1.67$$
$P(\bar{x} > 71) = P(z > 1.67) \approx \mathbf{0.0478}$.
There is about a 4.78% chance that the sample mean height exceeds 71 inches.
Example 9.3 — Skewed Population, CLT in Action
Monthly electric bills in a city have $\mu = \$120$ and $\sigma = \$30$. The distribution is right-skewed. A random sample of $n = 36$ bills is taken.
By the CLT (since $n = 36 \geq 30$): $\bar{x} \approx N\!\left(120,\, \dfrac{30}{\sqrt{36}}\right) = N(120,\, 5)$.
Find $P(\bar{x} < 115)$:
$$z = \frac{115 - 120}{5} = \frac{-5}{5} = -1$$
$P(\bar{x} < 115) = P(z < -1) \approx \mathbf{0.1587}$.
There is about a 15.87% chance the sample mean bill is below $115.
SAT scores have $\mu = 1050$ and $\sigma = 200$. A random sample of $n = 100$ students is taken. What is $P(\bar{x} > 1070)$?
Show Answer
$z = (1070 - 1050)/20 = 20/20 = 1.00$.
$P(\bar{x} > 1070) = P(z > 1.00) \approx \mathbf{0.1587}$.
AP Exam Tip: Always verify three conditions before using the Normal sampling distribution for $\bar{x}$: (1) Random — was the sample randomly selected? (2) Normal/Large Sample — is the population Normal, or is $n \geq 30$ for the CLT? (3) Independence — is $n < 10\%$ of the population? Missing any condition costs points on the AP exam.
As $n$ increases, the sampling distribution of $\bar{x}$ narrows and becomes more Normal. The three curves show the effect of increasing sample size (gray: $n=1$, blue: $n=10$, green: $n=30$).
Figure 9.1 — Central Limit Theorem: Effect of Sample Size on $\bar{x}$ Distribution
9.3 Sampling Distribution of $\hat{p}$
When our variable of interest is categorical (success/failure), the relevant statistic is the sample proportion:
$$\hat{p} = \frac{X}{n} = \frac{\text{number of successes}}{\text{sample size}}$$
Theorem: Sampling Distribution of $\hat{p}$
When the following conditions are met, $\hat{p}$ has an approximately Normal distribution:
- Random: Data come from a random sample or randomized experiment.
- Large Counts: $np \geq 10$ AND $n(1-p) \geq 10$.
- Independence: $n < 10\%$ of the population (10% condition).
When conditions are met:
- Mean: $\mu_{\hat{p}} = p$
- Standard deviation: $\sigma_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}}$
Example 9.4 — Sampling Distribution of $\hat{p}$: Water Bottles
Suppose 35% of students at a large school carry a water bottle ($p = 0.35$). A random sample of $n = 80$ students is selected.
Check conditions:
- Random: specified ✓
- Large Counts: $np = 80(0.35) = 28 \geq 10$ ✓ and $n(1-p) = 80(0.65) = 52 \geq 10$ ✓
- Independence: assume 80 < 10% of large school ✓
$\hat{p} \sim N\!\left(0.35,\, \sqrt{\dfrac{0.35 \cdot 0.65}{80}}\right) = N(0.35,\, \sqrt{0.002844}) = N(0.35,\, 0.0533)$
Find $P(\hat{p} > 0.40)$:
$$z = \frac{0.40 - 0.35}{0.0533} \approx \frac{0.05}{0.0533} \approx 0.938$$
$P(\hat{p} > 0.40) = P(z > 0.938) \approx \mathbf{0.174}$. There is about a 17.4% chance more than 40% of the sampled students carry water bottles.
Example 9.5 — Sampling Distribution of $\hat{p}$: Voter Support
Suppose 60% of registered voters in a city support a ballot measure ($p = 0.60$). A random sample of $n = 200$ voters is polled.
$\sigma_{\hat{p}} = \sqrt{\dfrac{0.60 \cdot 0.40}{200}} = \sqrt{0.0012} \approx 0.0346$
Find $P(\hat{p} < 0.55)$:
$$z = \frac{0.55 - 0.60}{0.0346} \approx \frac{-0.05}{0.0346} \approx -1.44$$
$P(\hat{p} < 0.55) = P(z < -1.44) \approx \mathbf{0.0749}$. There is about a 7.49% chance fewer than 55% of the sample supports the measure.
