6.4 Point & Interval Estimation

6.4 Point & Interval Estimation

Statistical estimation enables quality engineers to make quantitative inferences about population parameters (such as mean $\mu$, variance $\sigma^2$, or proportion $p$) based on sample data. Estimation is divided into two primary approaches: point estimation (a single calculated value) and interval estimation (a range of plausible values bounded by a confidence level).

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Point Estimation & Desirable Estimator Properties

A point estimator $\hat{\theta}$ is a sample statistic used to estimate an unknown population parameter $\theta$. An ideal point estimator satisfies four mathematical criteria:

1. Unbiasedness: The expected value of the estimator equals the true parameter value:

   $$E[\hat{\theta}] = \theta$$
   • Sample mean $\bar{x}$ is an unbiased estimator of $\mu$.
   • Sample variance $s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$ is an unbiased estimator of $\sigma^2$.

2. Efficiency: Among all unbiased estimators, the efficient estimator has the minimum variance:

   $$Var(\hat{\theta}_1) < Var(\hat{\theta}_2)$$

3. Consistency: As sample size $n \to \infty$, the estimator converges in probability to the true parameter value, causing $Var(\hat{\theta}) \to 0$.

4. Sufficiency: The estimator uses all information contained in the sample concerning the parameter.

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Confidence Intervals for Population Mean ($\mu$)

A Confidence Interval (CI) provides a range $[LBL, UBL]$ within which the true population parameter is expected to fall with a specified confidence level $1 - \alpha$ (typically $90\%$, $95\%$, or $99\%$).

$$\text{CI} = \text{Point Estimate} \pm \text{Margin of Error} = \hat{\theta} \pm (\text{Critical Value} \times \text{Standard Error})$$

Case 1: Mean $\mu$ with Known Population Standard Deviation $\sigma$

When population standard deviation $\sigma$ is known (or sample size $n \ge 30$), use the Standard Normal (Z) Distribution:

$$\text{CI}_{1-\alpha}(\mu) = \bar{x} \pm Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right)$$

Common Z Critical Values ($Z_{\alpha/2}$):

• $90\%$ Confidence ($\alpha = 0.10$): $Z_{0.05} = 1.645$
• $95\%$ Confidence ($\alpha = 0.05$): $Z_{0.025} = 1.960$
• $99\%$ Confidence ($\alpha = 0.01$): $Z_{0.005} = 2.576$

Case 2: Mean $\mu$ with Unknown $\sigma$ (Small Samples $n < 30$)

When population standard deviation $\sigma$ is unknown and must be estimated by sample standard deviation $s$, use Student's t-Distribution with $\nu = n - 1$ degrees of freedom:

$$\text{CI}_{1-\alpha}(\mu) = \bar{x} \pm t_{\alpha/2, \, n-1} \left( \frac{s}{\sqrt{n}} \right)$$

Properties of t-distribution: Flatter peak and wider tails than normal distribution to account for extra uncertainty in estimating $s$; approaches Z-distribution as $n \to \infty$.

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Confidence Intervals for Population Variance ($\sigma^2$) and Standard Deviation ($\sigma$)

Because variance is bounded at zero and squared deviations follow a skewed distribution, confidence intervals for variance rely on the Chi-Square ($\chi^2$) Distribution with $\nu = n - 1$ degrees of freedom.

Two-Sided Confidence Interval for Variance ($\sigma^2$)

$$\frac{(n - 1)s^2}{\chi^2_{\alpha/2, \, n-1}} \le \sigma^2 \le \frac{(n - 1)s^2}{\chi^2_{1 - \alpha/2, \, n-1}}$$

Two-Sided Confidence Interval for Standard Deviation ($\sigma$)

Take the square root of all upper and lower bounds:

$$\sqrt{\frac{(n - 1)s^2}{\chi^2_{\alpha/2, \, n-1}}} \le \sigma \le \sqrt{\frac{(n - 1)s^2}{\chi^2_{1 - \alpha/2, \, n-1}}}$$

Note on Asymmetry: Unlike Z or t intervals, Chi-Square intervals are asymmetric around point estimate $s^2$ because the $\chi^2$ distribution is positively skewed.

