6.3 Sampling Distributions & Central Limit Theorem

6.3 Sampling Distributions & Central Limit Theorem

In quality engineering, measuring an entire population of manufactured product is often cost-prohibitive or destructive. Consequently, quality decisions rely on statistics calculated from samples. A sampling distribution is the probability distribution of a sample statistic (such as sample mean $\bar{X}$ or sample proportion $\hat{p}$) obtained from all possible random samples of size $n$ drawn from a population.

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The Central Limit Theorem (CLT)

The Central Limit Theorem is one of the most powerful principles in statistics. It explains why normal distribution methods can be applied to real-world processes even when the underlying population is non-normal.

Formal Statement of CLT

Let $X_1, X_2, \dots, X_n$ be a random sample of size $n$ drawn from any population with finite mean $\mu$ and finite variance $\sigma^2$. As the sample size $n$ increases, the sampling distribution of the sample mean $\bar{X}$ approaches a normal distribution with:

• Expected Value (Mean): $E[\bar{X}] = \mu_{\bar{x}} = \mu$
• Variance: $Var(\bar{X}) = \sigma_{\bar{x}}^2 = \frac{\sigma^2}{n}$
• Standard Deviation (Standard Error): $SE_{\bar{x}} = \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

$$\bar{X} \xrightarrow{d} N\left(\mu, \frac{\sigma^2}{n}\right) \quad \text{as } n \to \infty$$

Key Rules of Thumb for CLT Application

1. Symmetric Populations: If the underlying population is already normal, $\bar{X}$ is perfectly normally distributed for any sample size ($n \ge 1$).
2. Moderately Skewed Populations: $\bar{X}$ becomes approximately normal for $n \ge 15$ to $20$.
3. Severely Skewed or Bimodal Populations: $n \ge 30$ is the standard ASQ benchmark rule required to safely assume normality of sample means.

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Standard Error of the Mean ($SE_{\bar{x}}$)

The Standard Error of the Mean quantifies the dispersion of sample means around the true population mean $\mu$.

$$SE_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \quad \text{(or } \frac{s}{\sqrt{n}} \text{ when } \sigma \text{ is unknown)}$$

The Square Root Rule

Because $SE$ is inversely proportional to $\sqrt{n}$, reducing standard error by half requires quadrupling the sample size $n$.

• Increasing $n$ from 25 to 100 reduces $SE$ by $50\%$.
• Increasing $n$ from 100 to 400 reduces $SE$ by another $50\%$.

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Sampling Distribution of Proportions (Attribute Data)

When evaluating attribute quality data (conforming vs. non-conforming), the sample proportion is $\hat{p} = \frac{X}{n}$, where $X$ is the number of non-conforming items in a sample of size $n$.

Normal Approximation for Proportions

By the Central Limit Theorem, the distribution of $\hat{p}$ approaches a normal distribution $N\left(p, \frac{p(1-p)}{n}\right)$ provided the sample size satisfies the two conditions:

$$np \ge 5 \quad \text{and} \quad n(1-p) \ge 5$$

• Expected Mean of Proportion: $\mu_{\hat{p}} = p$

• Standard Error of Proportion:

  $$SE_{\hat{p}} = \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$$

• Standard Normal Z-transformation for Proportions:

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

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Finite Population Correction (FPC) Factor

When sampling without replacement from a finite population of size $N$, individual draws are not independent. If the sample size $n$ represents more than $5\%$ of the population ($n/N > 0.05$), the standard error formula must be multiplied by the Finite Population Correction (FPC) factor:

$$\text{FPC} = \sqrt{\frac{N - n}{N - 1}}$$

Adjusted Standard Error Formulas

• For Sample Means: $$SE_{\bar{x},\text{adj}} = \left(\frac{\sigma}{\sqrt{n}}\right) \sqrt{\frac{N - n}{N - 1}}$$
• For Sample Proportions: $$SE_{\hat{p},\text{adj}} = \left(\sqrt{\frac{p(1-p)}{n}}\right) \sqrt{\frac{N - n}{N - 1}}$$

Notice that as population size $N \to \infty$, the FPC factor approaches $1.0$, rendering correction unnecessary for large production lots.

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Summary Table of Sampling Distributions & Standard Errors

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Worked Numerical Examples

Example 1: CLT Probability Calculation for Sample Means

A manufacturing process produces high-strength structural cables with mean tensile strength $\mu = 500\text{ N}$ and standard deviation $\sigma = 40\text{ N}$. The tensile strength distribution is positively skewed. A quality auditor selects a random sample of $n = 64$ cables.

What is the probability that the sample mean tensile strength $\bar{X}$ exceeds $510\text{ N}$?

1. Verify CLT Applicability: Sample size $n = 64 \ge 30$. $\bar{X}$ is approximately normally distributed.
2. Calculate Standard Error ($SE_{\bar{x}}$): $$SE_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{40}{\sqrt{64}} = \frac{40}{8} = 5.0\text{ N}$$
3. Calculate Z-score: $$Z = \frac{\bar{x} - \mu}{SE_{\bar{x}}} = \frac{510 - 500}{5.0} = \frac{10}{5.0} = 2.00$$
4. Find Probability: $$P(\bar{X} > 510) = P(Z > 2.00) = 1 - P(Z \le 2.00) = 1 - 0.9772 = 0.0228 \text{ (or } 2.28\%\text{)}$$

Example 2: Sampling Distribution of Proportions with FPC

An inventory batch consists of $N = 500$ precision valves. Historical data shows $p = 0.10$ defect rate. An inspector draws a sample of $n = 100$ valves without replacement.

Calculate the adjusted standard error of the sample proportion $SE_{\hat{p},\text{adj}}$.

1. Check FPC Threshold: $n / N = 100 / 500 = 0.20$ ($20\%$). Since $0.20 > 0.05$, FPC is required.
2. Compute Standard Error without Correction: $$SE_{\hat{p}} = \sqrt{\frac{0.10 \times 0.90}{100}} = \sqrt{0.0009} = 0.030$$
3. Compute FPC Multiplier: $$\text{FPC} = \sqrt{\frac{500 - 100}{500 - 1}} = \sqrt{\frac{400}{499}} = \sqrt{0.8016} \approx 0.8953$$
4. Calculate Adjusted Standard Error: $$SE_{\hat{p},\text{adj}} = 0.030 \times 0.8953 = 0.02686$$

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CQE Exam Tips & Strategy

• Sample Mean vs. Individual Value: Always check if a question asks for the probability of an individual item $X > k$ (use $\sigma$) or a sample mean $\bar{X} > k$ (use $\sigma / \sqrt{n}$).
• Impact of $n$ on Dispersion: The sampling distribution of $\bar{X}$ is always narrower (less variable) than the population distribution by a factor of $\sqrt{n}$.
• FPC Requirement Check: Remember to check $n/N > 0.05$ on finite batch questions before using basic $SE$ formulas.