A fair coin is flipped $n = 50$ times and $\hat{p}$ = proportion of heads. Find $\mu_{\hat{p}}$, $\sigma_{\hat{p}}$, and $P(\hat{p} \geq 0.6)$.
Show Answer
$\sigma_{\hat{p}} = \sqrt{0.5 \cdot 0.5 / 50} = \sqrt{0.005} \approx 0.0707$.
$z = (0.6 - 0.5)/0.0707 \approx 1.414$.
$P(\hat{p} \geq 0.6) = P(z \geq 1.414) \approx \mathbf{0.0786}$.
AP Exam Tip: The Large Counts condition ($np \geq 10$ and $n(1-p) \geq 10$) is required to use the Normal approximation for $\hat{p}$. Always state and verify this condition explicitly on the AP exam — failure to do so will cost points on free-response questions.
Normal distribution for $\hat{p}$ from Example 9.4: $\hat{p} \sim N(0.35, 0.0533)$. The shaded right tail shows $P(\hat{p} > 0.40) \approx 0.174$.
Figure 9.3 — Sampling Distribution of $\hat{p}$: Normal(0.35, 0.0533)
9.4 Bias and Variability in Sampling Distributions
Two properties determine the quality of a sampling distribution: bias (accuracy) and variability (precision).
Definition: Bias vs. Variability
Bias = accuracy of the center. A statistic is unbiased if the mean of its sampling distribution equals the true parameter. Both $\bar{x}$ and $\hat{p}$ are unbiased estimators: $\mu_{\bar{x}} = \mu$ and $\mu_{\hat{p}} = p$.
Variability = precision (spread). A smaller standard error means the statistic is more consistently close to the true parameter. Variability is controlled by sample size $n$ — larger $n$ produces smaller standard error.
Target analogy: Bias measures whether you are aiming at the right place (centered on the bullseye). Variability measures how tightly clustered your shots are.
Example 9.6 — Bias vs. Variability in Practice
Two sampling methods are used to estimate a population mean $\mu$:
- Method A: $\bar{x}$ is centered at $\mu$ (unbiased) but has high variability (small $n$).
- Method B: $\bar{x}$ is centered away from $\mu$ (biased) but has low variability (large $n$).
Method A is preferred. An unbiased estimator with high variability can be improved by increasing $n$. A biased estimator — even with low variability — will consistently miss the true value. Increasing $n$ reduces variability but does NOT fix bias.
A news website uses a voluntary response poll (readers self-select to participate) to estimate the proportion of Americans who support a policy. Is the resulting $\hat{p}$ likely to be biased or variable? Explain.
Show Answer
AP Exam Tip: The only way to reduce bias is to use better sampling methods — specifically, random sampling. Increasing $n$ reduces variability (standard error) but cannot fix bias. This distinction is frequently tested on the AP exam.
Four combinations of bias and variability — like four groups of shots at a target. Green (bottom-left) is the ideal: low bias, low variability.
Figure 9.2 — Bias and Variability: Four Combinations
Practice Problems
A population has $\mu = 50$ and $\sigma = 8$. Random samples of $n = 64$ are taken. Find $\mu_{\bar{x}}$, $\sigma_{\bar{x}}$, and $P(\bar{x} > 52)$.
Show Solution
$\sigma_{\bar{x}} = 8/\sqrt{64} = 8/8 = 1$.
$z = (52 - 50)/1 = 2.00$.
$P(\bar{x} > 52) = P(z > 2.00) \approx \mathbf{0.0228}$.
Is $\bar{x}$ an unbiased estimator of $\mu$? Explain what "unbiased" means in the context of sampling distributions.
Show Solution
A poll finds $\hat{p} = 0.58$ supports a ballot measure, based on $n = 400$. The true proportion is $p = 0.55$. Find $P(\hat{p} \geq 0.58)$ and interpret.
Show Solution
$z = (0.58 - 0.55)/0.0249 \approx 1.205$.
