2.3 Estimation, Confidence Intervals, and Sample Sizes

2.3 Estimation, Confidence Intervals, and Sample Sizes

Statistical inference enables engineers to draw valid conclusions about population parameters from sample measurements. This section covers the Central Limit Theorem, confidence interval construction, hypothesis testing methodology for population means, and simple linear regression and correlation analysis as specified in the NCEES FE Reference Handbook.

---

1. Central Limit Theorem (CLT)

The Central Limit Theorem is the foundation of inferential statistics. It states that for any independent, identically distributed (i.i.d.) random variables $X_1, X_2, \dots, X_n$ with population mean $\mu$ and finite variance $\sigma^2$, as sample size $n$ increases ($n \ge 30$ in practice), the distribution of the sample mean $\bar{X}$ approaches a normal distribution: $\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$

The standard deviation of the sampling distribution of $\bar{X}$ is the standard error: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

Regardless of whether the underlying population is skewed or non-normal, sample means drawn from large samples ($n \ge 30$) can be analyzed using standard normal $Z$-transformations: $Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$

2. Confidence Intervals (CI)

A confidence interval provides an estimated range likely to contain an unknown population parameter at a specified confidence level $1 - \alpha$ (such as 90%, 95%, or 99%).

A. Confidence Interval for Population Mean $\mu$ (Known $\sigma$)

When population standard deviation $\sigma$ is known (or $n \ge 30$ by CLT): $\text{CI: } \bar{x} \pm z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right)$ Where $z_{\alpha/2}$ is the standard normal critical value:

• 90% Confidence ($1 - \alpha = 0.90$): $z_{0.05} = 1.645$
• 95% Confidence ($1 - \alpha = 0.95$): $z_{0.025} = 1.960$
• 99% Confidence ($1 - \alpha = 0.99$): $z_{0.005} = 2.576$

B. Confidence Interval for Population Mean $\mu$ (Unknown $\sigma$, Small $n < 30$)

When population standard deviation $\sigma$ is unknown and sample size is small ($n < 30$), use Student's $t$-distribution with $df = n - 1$ degrees of freedom: $\text{CI: } \bar{x} \pm t_{\alpha/2, n-1} \left( \frac{s}{\sqrt{n}} \right)$

C. Confidence Interval for Population Proportion $p$

For sample proportion $\hat{p} = x / n$: $\text{CI: } \hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Worked Engineering Example: Small Sample $t$-Confidence Interval

Problem: A structural engineer tests $n = 16$ aluminum alloy samples, obtaining a sample mean yield strength $\bar{x} = 210.0\text{ MPa}$ and a sample standard deviation $s = 12.0\text{ MPa}$. Construct a 95% confidence interval for the true mean yield strength of the alloy. (Critical value $t_{0.025, 15} = 2.131$).

Solution: Step 1: Identify Parameters: $n = 16, \quad df = n - 1 = 15, \quad \bar{x} = 210.0\text{ MPa}, \quad s = 12.0\text{ MPa}, \quad t_{0.025, 15} = 2.131$

Step 2: Calculate Margin of Error (ME): $ME = t_{\alpha/2, df} \times \left( \frac{s}{\sqrt{n}} \right) = 2.131 \times \left( \frac{12.0}{\sqrt{16}} \right) = 2.131 \times 3.0 = 6.393\text{ MPa}$ (Wait, let me double check: 12 / 4 = 3.0; 2.131 * 3.0 = 6.393 MPa)

Step 3: Construct Confidence Interval bounds: $\text{Lower Limit} = 210.0 - 6.393 = 203.607\text{ MPa}$ $\text{Upper Limit} = 210.0 + 6.393 = 216.393\text{ MPa}$ $95\% \text{ CI: } [203.61, 216.39]\text{ MPa}$

Point Estimation and the Language of Estimators

NCEES lists "Estimation (e.g., point, confidence intervals)" as the first Probability and Statistics sub-topic, so the vocabulary itself is examinable.

Why the sample variance divides by $n-1$. Dividing by $n$ systematically underestimates the population variance, because deviations are measured from $\bar{x}$ — itself computed from the same data — rather than from the unknown true mean $\mu$. The $n-1$ divisor (Bessel's correction) removes that bias, and $n-1$ is the degrees of freedom: one is consumed by estimating the mean.

$s^2 = \frac{\sum(x_i-\bar{x})^2}{n-1} \qquad\text{(unbiased)}$

Choosing Between z and t

$\bar{x} \pm z\frac{\sigma}{\sqrt{n}} \qquad\qquad \bar{x} \pm t_{\alpha/2,\,n-1}\frac{s}{\sqrt{n}}$

The $t$ distribution has heavier tails than the normal, so its critical values are always larger — the interval is wider, which is the honest price of not knowing $\sigma$. At $n = 5$, $t_{0.025,4} = 2.776$ against $z = 1.960$: a 42% wider interval. By $n = 30$, $t_{0.025,29} = 2.045$, only 4% wider.

What a Confidence Level Actually Means

A 95% confidence interval means that if the sampling procedure were repeated many times, 95% of the intervals so constructed would contain the true parameter. It does not mean there is a 95% probability that the parameter lies inside this particular interval — the parameter is a fixed constant, and this interval either contains it or does not.

This distinction is a favorite conceptual item. Two related traps:

• A confidence interval for the mean is much narrower than the spread of individual values. An interval of $[48.2, 51.8]$ for the mean does not imply that 95% of individual measurements fall in that range.
• Raising the confidence level widens the interval. You cannot become more confident and more precise at the same time without collecting more data.

Sample Size Determination

Setting the margin of error $E = z\dfrac{\sigma}{\sqrt{n}}$ and solving:

$n = \left(\frac{z\sigma}{E}\right)^2 \qquad\text{(for a mean)}, \qquad n = \frac{z^2p(1-p)}{E^2} \qquad\text{(for a proportion)}$

Always round up. Rounding down would leave the margin of error larger than specified.

Because $n$ depends on $E^{-2}$:

For a proportion with no prior estimate, use $p = 0.5$: it maximizes $p(1-p)$ at 0.25 and therefore gives the conservative (largest) required sample size, guaranteeing the target margin whatever the true proportion turns out to be.