6.6 Specific Statistical Tests

6.6 Specific Statistical Tests

Selecting and properly executing the correct statistical test is a fundamental competency for Certified Quality Engineers. The choice of test depends on the nature of the data (continuous vs. discrete), the number of samples being compared, whether samples are independent or paired, and whether population parameters (such as $\sigma$) are known.

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1. Summary of Standard Statistical Tests

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2. One-Sample and Two-Sample $Z$-Tests and $t$-Tests for Means

One-Sample $t$-Test

When population standard deviation $\sigma$ is unknown, it is estimated using sample standard deviation $s$, introducing extra sampling variability. The test statistic follows Student's $t$-distribution with $df = n - 1$.

Two-Sample Independent $t$-Test (Equal Variances)

When comparing means from two independent groups assuming equal variances (verified via an $F$-test for variances), sample variances are combined into a pooled variance ($s_p^2$):

$$s_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$$ $$\text{Test Statistic:} \quad t_{calc} = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \quad \text{with } df = n_1 + n_2 - 2$$

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3. Paired $t$-Test for Dependent Samples

When data consists of paired measurements on the same experimental units (e.g., measuring roughness before and after polishing on $n$ identical parts), subject-to-subject variation is eliminated by analyzing individual differences $d_i = x_{1i} - x_{2i}$.

• Mean difference: $\bar{d} = \frac{\sum d_i}{n}$
• Sample standard deviation of differences: $s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}}$
• Test statistic: $$t_{calc} = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}} \quad \text{with } df = n - 1$$

CQE Tip: The paired $t$-test has $n - 1$ degrees of freedom (where $n$ is the number of pairs), which is half the degrees of freedom of an independent two-sample test ($2n - 2$). However, by removing background variance between units, pairing dramatically reduces error variance and increases test sensitivity.

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4. Chi-Square ($\chi^2$) Tests: Goodness-of-Fit and Independence

Chi-square tests evaluate non-parametric categorical count data.

$$\text{Chi-Square Test Statistic:} \quad \chi^2_{calc} = \sum_{i=1}^{k} \frac{(O_i - E_i)^2}{E_i}$$

where $O_i$ is the Observed frequency and $E_i$ is the Expected frequency.

Goodness-of-Fit Test

Evaluates whether observed sample counts fit a theoretical distribution (e.g., Poisson, Normal, Binomial). Expected count $E_i = n \cdot p_i$. Degrees of freedom $df = k - 1 - p$, where $k$ is the number of categories and $p$ is the number of parameters estimated from sample data.

Test of Independence (Contingency Tables)

Evaluates whether two categorical variables (e.g., defect type vs. manufacturing shift) are independent in an $r \times c$ table. Expected frequencies are computed as:

$$E_{ij} = \frac{(\text{Row } i \text{ Total}) \times (\text{Column } j \text{ Total})}{\text{Grand Total}}, \quad df = (r - 1)(c - 1)$$

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5. One-Way Analysis of Variance (ANOVA) $F$-Test

When comparing $k \ge 3$ group means, running multiple pairwise $t$-tests causes alpha inflation: overall Type I error risk rises to $\alpha_{overall} = 1 - (1 - \alpha)^C$, where $C = \frac{k(k-1)}{2}$. One-Way ANOVA avoids this by testing $H_0: \mu_1 = \mu_2 = \dots = \mu_k$ simultaneously.

ANOVA partitions total sum of squares ($SST$) into between-group variation ($SSB$) and within-group error variation ($SSW$):

$$SST = SSB + SSW$$

• Between-Group Mean Square ($MSB$): $MSB = \frac{SSB}{k - 1}$
• Within-Group Mean Square ($MSW$): $MSW = \frac{SSW}{N - k}$ (where $N = \sum n_i$)
• $F$-Statistic: $F_{calc} = \frac{MSB}{MSW}$ with $df_1 = k - 1$ and $df_2 = N - k$

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6. Worked Example: One-Way ANOVA for Machining Line Performance

Scenario: A quality engineer compares hole diameter measurements across $k = 3$ CNC machining lines. Sample size is $n_i = 5$ parts per line (Total $N = 15$).

• Line A mean $= 10.2\text{ mm}$
• Line B mean $= 12.4\text{ mm}$
• Line C mean $= 13.4\text{ mm}$
• Overall Grand Mean $\bar{x}_{..} = 12.0\text{ mm}$
• Calculated $SSB = 26.8$, and calculated $SSW = 18.0$.

Test at $\alpha = 0.05$ ($F_{crit, 0.05, 2, 12} = 3.89$) whether machine line means differ significantly.

Step-by-Step ANOVA Solution:

1. Degrees of Freedom:
   • $df_{between} = k - 1 = 3 - 1 = 2$
   • $df_{within} = N - k = 15 - 3 = 12$
   • $df_{total} = N - 1 = 14$
2. Mean Squares:
   • $MSB = \frac{SSB}{df_{between}} = \frac{26.8}{2} = 13.4$
   • $MSW = \frac{SSW}{df_{within}} = \frac{18.0}{12} = 1.5$
3. Calculate $F$-Statistic: $$F_{calc} = \frac{MSB}{MSW} = \frac{13.4}{1.5} = 8.933$$
4. Decision: Since $F_{calc} = 8.933 > F_{crit} = 3.89$, reject $H_0$. There is statistically significant evidence at $\alpha = 0.05$ that at least one CNC machining line mean differs from the others.