23-Ind-B1 Reliability and Maintainability · December 2018

Question 2 of 9: Two-Sample $t$-Test for a Process Design Change

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 2 (Section A.2): Two-Sample $t$-Test for a Process Design Change (10 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.

Sample	$n$	$\bar{x}$	$s$
Before (1)	5	10	3
After (2)	5	4	5

Find. (a) Two-sided test of $\mu_1=\mu_2$ at $\alpha=0.05$; (b) one-sided test of $\mu_1>\mu_2$ at $\alpha=0.05$; (c) reconciliation, plus a test of whether the improvement $\mu_1-\mu_2$ is at least 1.

Approach. Both samples are small ($n=5$) and normally distributed with unknown, comparable-magnitude population variances, so use the pooled two-sample $t$-test: pool the sample variances into $s_p^2$, form the pooled standard error, then compare $t=(\bar{x}_1-\bar{x}_2-\delta_0)/SE$ against the appropriate critical value.

Pool the variances and the standard error. $$s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}=\frac{4(3^2)+4(5^2)}{8}=\frac{36+100}{8}=17.0,\qquad s_p=4.123$$ $$SE=s_p\sqrt{\tfrac{1}{n_1}+\tfrac{1}{n_2}}=4.123\sqrt{0.4}=2.608,\qquad df=n_1+n_2-2=8$$
(a) Two-sided test, $\alpha=0.05$. $$t=\frac{\bar{x}_1-\bar{x}_2}{SE}=\frac{10-4}{2.608}=\boxed{2.301}$$ Critical value $t_{0.025,8}=2.306$. Since $|t|=2.301<2.306$ (just barely), fail to reject $H_0$ at the two-sided 5% level ($p=0.0504$, a hair above 0.05).
(b) One-sided test, $\alpha=0.05$. Same $t=2.301$; critical value $t_{0.05,8}=1.860$. Since $t=2.301>1.860$, reject $H_0$ in favour of $H_1:\mu_1>\mu_2$ ($p=0.0252$).
(c) Reconcile and test the ≥1 improvement claim. Parts (a) and (b) are not contradictory — they test different alternative hypotheses against different rejection regions. The two-sided test in (a) asks only whether the means differ at all and lands just short of significance; the one-sided test in (b), justified once the direction of the expected change (a design change intended to reduce the process value) is fixed in advance, uses a smaller critical value and does detect the difference. To directly answer management's question — is the improvement at least 1, where lower is better — shift the null to that threshold: $H_0:\mu_1-\mu_2\le1$ vs. $H_1:\mu_1-\mu_2>1$. $$t_c=\frac{(\bar{x}_1-\bar{x}_2)-1}{SE}=\frac{6-1}{2.608}=\boxed{1.917}$$ Critical value (one-sided, $\alpha=0.05$, $df=8$) is again $1.860$; since $t_c=1.917>1.860$ ($p=0.0457$), reject $H_0$: there is sufficient evidence at the 5% level that the process improved by at least 1 unit.

Final Results
Test	Statistic	Critical value	Conclusion
(a) two-sided, $\mu_1=\mu_2$	$t=2.301$	$\pm2.306$	Fail to reject ($p=0.050$)
(b) one-sided, $\mu_1>\mu_2$	$t=2.301$	$1.860$	Reject ($p=0.025$)
(c) one-sided, improvement $\ge1$	$t_c=1.917$	$1.860$	Reject ($p=0.046$) — improvement of $\ge1$ is significant

Check Only sample means/SDs are given, not raw data, and the exam does not state whether the two population variances should be treated as equal — the pooled-variance $t$-test (equal-variance assumption, the standard default for two comparable small samples with no stated evidence otherwise) is used throughout part (c)'s guidance.

Topic: Two-sample $t$-test for a mean difference (one- and two-sided), pooled variance.

Key relations: $s_p^2=\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$; $t=\dfrac{(\bar{x}_1-\bar{x}_2)-\delta_0}{s_p\sqrt{1/n_1+1/n_2}}\sim t_{n_1+n_2-2}$.

Why this works: A one-sided alternative concentrates the whole 5% rejection region in one tail, so it detects a difference in the hypothesized direction with a smaller critical value than a two-sided test, which must split 5% across both tails; shifting $\delta_0$ from 0 to 1 directly tests a business-meaningful threshold rather than mere non-equality.

Common pitfall: Treating a "fail to reject" two-sided result at $p=0.050$ as proof of no difference, then being surprised the one-sided test rejects — the two tests answer different questions, not the same question twice.

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th ed., §9.3 (inference on two population means, pooled $t$-test).

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