23-Ind-B1 Reliability and Maintainability · December 2014

Question 6 of 9: Sample-Size, One-Sample $t$-Test, and Confidence Intervals for IBU

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

Notes on this paper

National Exams — December 2014 — 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: four sections — Section A: do 2 of 3 questions (20 marks); Section B: do 1 of 2 (25 marks); Section C: do 1 of 2 (25 marks); Section D: do 1 of 2 (30 marks) — a 5-question, 100-mark paper as printed. All nine questions across the four sections are solved below for completeness. Page 1's own NOTES list is mis-numbered (two items both labelled "4."), transcribed as printed.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — point/interval estimation (ch. 8–9), hypothesis testing (ch. 9–10), simple/multiple linear regression (ch. 11–12), single- and two-factor ANOVA (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).

Question 6 (Section C.1): Sample-Size, One-Sample $t$-Test, and Confidence Intervals for IBU (25 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. Baseline yield $\mu_0=80$, $\sigma=5$ IBU (known, from the unmodified plant); target detectable mean $\mu_1=82$; $\alpha=0.05$. Sample IBU readings ($n=5$): 88.5, 91.0, 97.7, 85.0, 89.7.

Find. (a) minimum sample size $n$; (b) name the calculation; (c) test $H_0:\mu=85$ vs. $H_1:\mu>85$; (d) 90% CI for $\mu$; (e) 90% CI for $\sigma^2$.

Approach. (a)–(b) use the one-sided $z$-test sample-size formula against $\sigma$; (c) uses a one-sample $t$-test since $\sigma$ is not known for the modified plant; (d)–(e) use the $t$- and $\chi^2$-based interval formulas at $df=n-1$.

Part (a) — minimum sample size. Treating $\sigma=5$ as known (carried over from the unmodified plant's stated distribution), a one-sided $z$-test rejects $H_0:\mu=80$ when $\bar x > \mu_0+z_{0.05}\sigma/\sqrt n$. The smallest sample that can still detect a true mean of exactly $\mu_1=82$ is the $n$ for which this critical value coincides with $\mu_1$ itself: $$\mu_0+z_{0.05}\frac{\sigma}{\sqrt n}=\mu_1 \ \Longrightarrow\ n=\left(\frac{z_{0.05}\,\sigma}{\mu_1-\mu_0}\right)^2 = \left(\frac{1.645\times5}{2}\right)^2 = 16.9 \ \Longrightarrow\ \boxed{n=17}.$$
Part (b) — naming the calculation. This is a sample-size (power) determination for a one-sample hypothesis test — specifically, since no target power/$\beta$ is stated in the question, the $n$ found is the smallest sample for which the test's rejection threshold reaches the alternative mean at all, i.e. the sample size that gives exactly 50% power to detect $\mu_1=82$. A complete power analysis would separately fix a target power (e.g. 80% or 90%) and solve the two-term formula $n=\left(\dfrac{(z_\alpha+z_\beta)\sigma}{\mu_1-\mu_0}\right)^2$; with no $\beta$ specified here, $n=17$ is the minimum defensible answer, not a full power-analysis result.
Part (c) — one-sample $t$-test. From the 5 IBU readings, $\bar x=90.38$, $s=4.661$. Since $\sigma$ is unknown for the modified plant, test with $$t_0=\frac{\bar x-85}{s/\sqrt n}=\frac{90.38-85}{4.661/\sqrt5}=\boxed{2.581}.$$ The one-sided critical value is $t_{0.05,4}=2.132$. Since $t_0=2.581>2.132$, reject $H_0$: the sample provides significant evidence at $\alpha=0.05$ that the modified plant's mean IBU exceeds 85.
Part (d) — 90% CI for the mean. With $t_{0.05,4}=2.132$ (two-sided 90%, $df=4$): $$\bar x \pm t_{0.05,4}\frac{s}{\sqrt n} = 90.38 \pm 2.132\left(\frac{4.661}{\sqrt5}\right) = \boxed{(85.9,\ 94.8)\ \text{IBU}}.$$
Part (e) — 90% CI for the variance. With $s^2=21.73$, $df=4$, and $\chi^2_{0.05,4}=9.488$, $\chi^2_{0.95,4}=0.711$: $$\left(\frac{(n-1)s^2}{\chi^2_{0.05,4}},\ \frac{(n-1)s^2}{\chi^2_{0.95,4}}\right) = \left(\frac{4(21.73)}{9.488},\ \frac{4(21.73)}{0.711}\right) = \boxed{(9.16,\ 122.3)}.$$

Part	Result
(a) Minimum $n$	17
(b) Calculation type	sample-size / power determination (50% power, since $\beta$ not specified)
(c) $t_0$ vs $t_{0.05,4}$	2.581 > 2.132 — reject $H_0$
(d) 90% CI for $\mu$	(85.9, 94.8) IBU
(e) 90% CI for $\sigma^2$	(9.16, 122.3)

Check: part (a) has no stated target power/$\beta$, only $\alpha=0.05$ — the minimum defensible reading is the smallest $n$ for which the test can reject at all when $\mu=82$ (50% power), used above and named explicitly in part (b) rather than silently assumed.

Concept & Theory

Topic. Sample-size determination, one-sample $t$-test, and $t$-/$\chi^2$-based confidence intervals.

Key relations. $n=\left(z_\alpha\sigma/\Delta\right)^2$ (50% power case); $t_0=(\bar x-\mu_0)/(s/\sqrt n)$; CI for mean $\bar x\pm t_{\alpha/2,n-1}s/\sqrt n$; CI for variance $\left(\tfrac{(n-1)s^2}{\chi^2_{\alpha/2,n-1}},\tfrac{(n-1)s^2}{\chi^2_{1-\alpha/2,n-1}}\right)$.

Why this works. The variance CI is asymmetric because $(n-1)s^2/\sigma^2$ follows a right-skewed $\chi^2$ distribution, not a symmetric one like $\bar x$'s $t$-distribution — the lower bound divides by the upper-tail $\chi^2$ critical value, and vice versa, which is a common sign-swap error.

Common pitfall. Using $z$ instead of $t$ for parts (c)–(d) once $\sigma$ is no longer assumed known (it is only known for the unmodified plant, used in part (a); the modified plant's $\sigma$ must be estimated from the 5-sample $s$).

Source. Montgomery & Runger, Applied Statistics and Probability for Engineers, ch. 9 (hypothesis testing on a single sample), ch. 8 (interval estimation, incl. variance).

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