7.8 Tests for Means, Variances, and Proportions

Key Takeaways

  • 2-sample t-tests assess mean differences between independent groups; equal variance assumption uses pooled variance s_p^2, while unequal variance uses Welch's t-test with Satterthwaite df.

  • Paired t-tests reduce nuisance variability by analyzing paired differences d_i = x_1i - x_2i with df = n - 1 pairs, increasing statistical power for matched designs.

  • One-Way ANOVA decomposes total variance into between-group (SSB) and within-group (SSW) variance, evaluating F0 = MSB / MSW across k >= 3 factor levels.

  • Tukey's Honestly Significant Difference (HSD) post-hoc test controls the family-wise Type I error rate (alpha_FW) when conducting all pairwise comparisons following a significant ANOVA.

  • Chi-Square tests evaluate categorical data: Goodness-of-Fit tests categorical distribution fit (df = k - 1), while Independence tests evaluate contingency table associations (df = (r-1)(c-1)).

Last updated: August 2026

Comparative hypothesis testing evaluates process parameter differences across two or more experimental conditions, treatments, shifts, or operational lines. Choosing the correct parametric or non-parametric test depends on data type (continuous vs. discrete), sample independence (paired vs. independent), and variance homogeneity.


Two-Sample Hypothesis Tests for Means and Proportions

1. Independent Two-Sample tt-Test

Compares the means of two independent continuous populations (μ1\mu_1 vs. μ2\mu_2).

Case A: Equal Variances (σ12=σ22\sigma_1^2 = \sigma_2^2) — Pooled tt-Test

  • Pooled Variance (sp2s_p^2): sp2=(n1−1)s12+(n2−1)s22n1+n2−2s_p^2 = \frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2}{n_1 + n_2 - 2}
  • Test Statistic: t0=xˉ1−xˉ2sp2(1n1+1n2)t_0 = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}
  • Degrees of Freedom: df=n1+n2−2\text{df} = n_1 + n_2 - 2.

Case B: Unequal Variances (σ12≠σ22\sigma_1^2 \neq \sigma_2^2) — Welch's tt-Test

Uses unpooled standard errors with Welch-Satterthwaite degrees of freedom approximation (ν\nu):

ν=(s12n1+s22n2)2(s12/n1)2n1−1+(s22/n2)2n2−1\nu = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1 - 1} + \frac{(s_2^2/n_2)^2}{n_2 - 1}}

2. Paired Two-Sample tt-Test

Used when observations are naturally paired or matched (e.g., before-and-after measurements on the exact same production parts, tools, or operators).

  • Difference Variable: di=x1i−x2id_i = x_{1i} - x_{2i}
  • Test Statistic: t0=dˉ−μdsdnt_0 = \frac{\bar{d} - \mu_d}{\frac{s_d}{\sqrt{n}}}
  • Degrees of Freedom: df=n−1\text{df} = n - 1 (where nn is the number of matched pairs).
  • Operational Advantage: Eliminates unit-to-unit baseline variability, greatly reducing residual error variance and maximizing test sensitivity/power.

3. Two-Sample Proportion zz-Test

Compares two independent binomial success proportions (p1p_1 vs. p2p_2).

  • Pooled Proportion (pˉ\bar{p}): pˉ=x1+x2n1+n2\bar{p} = \frac{x_1 + x_2}{n_1 + n_2}
  • Test Statistic: z0=p^1−p^2pˉ(1−pˉ)(1n1+1n2)z_0 = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\bar{p}(1-\bar{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}

4. FF-Test for Equality of Two Variances

Evaluates whether two independent population variances are equal (H0:σ12=σ22H_0: \sigma_1^2 = \sigma_2^2).

  • Test Statistic: F0=s12s22(placing larger sample variance in numerator, s12≥s22)F_0 = \frac{s_1^2}{s_2^2} \quad \text{(placing larger sample variance in numerator, } s_1^2 \ge s_2^2 \text{)}
  • Degrees of Freedom: df1=n1−1\text{df}_1 = n_1 - 1, df2=n2−1\text{df}_2 = n_2 - 1.
  • Sensitivity Note: The FF-test is highly sensitive to non-normality. Levene's test or Bartlett's test are preferred robust alternatives.

