23-Ind-B1 Reliability and Maintainability · December 2016

Question 5 of 11: Blood-Collection Shakers — Variance, Mean, and One-Sample Tests, CI, Anomaly Check

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

Notes on this paper

National Exams — December 2016 — 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 (30 marks); Section B: do 2 of 3 (30 marks); Section C: do 2 of 4 (40 marks) — a 6-question, 100-mark paper as printed. All eleven 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 and covariance (ch. 5), point/interval estimation (ch. 8), hypothesis testing incl. two-sample and goodness-of-fit tests (ch. 9–10), simple linear regression (ch. 11), design and analysis of single-factor and factorial experiments (ch. 13–14). 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) — two-way factorial ANOVA and $2^k$ designs (ch. 5, 6–7).

Question 5 (Section B.1): Blood-Collection Shakers — Variance, Mean, and One-Sample Tests, CI, Anomaly Check (15 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. $n=10$ per shaker, target $=480$ ml:

Shaker A	486	474	471	473	471	478	476	486	466	483
Shaker B	494	479	486	477	483	484	487	484	488	482

Find. (a) equal variances?; (b) means differ?; (c) is $\mu_A\lt 480$?; (d) 95% CI for $\mu_B$; (e) is 502 ml anomalous?; (f) qualitative normality argument.

Approach. $F$-test for (a), then a pooled two-sample $t$-test for (b), a one-sample one-sided $t$-test for (c), a $t$-based CI for (d), and a prediction interval for (e).

Sample statistics. $\bar x_A=476.4$, $s_A^2=46.044$ ($s_A=6.786$). $\bar x_B=484.4$, $s_B^2=22.933$ ($s_B=4.789$).
(a) Equal-variance test. $$F=\frac{s_A^2}{s_B^2}=\frac{46.044}{22.933}=\boxed{2.008}$$ Two-tailed $F_{0.025,9,9}=4.026$. Since $2.008\lt4.026$, fail to reject $H_0:\sigma_A^2=\sigma_B^2$ — the variances are not significantly different.
(b) Do the means differ? Pooled $s_p^2=\dfrac{9(46.044)+9(22.933)}{18}=34.489$, $se=\sqrt{34.489(2/10)}=2.626$. $$t=\frac{476.4-484.4}{2.626}=\boxed{-3.046}$$ $df=18$, $t_{0.025,18}=2.101$; since $|t|\gt t_{crit}$ ($p=0.0070$), reject $H_0$: the two shakers collect significantly different volumes (Shaker B collects more, on average).
(c) Is Shaker A's mean below 480 ml? $H_0:\mu_A\ge 480$ vs. $H_a:\mu_A\lt 480$. $se_A=6.786/\sqrt{10}=2.146$. $$t=\frac{476.4-480}{2.146}=\boxed{-1.678}$$ $t_{0.05,9}=1.833$ (one-tailed); since $|{-1.678}|\lt1.833$ ($p=0.064$), fail to reject — although the sample mean sits below target, there is not quite enough evidence at $\alpha=0.05$ to conclude Shaker A under-fills on average.
(d) 95% CI for $\mu_B$. $se_B=4.789/\sqrt{10}=1.5145$, $t_{0.025,9}=2.262$. $$484.4\pm2.262(1.5145)=\boxed{(480.97,\ 487.83)\ \text{ml}}$$
(e) Is a 502 ml unit from Shaker B anomalous? Treat this as a single new draw from Shaker B's population and use a 95% prediction interval: $484.4\pm2.262(4.789)\sqrt{1+\tfrac{1}{10}}=\boxed{(473.04,\ 495.76)\ \text{ml}}$. Since $502\text{ ml}$ falls outside this interval, $$\boxed{\text{yes, 502 ml is a statistically unusual (anomalous) reading}}$$

Summary
Part	Result
(a) $F$, verdict	2.008 vs. crit. 4.026 — variances equal
(b) $t$, verdict	−3.046 — means differ ($p=0.007$)
(c) $t$, verdict	−1.678 — not sig. below 480 ($p=0.064$)
(d) 95% CI, $\mu_B$	(480.97, 487.83) ml
(e) 95% PI, verdict	(473.04, 495.76) ml — 502 ml is an anomaly

(f) Normality argument (no calculation). Collected blood volume is the net result of many small, roughly independent physiological and mechanical influences (donor flow-rate variation, shaker timing/calibration jitter, tubing/needle friction), each contributing a small additive perturbation around a stable target — by the Central Limit Theorem, a measurement built up from many small additive effects tends toward a Normal shape. The values also cluster symmetrically around a central value with no natural hard floor or ceiling nearby (unlike, say, a count or a proportion), which is exactly the physical signature a Normal model describes well.

Topic: $F$-test for equal variances, pooled two-sample and one-sample $t$-tests, CI, and a prediction-interval anomaly check.

Key relations: $F=s_1^2/s_2^2$ vs. $F_{\alpha/2,n_1-1,n_2-1}$; pooled $t=(\bar x_1-\bar x_2)/(s_p\sqrt{1/n_1+1/n_2})$; PI for a new observation $=\bar x\pm t_{\alpha/2,n-1}s\sqrt{1+1/n}$.

Why this works: Confirming equal variances first justifies using the pooled (rather than Welch) $t$-test in (b); treating the single new reading in (e) as drawn from the same population makes the PI (not the narrower CI for the mean) the correct tool for judging whether it is unusual.

Common pitfall: Using the CI for $\mu_B$ (part d) to judge whether the single new 502 ml reading is anomalous — a CI describes uncertainty about the mean, not the spread of individual future observations; only the wider PI is appropriate for that question.

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed.), ch. 8–10.

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