23-Ind-B1 Reliability and Maintainability · December 2018

Question 9 of 9: $2^2$ Factorial Design — Hospital Patient-Flow Screening Experiment

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

Notes on this paper

National Exams — December 2018 — 17-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 (20 marks); Section B: do 2 of 3 (40 marks); Section C: do 1 of 2 (20 marks) — a 9-question, printed-total-80-mark paper as actually structured (the front page's own summary line states “Exam: 5 Questions. Total marks: 100”, but 20+40+20=80 by the section instructions and per-question mark values printed beside each question — the front page's 100 is internally inconsistent with its own section table; the 80-mark reading is used throughout, consistent with the section-by-section split printed on pages 2, 4 and 6). All nine questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — normal distribution and sums of normals (ch. 4–5), point/interval estimation (ch. 8), two-sample hypothesis testing (ch. 9–10), simple linear regression (ch. 11), single-factor ANOVA and multiple comparisons (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — Bartlett's test, $2^k$ factorial designs (ch. 3, 6).

Question 9 (Section C.2): $2^2$ Factorial Design — Hospital Patient-Flow Screening Experiment (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. $2^2$ factorial in Factor A (off-servicing) and Factor B (discharge planning), $n=5$ replications per run; average beds occupied per run:

Run	A	B	Rep 1	Rep 2	Rep 3	Rep 4	Rep 5	Average
1	−1	−1	73.93	70.02	72.97	70.13	71.71	71.75
2	+1	−1	69.53	68.80	71.40	71.52	71.53	70.56
3	−1	+1	71.08	73.39	73.78	72.23	73.16	72.73
4	+1	+1	68.51	69.16	71.96	70.88	69.58	70.02

Find. (a) Complete the design matrix (add the $AB$ column) and interpret the coded signs; (b) ANOVA — which factor(s) significantly affect average beds occupied?

Approach. Add the interaction column as the row-wise product of $A$ and $B$; then use the standard $2^2$-with-replication ANOVA decomposition, computing each effect's contrast from the run totals and testing against $MSE$ from the within-run replicate variation.

(a) Completed design matrix and interpretation.

Run	A	B	$AB$
1	−1	−1	+1
2	+1	−1	−1
3	−1	+1	−1
4	+1	+1	+1

The $AB$ column is the row-wise product of the $A$ and $B$ columns. Coded $-1$ means the factor is at its low (not-implemented) setting and $+1$ means its high (implemented) setting. The $A$ and $B$ columns estimate each factor's own average, independent effect on beds occupied. The $AB$ column is $+1$ whenever $A$ and $B$ are at the same setting (both off or both on) and $-1$ whenever they are at opposite settings; it estimates the interaction — whether off-servicing's effect on beds occupied depends on whether discharge planning is also active (and vice versa), rather than the two policies acting independently of one another.

(b) ANOVA, $\alpha=0.05$, $N=20$ observations. Recomputing directly from the 20 raw replicate values gives $SST=53.17$, matching the value the question supplies — the raw table is confirmed reliable. Run totals: $T_1=358.76,\ T_2=352.78,\ T_3=363.64,\ T_4=350.09$. Effect contrasts (using the $\pm1$ signs above, $n=5$ reps/run): $$Contrast_A=-T_1+T_2-T_3+T_4=-19.53,\quad Contrast_B=-T_1-T_2+T_3+T_4=2.19,\quad Contrast_{AB}=T_1-T_2-T_3+T_4=-7.57$$ $$SS_A=\frac{Contrast_A^2}{4n}=\frac{(-19.53)^2}{20}=19.07,\qquad SS_B=\frac{2.19^2}{20}=0.24,\qquad SS_{AB}=\frac{(-7.57)^2}{20}=2.87$$ $$SSE=SST-SS_A-SS_B-SS_{AB}=53.17-19.07-0.24-2.87=\boxed{30.99},\qquad df_E=N-4=16,\qquad MSE=\frac{30.99}{16}=1.937$$ $$F_A=\frac{SS_A/1}{MSE}=\boxed{9.84},\qquad F_B=\frac{SS_B/1}{MSE}=\boxed{0.124},\qquad F_{AB}=\frac{SS_{AB}/1}{MSE}=\boxed{1.479}$$ Critical value $F_{0.05,1,16}=4.494$. Only Factor A (off-servicing) is significant ($F_A=9.84>4.494$); Factor B (discharge planning, $F_B=0.124$) and the $AB$ interaction ($F_{AB}=1.479$) are not.

Final Results
Source	$SS$	$df$	$MS$	$F$	Significant at $\alpha=0.05$?
A (off-servicing)	19.07	1	19.07	9.84	Yes
B (discharge planning)	0.24	1	0.24	0.12	No
$AB$	2.87	1	2.87	1.48	No
Error	30.99	16	1.94	—	—
Total	53.17	19	—	—	—

Only off-servicing measurably affects average beds occupied in this screening experiment, reducing it by about 1.95 beds (the $A$ effect, $Contrast_A/2n$) when moved from not-implemented to implemented; since the interaction is not significant, this reduction can be read as roughly the same regardless of the discharge-planning setting. Discharge planning alone shows no detectable effect on this particular response at the tested sample size.

Topic: $2^2$ factorial design with replication — effect estimation and ANOVA.

Key relations: $Effect=Contrast/(2n)$; $SS=Contrast^2/(4n)$ for a $2^2$ design; $SSE=SST-\sum SS_{effects}$, $df_E=N-(\text{number of estimated effects}+1)$.

Why this works: Coding factors at $\pm1$ makes the design orthogonal — each effect's contrast isolates that factor's (or interaction's) influence independently of the others, so $SS_A$, $SS_B$, $SS_{AB}$ sum exactly with $SSE$ to $SST$ with no shared variance to apportion.

Common pitfall: Concluding a factor is “unimportant” from a non-significant screening-experiment result at $n=5$/run — a screening design like this one is meant to flag candidates for a larger follow-up study, not to make a final claim; Factor B's true effect could still be practically meaningful but too small to detect at this sample size.

Source: Montgomery, Design and Analysis of Experiments, 9th ed., §6.2–6.4 ($2^k$ factorial designs, effect estimation and ANOVA).

← Question 8 Back to the paper →