Non-Parametric Hypothesis Tests

When continuous data violates normality and transformation is unsuccessful, non-parametric tests based on data ranks must be deployed:

Parametric TestNon-Parametric EquivalentPrimary Metric Tested
1-Sample tt-Test1-Sample Wilcoxon Signed-RankMedian vs. Target
2-Sample tt-Test (Independent)Mann-Whitney UU Test (Wilcoxon Rank-Sum)Median Difference (2 Groups)
Paired tt-TestPaired Wilcoxon Signed-RankMedian Paired Difference
One-Way ANOVAKruskal-Wallis HH TestMedian Difference (k≥3k \ge 3 Groups)

Tests for Variances

The Body of Knowledge requires tests for means, variances, and proportions. Variance tests are the ones teams most often skip, and they matter for two reasons: variation reduction is frequently the project's actual objective, and equality of variance is an assumption of the two-sample t-test and of ANOVA.

One sample against a target variance

χ2=(n−1)s2σ02,df=n−1\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}, \qquad df = n - 1

Compare against the chi-square distribution. This test is very sensitive to non-normality and should not be used on visibly skewed data.

Two variances: the F test

F=s12s22,df1=n1−1, df2=n2−1F = \frac{s_1^2}{s_2^2}, \qquad df_1 = n_1 - 1,\ df_2 = n_2 - 1

By convention the larger variance goes in the numerator so F≥1F \ge 1, and the result is compared with Fα/2,df1,df2F_{\alpha/2, df_1, df_2} for a two-sided test.

Worked example. Line A: n1=16n_1 = 16, s1=2.4s_1 = 2.4. Line B: n2=13n_2 = 13, s2=1.5s_2 = 1.5.

F=2.421.52=5.762.25=2.56F = \frac{2.4^2}{1.5^2} = \frac{5.76}{2.25} = 2.56

with 15 and 12 degrees of freedom. The two-sided critical value F0.025,15,12≈3.18F_{0.025, 15, 12} \approx 3.18. Since 2.56<3.182.56 < 3.18, do not reject: the evidence does not establish a difference in variability at the 5% level, despite the sample standard deviations differing by 60%. Variance tests need substantial sample sizes.

Like the chi-square test, the F test for variances is highly sensitive to departures from normality, much more so than the t-test is for means.

Several variances

TestAssumptionUse
Bartlett's testRequires normality; very sensitive to violationData confidently normal
Levene's testUses absolute deviations from the group mean; robust to non-normalityThe general-purpose default
Brown-ForsytheLevene's using deviations from the median; still more robustSkewed data

Because Bartlett's test frequently rejects for non-normality rather than for unequal variance, Levene's test is the practical default when checking the equal-variance assumption before ANOVA.

When variances are unequal

SituationResponse
Comparing two means with unequal variancesUse Welch's t-test, which does not pool the variances and adjusts the degrees of freedom
Comparing several means with unequal variancesUse Welch's ANOVA, or transform the response
Variance itself increases with the meanApply a variance-stabilizing transformation: logarithm, square root, or Box-Cox
Data badly non-normalUse the non-parametric alternative: Mann-Whitney for two groups, Kruskal-Wallis for several

Why variance tests matter to a Black Belt

A project that reduces the mean of a cycle time while leaving the spread unchanged has improved the average experience and left the worst case where it was. Customers experience the tail, not the mean. In many projects the correct primary metric is the standard deviation or a high percentile, and the corresponding hypothesis test is a variance test rather than a t-test.

The same logic drives the capability indices: CpC_p is entirely a statement about variance, and a project targeting CpC_p rather than CpkC_{pk} is targeting variation rather than centring.

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Two-Sample and Multi-Sample Test Selection Guide
Test Your Knowledge

A quality analyst conducts a Chi-Square test for independence analyzing defect types (3 categories) across shift operations (4 shifts). How many degrees of freedom does the test statistic have?

A

6

B

7

C

11

D

12

Test Your Knowledge

Why is a paired t-test preferred over an independent two-sample t-test when evaluating part measurements before and after heat treatment processing?

A

It relaxes the requirement for continuous measurement data

B

It isolates unit-to-unit baseline variation, reducing error variance and increasing statistical power

C

It automatically guarantees equal variance across treatment groups

D

It eliminates the need for calculating degrees of freedom

Test Your Knowledge

A One-Way ANOVA comparing four machine filling lines yields a calculated F-statistic F0 = 8.45 with a p-value = 0.0002 (alpha = 0.05). What conclusion should the Black Belt draw?

A

All four machine lines have identical mean filling volumes

B

Machine line 1 is significantly higher than machine line 4

C

Reject H0; at least one machine line mean is significantly different from the others, requiring post-hoc testing

D

Fail to reject H0; differences are attributable solely to random sampling variation

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