23-Ind-B1 Reliability and Maintainability · May 2015

Question 9 of 10: Hop-Plant Yields — Bartlett's Test, One-Way ANOVA, Tukey's Test, and a Single-df Contrast

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

Notes on this paper

National Exams — May 2015 — 98-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: three sections — Section A: do 2 of 4 questions (10 marks); Section B: do 2 of 3 (20 marks); Section C: do 1 of 3 (20 marks) — a 5-question, 50-mark paper as printed. All ten questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — joint distributions (ch. 5), binomial/geometric probability (ch. 3), normal distribution and sampling distributions (ch. 4–7), point/interval estimation (ch. 8–9), hypothesis testing incl. sample-size design (ch. 9–10), simple linear regression and ANOVA (ch. 11–13), Bartlett's and Tukey's tests, single-degree-of-freedom contrasts and 2³ factorial designs (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).

Question 9 (Section C.2): Hop-Plant Yields — Bartlett's Test, One-Way ANOVA, Tukey's Test, and a Single-df Contrast (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. Five annual yields (g) per plant, $n_i=5$ each ($N=15$):

Williamette (W)	118	118	135	126	130
Fuggles (F)	158	155	152	167	157
Centennial (C)	227	215	215	221	217

Find. (a) Bartlett's $\chi^2$; (b) one-way ANOVA $F$; (c) Tukey pairwise comparisons; (d) contrast $C$ vs. $(W+F)/2$.

Approach. Compute each group's mean/variance, run Bartlett's test to check the ANOVA's equal-variance assumption, then the one-way ANOVA $F$-test, Tukey's HSD for all pairs, and finally a targeted single-df contrast.

Group statistics. $\bar x_W=125.4,\ s_W^2=55.8$; $\bar x_F=157.8,\ s_F^2=31.7$; $\bar x_C=219.0,\ s_C^2=26.0$ (all $n_i=5$).
(a) Bartlett's test, $H_0:\sigma_W^2=\sigma_F^2=\sigma_C^2$. Pooled variance $s_p^2=\dfrac{4(55.8)+4(31.7)+4(26.0)}{12}=37.833$. $$q=(N-k)\ln s_p^2-\sum(n_i-1)\ln s_i^2=0.6535,\qquad C=1+\frac{\sum\frac1{n_i-1}-\frac1{N-k}}{3(k-1)}=1.1111$$ $$\chi^2=q/C=\boxed{0.588}$$ $\chi^2_{0.05,2}=5.991$; since $0.588\lt 5.991$, fail to reject $H_0$ — the three variances are not significantly different (the ANOVA's equal-variance assumption is supported).
(b) One-way ANOVA. Directly summing $\sum(y_{ij}-\bar y_i)^2$ within each group gives $SSE=454.0$ (matching the exam's "approximately 460" closely). Using the given $SST=23047.6$: $SS_{Tr}=23047.6-454.0=22{,}593.6$, $df_{Tr}=2$, $df_E=12$. $$F=\frac{SS_{Tr}/2}{SSE/12}=\frac{11{,}296.8}{37.83}=\boxed{298.6}$$ $F_{0.05,2,12}=3.885$; since $298.6\gg 3.885$, strongly reject $H_0$ — plant type has a highly significant effect on yield.
(c) Tukey's HSD, $\alpha=0.05$. $q_{0.05,3,12}=3.773$, $HSD=q\sqrt{MSE/n}=3.773\sqrt{37.83/5}=\boxed{10.38\text{ g}}$. Pairwise mean differences: $|\bar x_W-\bar x_F|=32.4$, $|\bar x_W-\bar x_C|=93.6$, $|\bar x_F-\bar x_C|=61.2$ — all three exceed $HSD=10.38$, so every pair differs significantly (Centennial > Fuggles > Williamette).
(d) Single-df contrast: Centennial vs. average of Williamette & Fuggles. $L=\bar x_C-\tfrac12(\bar x_W+\bar x_F)$ has coefficient set $(-1,-1,2)$ on $(W,F,C)$: $L=154.8$, $\sum c_i^2=6$. $$SS_{contrast}=\frac{n\,L^2}{\sum c_i^2}=\frac{5(154.8)^2}{6}=19{,}969.2,\qquad F=\frac{SS_{contrast}}{MSE}=\frac{19{,}969.2}{37.83}=\boxed{527.8}$$ $F_{0.05,1,12}=4.747$; since $527.8\gg4.747$, reject $H_0$ — Centennial's yield is significantly different from (higher than) the average of Williamette and Fuggles.

Summary
Part	Result
(a) Bartlett $\chi^2$	0.588 < 5.991; variances equal
(b) ANOVA $F$	298.6 > 3.885; plant type significant
(c) Tukey HSD	10.38 g; all 3 pairs differ
(d) contrast $F$	527.8 > 4.747; Centennial differs from W&F avg.

Topic: Bartlett's test, one-way ANOVA, Tukey's HSD, and orthogonal single-degree-of-freedom contrasts.

Key relations: Bartlett $\chi^2=q/C$; ANOVA $F=MS_{Tr}/MS_E$; Tukey $HSD=q_{\alpha,k,N-k}\sqrt{MSE/n}$; contrast $SS=nL^2/\sum c_i^2$ tested against $MSE$ with 1 df.

Why this works: Bartlett's test validates the equal-variance assumption the ANOVA $F$-test relies on; Tukey's HSD controls the family-wise error rate across all $\binom{3}{2}=3$ pairwise comparisons; the $(-1,-1,2)$ contrast isolates exactly the "Centennial vs. the other two, combined" comparison the question asks for, using only 1 of the treatment's 2 degrees of freedom.

Common pitfall: Running Tukey's test or a contrast when the omnibus ANOVA $F$-test was not significant — here it is safe because $F=298.6$ is overwhelming, but the sequence (omnibus test, then pairwise/contrast follow-up) matters in general.

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed.), ch. 13 (single-factor ANOVA, Bartlett's test, multiple comparisons, contrasts).

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