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Confidence Intervals for Population Proportion ($p$)

For attribute data where $n\hat{p} \ge 5$ and $n(1-\hat{p}) \ge 5$, the normal approximation Wald confidence interval is:

$$\text{CI}_{1-\alpha}(p) = \hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$$

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Sample Size Determination Formulas

Quality engineers frequently determine the sample size required to estimate parameters within a specified margin of error $E$.

Sample Size for Estimating Population Mean $\mu$

$$n = \left( \frac{Z_{\alpha/2} \cdot \sigma}{E} \right)^2$$

Sample Size for Estimating Population Proportion $p$

$$n = \frac{Z_{\alpha/2}^2 \cdot p(1-p)}{E^2}$$

(If historical $p$ is unknown, use worst-case planning value $p = 0.50$ to yield maximum sample size $n = \frac{Z_{\alpha/2}^2 \cdot 0.25}{E^2}$.)

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Summary Matrix of Confidence Interval Formulas

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Worked Numerical Examples

Example 1: Student's t Confidence Interval for Mean

A quality engineer inspects $n = 16$ machined parts. The sample mean shaft width is $\bar{x} = 25.40\text{ mm}$ with sample standard deviation $s = 0.12\text{ mm}$. Construct a $95\%$ confidence interval for true mean width $\mu$.

1. Select Distribution: Population $\sigma$ is unknown, $n = 16 < 30$. Use Student's t distribution with $\nu = 16 - 1 = 15$ degrees of freedom.
2. Critical Value: For $95\%$ confidence ($\alpha = 0.05$), $t_{0.025, 15} = 2.131$.
3. Calculate Standard Error: $$SE = \frac{s}{\sqrt{n}} = \frac{0.12}{\sqrt{16}} = \frac{0.12}{4} = 0.030\text{ mm}$$
4. Calculate Margin of Error: $$ME = 2.131 \times 0.030 = 0.06393\text{ mm}$$
5. Construct Interval: $$25.40 \pm 0.0639 = [25.3361\text{ mm}, \; 25.4639\text{ mm}]$$

Example 2: Chi-Square Confidence Interval for Variance

Using the same sample data ($n = 16$, $s = 0.12\text{ mm}$, $s^2 = 0.0144\text{ mm}^2$, $\nu = 15$), construct a $95\%$ confidence interval for population variance $\sigma^2$.

1. Find Chi-Square Critical Values for $\nu = 15$:
   • Upper tail critical value ($\alpha/2 = 0.025$): $\chi^2_{0.025, 15} = 27.488$
   • Lower tail critical value ($1 - \alpha/2 = 0.975$): $\chi^2_{0.975, 15} = 6.262$
2. Calculate Lower Bound for $\sigma^2$: $$LBL = \frac{(16 - 1) \times 0.0144}{27.488} = \frac{15 \times 0.0144}{27.488} = \frac{0.216}{27.488} = 0.007858\text{ mm}^2$$
3. Calculate Upper Bound for $\sigma^2$: $$UBL = \frac{(16 - 1) \times 0.0144}{6.262} = \frac{0.216}{6.262} = 0.034494\text{ mm}^2$$
4. Variance Confidence Interval: $$0.007858 \text{ mm}^2 \le \sigma^2 \le 0.034494 \text{ mm}^2$$
5. Standard Deviation Confidence Interval: $$\sqrt{0.007858} \le \sigma \le \sqrt{0.034494} \implies [0.0886\text{ mm}, \; 0.1857\text{ mm}]$$

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CQE Exam Tips & Strategy

• Degrees of Freedom: Always remember $\nu = n - 1$ for single sample t and Chi-Square distribution lookups.
• Chi-Square Denominator Inversion: Pay close attention to Chi-Square formulas: the larger critical value $\chi^2_{\alpha/2}$ goes into the lower bound denominator, while the smaller critical value $\chi^2_{1-\alpha/2}$ goes into the upper bound denominator.
• Sample Size Rounding: Always round sample size calculations UP to the next integer ($n = 96.12 \implies n = 97$) to maintain the specified error margin.