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23-Ind-B1 Reliability and Maintainability · December 2017

Question 6 of 10: Deflection Temperature of Plastic Pipe — Variance Test, Two-Sample t -Test, Sample-Size Design

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Notes on this paper

National Exams — December 2017 — 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 3 (40 marks) — a 6-question, 100-mark paper as printed. All ten questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — discrete/continuous distributions (ch. 3–4), joint distributions (ch. 5), point/interval estimation and sample size (ch. 8), hypothesis testing incl. two-sample tests (ch. 9–10), simple linear regression (ch. 11), design and analysis of single-factor experiments (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — multi-factor and $2^k$ factorial designs (ch. 5–6).

Question 6 (Section B.2): Deflection Temperature of Plastic Pipe — Variance Test, Two-Sample t-Test, Sample-Size Design (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. Two independent samples, $n_1=n_2=10$, listed above.

Find. (a) $F$-test for equal variances; (b) one-sided two-sample $t$-test; (c) required $n$ per group for 90% power at $\delta=2^{\circ}F$.

Approach. Compute both sample means/variances, test equal variances first (to decide pooled vs. Welch $t$), then run the one-sided pooled $t$-test and the standard two-sample power/sample-size formula.

  1. Sample statistics. Type 1: $\bar x_1=196.7$, $s_1^2=101.57$. Type 2: $\bar x_2=191.1$, $s_2^2=117.43$ (both $n=10$).
  2. (a) $F$-test for equal variances. $H_0:\sigma_1^2=\sigma_2^2$. $$F=\frac{s_2^2}{s_1^2}=\frac{117.43}{101.57}=\boxed{1.156}$$ Two-tailed $\alpha=0.05$, $F_{0.025,9,9}=4.026$; since $1.156\lt 4.026$, fail to reject $H_0$ — the two variances are not significantly different, so pooling is justified.
  3. (b) One-sided pooled $t$-test, $H_0:\mu_1\le\mu_2$ vs. $H_a:\mu_1>\mu_2$. Pooled variance $s_p^2=\dfrac{9(101.57)+9(117.43)}{18}=109.5$. $$t=\frac{196.7-191.1}{\sqrt{109.5(\tfrac1{10}+\tfrac1{10})}}=\boxed{1.197}$$ $t_{0.05,18}=1.734$; since $1.197\lt 1.734$ ($p=0.123$), fail to reject $H_0$ — the data do not provide significant evidence that Type 1's deflection temperature exceeds Type 2's, despite the higher sample mean.
  4. (c) Sample size for 90% power at $\delta=2^{\circ}F$. Using the pooled standard deviation $s_p=\sqrt{109.5}=10.464$ from part (a)-(b), one-sided $\alpha=0.05$ ($z_{0.05}=1.645$), power $=0.90$ ($z_{0.10}=1.282$): $$n=\frac{2(z_\alpha+z_\beta)^2 s_p^2}{\delta^2}=\frac{2(1.645+1.282)^2(109.5)}{2^2}=468.9$$ Rounded to the nearest 10: $\boxed{n\approx 470\text{ pipes per type}}$.
Summary
PartResult
(a) $F$, verdict1.156; variances not different
(b) $t$, verdict1.197; no significant excess ($p=0.123$)
(c) required $n$/group≈470
Check — large required sample size The naturally large $n\approx 470$ in (c) follows directly from the data: the pipe-to-pipe standard deviation ($\approx 10.5^{\circ}$F) is roughly 5× the 2°F difference the study wants to detect, so a large sample is genuinely needed — this is not a calculation error.