23-Ind-B1 Reliability and Maintainability · December 2016
Question 4 of 11: Aeration and Fermentation Time — Two-Sample t -Test Against a Claimed Reduction
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 4 (Section A.4): Aeration and Fermentation Time — Two-Sample t-Test Against a Claimed Reduction (15 marks)
Given. Original-setup fermentation times (Q2): 19, 18, 19, 20, 18, 19 days. New-setup fermentation times: 17, 20, 18, 16, 14, 17 days. Claim: mean drop of 1.5 days.
Find. Whether the data support (are consistent with) a true 1.5-day reduction in mean fermentation time.
Approach. Two independent samples of $n=6$ each: first an $F$-test decides pooled vs. Welch variance handling, then a two-sample $t$-test of $H_0:\mu_{\text{orig}}-\mu_{\text{new}}=1.5$ directly addresses the claim; a supporting one-sided test against $0$ checks whether any real reduction exists at all.
Equal-variance check. $$F=\frac{s_2^2}{s_1^2}=\frac{4.0}{0.5667}=7.059$$ Two-tailed $F_{0.025,5,5}=7.146$. Since $7.059\lt 7.146$, fail to reject equal variances (barely) — proceed with the pooled-variance $t$-test.
Test $H_0:\mu_1-\mu_2=1.5$ (does the data match the claim?). Pooled variance $s_p^2=\dfrac{5(0.5667)+5(4.0)}{10}=2.283$, so $s_p=1.511$. $$se=s_p\sqrt{\tfrac16+\tfrac16}=0.8724,\qquad t=\frac{(18.833-17.0)-1.5}{0.8724}=\boxed{0.382}$$ $df=10$, $t_{0.025,10}=2.228$; since $|t|\ll t_{crit}$, fail to reject $H_0$: the data are consistent with the claimed 1.5-day reduction — there is no evidence the true reduction differs from 1.5 days. A 95% CI for the true mean reduction is $1.833\pm2.228(0.8724)=\boxed{(-0.111,\ 3.777)\ \text{days}}$, which comfortably contains 1.5.
Summary
Quantity
Value
$\bar x_1-\bar x_2$ (observed drop)
1.833 days
$F$ (equal-var check)
7.059 vs. crit. 7.146 — equal
$t$ ($H_0$: drop $=1.5$)
0.382, $df=10$ — fail to reject
95% CI for true mean drop
(−0.111, 3.777) days
Verdict
Data support the 1.5-day claim
Check — a real reduction does exist, it just cannot be pinned to exactly 1.5 days A one-sided test of $H_0:\mu_1-\mu_2\le 0$ vs. $H_a:\mu_1-\mu_2\gt 0$ gives $t=1.833/0.8724=2.101$ against $t_{0.05,10}=1.812$ ($p=0.031$) — a statistically significant improvement exists. With only $n=6$ per group the CI for the size of that improvement is wide (roughly 0 to 3.8 days), so the data cannot distinguish a 1.5-day effect from, say, a 1.0- or 2.5-day one; they simply do not contradict the claim.