$P(\hat{p} \geq 0.58) = P(z \geq 1.205) \approx \mathbf{0.114}$.
If the true proportion is 0.55, there is about an 11.4% chance that a sample of 400 would show 58% or more in support. This is not unusually rare.
Heights follow a skewed distribution: $\mu = 160$ cm, $\sigma = 25$ cm. For $n = 100$, describe the sampling distribution of $\bar{x}$. Can you use Normal calculations? Why?
Show Solution
Yes, Normal calculations can be used because the large sample size satisfies the CLT condition.
A school has 30% left-handed students. A sample of $n = 90$ is taken. (a) Verify the Large Counts condition. (b) Find $P(\hat{p} > 0.35)$.
Show Solution
(b) $\sigma_{\hat{p}} = \sqrt{0.30 \cdot 0.70/90} = \sqrt{0.002\overline{3}} \approx 0.0483$.
$z = (0.35 - 0.30)/0.0483 \approx 1.035$.
$P(\hat{p} > 0.35) \approx P(z > 1.035) \approx \mathbf{0.150}$.
Doubling the sample size from $n = 100$ to $n = 200$: by what factor does $\sigma_{\bar{x}}$ decrease?
Show Solution
Ratio $= \dfrac{\sigma/\sqrt{200}}{\sigma/\sqrt{100}} = \dfrac{\sqrt{100}}{\sqrt{200}} = \sqrt{\dfrac{100}{200}} = \sqrt{0.5} = \dfrac{1}{\sqrt{2}} \approx 0.707$.
$\sigma_{\bar{x}}$ decreases by a factor of $\sqrt{2} \approx 1.414$, or decreases to about 70.7% of its original value. To cut standard error in half, you must quadruple $n$.
A population is highly skewed right. For which sample size is the CLT most likely to apply: $n = 5$, $n = 15$, or $n = 40$? Explain.
Show Solution
AP FRQ: A company claims 20% of its light bulbs fail within 1 year. A quality inspector tests $n = 150$ bulbs and finds $\hat{p} = 0.27$. (a) State the parameter and statistic. (b) Find $\mu_{\hat{p}}$ and $\sigma_{\hat{p}}$ assuming $p = 0.20$. (c) Is $\hat{p} = 0.27$ surprising? Find $P(\hat{p} \geq 0.27)$ and interpret.
Show Solution
(b) Check Large Counts: $np = 150(0.20) = 30 \geq 10$ ✓; $n(1-p) = 150(0.80) = 120 \geq 10$ ✓.
$\mu_{\hat{p}} = 0.20$.
$\sigma_{\hat{p}} = \sqrt{0.20 \cdot 0.80/150} = \sqrt{0.001\overline{06}} \approx 0.0327$.
(c) $z = (0.27 - 0.20)/0.0327 \approx 2.14$.
$P(\hat{p} \geq 0.27) = P(z \geq 2.14) \approx \mathbf{0.016}$.
If the true failure rate is 20%, there is only about a 1.6% chance of observing $\hat{p} \geq 0.27$ in a sample of 150. This is quite surprising and provides evidence that the company's claimed 20% failure rate may be too low.
📋 Chapter Summary
Sampling Distribution of $\bar{x}$
$\mu_{\bar{x}} = \mu$ — the sampling distribution of $\bar{x}$ is centered at the population mean. The sample mean is an unbiased estimator.
$\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}}$ — spread decreases as sample size grows. Larger $n$ → more precise estimates.
For large $n$ (usually $n \geq 30$), $\bar{x}$ is approximately Normal regardless of the population shape.
$\mu_{\hat{p}} = p$, $\sigma_{\hat{p}} = \sqrt{p(1-p)/n}$. Normal when $np \geq 10$ and $n(1-p) \geq 10$.
Key Conditions
- Random — sample must be randomly selected (SRS or equivalent)
- 10% Condition — $n \leq 0.10N$ to treat observations as independent
- Normal/Large Sample — $n \geq 30$ (CLT) or population is Normal
- Success/Failure (for proportions) — $np \geq 10$ and $n(1-p) \geq 10$