5.3 Engineering Data Analysis II — Confidence Intervals, Hypothesis Testing & Regression

5.3 Engineering Data Analysis II — Confidence Intervals, Hypothesis Testing & Regression

The second half of the PRC Enhanced TOS Engineering Data Analysis topic moves from describing a data set to drawing defensible conclusions from it. These are the tools behind load-forecasting studies, meter-accuracy audits, insulation-test acceptance decisions and reliability benchmarking under the Philippine Distribution Code.

---

1. Sampling Distributions & the Central Limit Theorem

If repeated samples of size $n$ are drawn from a population with mean $\mu$ and standard deviation $\sigma$, the sample means themselves form a distribution with

$\mu_{\bar{x}} = \mu \qquad\qquad \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

The quantity $\sigma/\sqrt{n}$ is the standard error of the mean. The Central Limit Theorem states that for sufficiently large $n$ (conventionally $n \ge 30$) the distribution of $\bar{x}$ is approximately normal regardless of the shape of the parent population. This is why normal-based methods work on decidedly non-normal quantities such as outage durations.

Note the $\sqrt{n}$: to halve the standard error you must quadruple the sample size.

---

2. Confidence Intervals

When $\sigma$ is known (or $n > 30$)

$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

When $\sigma$ is unknown and $n$ is small

$\bar{x} \pm t_{\alpha/2,\;\nu} \frac{s}{\sqrt{n}}, \qquad \nu = n - 1 \text{ degrees of freedom}$

The $t$ distribution is wider than the normal, converging to it as $\nu \to \infty$. Choosing $z$ when the problem states "sample standard deviation" and $n$ is small is a graded error.

---

3. Hypothesis Testing

A hypothesis test compares a null hypothesis $H_0$ (the status quo — "the meters meet the accuracy class") against an alternative $H_1$.

Test statistic for a mean

$z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \qquad\qquad t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$

Decision rule

Reject $H_0$ when the test statistic falls in the rejection region, equivalently when $p < \alpha$.

The two error types

In electrical acceptance testing, a Type I error rejects a good batch (producer's risk); a Type II error accepts a defective batch (consumer's risk). Reducing $\alpha$ without increasing $n$ necessarily increases $\beta$.

One-tailed vs. two-tailed

"Is the mean different from 220 V?" is two-tailed. "Is the loss greater than 150 W?" is one-tailed, and the whole of $\alpha$ sits in one tail, so the critical value is smaller (1.645 rather than 1.96 at 95%).

---

4. Linear Regression & Correlation

The least-squares line $y = a + bx$ minimises $\sum (y_i - \hat{y}_i)^2$:

$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^{2} - \left(\sum x\right)^{2}} \qquad\qquad a = \bar{y} - b\bar{x}$

The Pearson correlation coefficient measures linear association:

$r = \frac{n\sum xy - \sum x \sum y}{\sqrt{\left[n\sum x^{2} - (\sum x)^{2}\right]\left[n\sum y^{2} - (\sum y)^{2}\right]}}, \qquad -1 \le r \le 1$

The coefficient of determination $r^{2}$ is the proportion of the variation in $y$ explained by the regression on $x$. An $r = 0.9$ means $r^{2} = 0.81$, so 81% of the variation is explained — not 90%.

Correlation is not causation, and $r$ measures only linear association: a perfect parabola can have $r \approx 0$.

---

5. Designed Experiments & Statistical Process Control

Control limits are not specification limits: control limits describe what the process actually does, while specification limits state what the customer requires.

---

Solved Board Exam Examples

Example 1: Confidence Interval on Feeder Voltage

Thirty-six voltage readings on a distribution feeder average 228.4 V with a known population standard deviation of 3.6 V. Construct the 95% confidence interval for the true mean voltage.

Solution. With $\sigma$ known and $n = 36$, use $z_{0.025} = 1.96$:

$\text{ME} = 1.96 \times \frac{3.6}{\sqrt{36}} = 1.96 \times 0.60 = 1.176 \text{ V}$

$228.4 \pm 1.176 \implies \boxed{227.22 \text{ V} \le \mu \le 229.58 \text{ V}}$

The whole interval lies inside the $\pm 10%$ service-voltage band, so the feeder is compliant.

Example 2: Hypothesis Test on Transformer Efficiency

A manufacturer claims a mean efficiency of at least 98.5%. Sixteen units average 98.2% with a sample standard deviation of 0.4%. Test the claim at $\alpha = 0.05$.

Solution. $H_0: \mu \ge 98.5%$ against $H_1: \mu < 98.5%$ (one-tailed, left). With $\sigma$ unknown and $n = 16$, use $t$ with $\nu = 15$:

$t = \frac{98.2 - 98.5}{0.4/\sqrt{16}} = \frac{-0.3}{0.1} = -3.00$

The one-tailed critical value is $t_{0.05,15} = -1.753$. Since $-3.00 < -1.753$, reject $H_0$: at the 5% level the evidence contradicts the manufacturer's efficiency claim.

Example 3: Least-Squares Load Growth

Peak demand over four years is: year 1, 12.0 MW; year 2, 12.9 MW; year 3, 13.6 MW; year 4, 14.7 MW. Fit a least-squares line and forecast year 6.

Solution. With $n = 4$: $\sum x = 10$, $\sum y = 53.2$, $\sum xy = 138.3$, $\sum x^2 = 30$.

$b = \frac{4(138.3) - (10)(53.2)}{4(30) - (10)^{2}} = \frac{553.2 - 532}{120 - 100} = \frac{21.2}{20} = 1.06 \text{ MW/yr}$

$a = \frac{53.2}{4} - 1.06\left(\frac{10}{4}\right) = 13.30 - 2.65 = 10.65 \text{ MW}$

$\hat{y}(6) = 10.65 + 1.06(6) = \boxed{17.01 \text{ MW}}$