23-Ind-B1 Reliability and Maintainability · December 2014

Question 9 of 9: One-Way ANOVA, Bartlett's Test, and Tukey's Test — Packing Times

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

Question 9 (Section D.2): One-Way ANOVA, Bartlett's Test, and Tukey's Test — Packing Times (30 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. Cycle times (s) at four workstations. Note: the prose states "a sample of three cycles," while the printed table lists four cycle-time values per station — the analysis below uses all four printed values per workstation (see check callout).

Find. (a) one-way ANOVA at $\alpha=0.05$; (b) Bartlett's test at $\alpha=0.01$, related to (a); (c) Tukey's pairwise comparisons at $\alpha=0.05$, related to (a).

Approach. Partition total variability into between-workstation and within-workstation sums of squares for the $F$-test; use Bartlett's statistic to check the equal-variance assumption the ANOVA relies on; then use Tukey's HSD to locate exactly which pairs of stations differ.

1. Part (a) — one-way ANOVA. With $k=4$ stations, $n_i=4$ each, $N=16$: station means are $\bar y_1=26.00$, $\bar y_2=26.15$, $\bar y_3=23.55$, $\bar y_4=31.725$, grand mean $\bar{\bar y}=26.86$. $$SS_{Tr}=\sum n_i(\bar y_i-\bar{\bar y})^2=143.47,\qquad SST=\sum(y_{ij}-\bar{\bar y})^2=190.02,\qquad SSE=SST-SS_{Tr}=46.55.$$ With $df_{Tr}=3$, $df_E=12$: $MS_{Tr}=143.47/3=47.82$, $MSE=46.55/12=3.879$, $$F_0=\frac{MS_{Tr}}{MSE}=\boxed{12.33}.$$ Against $F_{0.05,3,12}=3.49$: $12.33\gg3.49$, so reject $H_0$ — at least one workstation's mean packing time differs from the others.
2. Part (b) — Bartlett's test. The four sample variances are $s_1^2=2.233$, $s_2^2=3.003$, $s_3^2=6.737$, $s_4^2=3.542$ ($df=3$ each). Bartlett's statistic (bias-corrected) is $$B_0 = \boxed{0.913}$$ computed from the pooled variance and the log-variance weighted sum in the usual way, and is compared against $\chi^2_{0.01,3}=11.34$. Since $0.913\ll11.34$, fail to reject $H_0:\sigma_1^2=\sigma_2^2=\sigma_3^2=\sigma_4^2$ — the four variances are statistically indistinguishable at $\alpha=0.01$. This supports the ANOVA in part (a): the equal-variance assumption the $F$-test relies on is not contradicted by the data, so the significant $F_0=12.33$ finding is not an artifact of unequal spread.
3. Part (c) — Tukey's HSD. With $q_{0.05,4,12}=4.199$ and $MSE=3.879$, $n_i=4$: $$HSD = q_{0.05,4,12}\sqrt{\frac{MSE}{n_i}} = 4.199\sqrt{\frac{3.879}{4}} = \boxed{4.135\text{ s}}.$$ Comparing all six pairwise mean differences against this HSD: $$|\bar y_1-\bar y_2|=0.15\ (ns),\ |\bar y_1-\bar y_3|=2.45\ (ns),\ |\bar y_1-\bar y_4|=5.73\ (\text{sig.}),$$ $$|\bar y_2-\bar y_3|=2.60\ (ns),\ |\bar y_2-\bar y_4|=5.58\ (\text{sig.}),\ |\bar y_3-\bar y_4|=8.18\ (\text{sig.}).$$ WS4 differs significantly from every other station (WS1, WS2, and WS3 alike); WS1, WS2, and WS3 are not significantly different from one another. This is exactly consistent with part (a): the overall significant $F$-test is being driven entirely by WS4's noticeably slower packing times, not by broad differences across all four stations.