23-Ind-B1 Reliability and Maintainability · December 2019

Question 7 of 9: One-Way ANOVA, Bartlett's Test, and Tukey's HSD

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 7 (Section B.3): One-Way ANOVA, Bartlett's Test, and Tukey's HSD (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Packing cycle times (s), 3 cycles at each of 4 workstations:

Find. (a) One-way ANOVA across the 4 workstations; (b) Bartlett's test for equal variances, and reconcile with (a); (c) Tukey's HSD to find which pair(s) differ; (d) the number of pairwise tests and the per-comparison confidence level needed for a family-wise $\alpha=0.05$.

Approach. Recompute all sums of squares directly from the 12 raw times, run the standard one-way ANOVA $F$-test, then Bartlett's test on the same four groups, then Tukey's HSD, then the Bonferroni pairwise-count/confidence-level calculation for (d).

1. Group and grand statistics (from the raw table). $$\bar x_1=25.933,\ \bar x_2=26.167,\ \bar x_3=23.233,\ \bar x_4=31.900,\qquad \bar{\bar x}=26.808$$ $$s_1^2=3.323,\ s_2^2=4.503,\ s_3^2=9.503,\ s_4^2=5.130$$
2. (a) One-way ANOVA, $\alpha=0.05$. $$SS_{Tr}=3\sum(\bar x_i-\bar{\bar x})^2=119.65,\quad df_{Tr}=3,\qquad SSE=\sum\sum(x_{ij}-\bar x_i)^2=44.92,\quad df_E=8$$ $$MS_{Tr}=\frac{119.65}{3}=39.88,\qquad MSE=\frac{44.92}{8}=5.615$$ $$F=\frac{MS_{Tr}}{MSE}=\boxed{7.10}$$ Critical value $F_{0.05,3,8}=4.066$. Since $F=7.10\gt4.066$ ($p=0.012$), reject $H_0$: at least one workstation's mean cycle time differs from the others.
3. (b) Bartlett's test for equal variances, $\alpha=0.05$. Using the pooled variance $S_p^2=MSE=5.615$ and each group's own variance: $$\chi^2=\frac{(N-k)\ln S_p^2-\sum(n_i-1)\ln s_i^2}{C},\quad C=1+\frac{\sum\frac{1}{n_i-1}-\frac1{N-k}}{3(k-1)}$$ Evaluating: $\chi^2=\boxed{0.512}$. Critical value $\chi^2_{0.05,3}=7.815$. Since $0.512\ll7.815$ ($p=0.916$), fail to reject $H_0$: the four workstations' cycle-time variances are homogeneous. Comment on (a): this is a reassuring result, not a contradiction — Bartlett's test confirms the equal-variance assumption that the one-way ANOVA in (a) itself depends on is well supported, so the significant mean difference found in (a) is a genuine effect and not an artifact of unequal within-group spread. The two tests answer independent questions (means vs. variances) and both are consistent here.
4. (c) Tukey's HSD, $\alpha=0.05$. $$q_{0.05,4,8}=4.529,\qquad HSD=q\sqrt{\frac{MSE}{n}}=4.529\sqrt{\frac{5.615}{3}}=\boxed{6.196}$$ Pairwise absolute mean differences vs. $HSD$:

Only WS3 vs. WS4 differ significantly — WS4's mean cycle time ($31.9$s) is significantly slower than WS3's ($23.2$s); no other pair is distinguishable at the 5% family-wise level.

5. (d) Full pairwise comparison count and per-comparison confidence level. The number of distinct pairs among 4 workstations is $$\binom{4}{2}=\boxed{6\text{ tests}}$$ To hold the overall (family-wise) error rate at $\alpha=0.05$ across all 6 independent comparisons, the Bonferroni correction allocates the error budget evenly: $$\alpha_{\text{each}}=\frac{0.05}{6}=0.00833\quad\Rightarrow\quad\text{each comparison's confidence level}=1-0.00833=\boxed{99.17\%}$$ This is markedly more conservative than testing each pair at a plain 95% level, and it is the reason Tukey's HSD (which controls the family-wise rate directly via the studentized range distribution, part (c)) is generally preferred over 6 separate Bonferroni-corrected $t$-tests — both approaches are valid, but Tukey's is less conservative for this specific “all pairwise comparisons” structure.