23-Ind-B1 Reliability and Maintainability · December 2016

Question 2 of 11: Beer Fermentation — Correlation, Confidence Interval, Prediction Interval

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 2 (Section A.2): Beer Fermentation — Correlation, Confidence Interval, Prediction Interval (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. Six batches, basement temperature (°C) and fermentation time (days):

Batch	1	2	3	4	5	6
Temp (°C)	14	18	21	19	16	17
Time (days)	19	18	19	20	18	19

Find. (a) mean/variance of each; (b) $\mathrm{Cov}$; (c) $r$ and its interpretation; (d) 95% CI for mean fermentation time; (e) 95% PI for basement temperature.

Approach. Sample moments from the six pairs, then the standard $t$-based CI (for the population mean) and PI (for one future individual observation), both with $n-1=5$ degrees of freedom.

1. (a) Means and variances. $\bar T=17.5\,{}^{\circ}\text{C}$, $s_T^2=5.9$ (so $s_T=2.429$). $\bar t=18.833$ days, $s_t^2=0.5667$ (so $s_t=0.7528$).
2. (b) Covariance. $$s_{Tt}=\frac{1}{n-1}\sum(T_i-\bar T)(t_i-\bar t)=\boxed{0.500}$$
3. (c) Correlation. $$r=\frac{s_{Tt}}{s_T s_t}=\frac{0.500}{2.429\times 0.7528}=\boxed{0.2735}$$ Testing $H_0:\rho=0$: $t=r\sqrt{(n-2)/(1-r^2)}=0.569$, $df=4$, $t_{0.025,4}=2.776$; since $|t|\ll t_{crit}$, the correlation is not statistically significant — on this sample, basement temperature is only weakly (and not demonstrably) associated with fermentation time.
4. (d) 95% CI for mean fermentation time. $t_{0.025,5}=2.5706$. $$\bar t\pm t_{0.025,5}\frac{s_t}{\sqrt n}=18.833\pm 2.5706\frac{0.7528}{\sqrt6}=\boxed{(18.043,\ 19.623)\ \text{days}}$$
5. (e) 95% PI for basement temperature. A prediction interval for one future individual temperature draw uses $\bar T\pm t_{0.025,5}\,s_T\sqrt{1+1/n}$: $$14.5+3.0\pm 2.5706\times2.429\sqrt{1+\tfrac16}=\boxed{(10.756,\ 24.244)\ {}^{\circ}\text{C}}$$

Summary

Quantity	Value
$\bar T$, $s_T^2$	17.5°C, 5.9
$\bar t$, $s_t^2$	18.833 d, 0.5667
$\mathrm{Cov}$, $r$	0.500, 0.2735 (n.s., $p=0.60$)
95% CI mean time	(18.043, 19.623) days
95% PI temp	(10.756, 24.244) °C

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