23-Ind-B1 Reliability and Maintainability · May 2015

Question 6 of 10: Class-Mark Variability at Université de Quimper — CI for Variance, Ratio of Variances, and a Mean Test

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

Notes on this paper

National Exams — May 2015 — 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 questions (10 marks); Section B: do 2 of 3 (20 marks); Section C: do 1 of 3 (20 marks) — a 5-question, 50-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) — joint distributions (ch. 5), binomial/geometric probability (ch. 3), normal distribution and sampling distributions (ch. 4–7), point/interval estimation (ch. 8–9), hypothesis testing incl. sample-size design (ch. 9–10), simple linear regression and ANOVA (ch. 11–13), Bartlett's and Tukey's tests, single-degree-of-freedom contrasts and 2³ factorial designs (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).

Question 6 (Section B.2): Class-Mark Variability at Université de Quimper — CI for Variance, Ratio of Variances, and a Mean Test (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. Recent 5-year averages: 71.24, 75.05, 79.12, 73.20, 78.50 ($n=5$). 1980's: $\bar x=72.6$, $s^2=6.02$, $n=10$.

Find. (a) 90% CI for the current variance; (b) 90% CI for the ratio of variances; (d) test $H_0:\mu=72.6$.

Approach. (a) uses the chi-square CI for a single variance; (b) the $F$-based CI for a ratio of two independent sample variances; (d) a one-sample $t$-test of the recent mean against the historical 1980's mean.

(a) 90% CI for the current 5-year variance. Sample mean $=75.422$, sample variance $s^2=11.4285$ (computed from the 5 values). With $\chi^2_{0.95,4}=0.7107$ and $\chi^2_{0.05,4}=9.4877$: $$\left(\frac{(n-1)s^2}{\chi^2_{0.05,4}},\ \frac{(n-1)s^2}{\chi^2_{0.95,4}}\right)=\left(\frac{4(11.4285)}{9.4877},\ \frac{4(11.4285)}{0.7107}\right)=\boxed{(4.82,\ 64.32)}$$
(b) 90% CI for $\sigma^2_{now}/\sigma^2_{80s}$. Ratio of sample variances $=11.4285/6.02=1.898$. With $F_{0.05,4,9}=3.633$ and $F_{0.05,9,4}=5.999$: $$\left(\frac{s^2_{now}/s^2_{80s}}{F_{0.05,4,9}},\ \frac{s^2_{now}}{s^2_{80s}}\times F_{0.05,9,4}\right)=\left(\frac{1.898}{3.633},\ 1.898\times 5.999\right)=\boxed{(0.52,\ 11.39)}$$
(c) Interpretation of the ratio CI. The 90% CI for $\sigma^2_{now}/\sigma^2_{80s}$ spans 0.52 to 11.39 — it comfortably contains 1, so the data give no evidence the year-to-year spread of class averages has changed between the two eras (the CI is very wide because both variance estimates rest on small samples, $n=5$ and $n=10$).
(d) Has the mean class average changed? $H_0:\mu=72.6$ vs. $H_a:\mu\ne 72.6$. $$t=\frac{75.422-72.6}{\sqrt{11.4285}/\sqrt5}=\frac{2.822}{1.512}=\boxed{1.867}$$ $t_{0.025,4}=2.776$; since $1.867\lt 2.776$ ($p=0.135$), fail to reject $H_0$: no significant evidence the mean has changed.
(e) Interpretation of (d). Because neither the mean nor the variance shows a statistically significant shift from the 1980's, the modest year-to-year swings observed in the 5 recent averages (71.2 to 79.1) are consistent with ordinary sampling variability among different student cohorts, not a systematic change in course difficulty, grading, or student population.

Summary
Part	Result
(a) 90% CI, variance	(4.82, 64.32)
(b) 90% CI, variance ratio	(0.52, 11.39)
(d) $t$, verdict	1.867; no significant mean change ($p=0.135$)

Topic: Chi-square CI for a variance, $F$-based CI for a ratio of variances, one-sample $t$-test.

Key relations: $\left(\dfrac{(n-1)s^2}{\chi^2_{\alpha/2,n-1}},\dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2,n-1}}\right)$ for a variance; $\left(\dfrac{s_1^2/s_2^2}{F_{\alpha/2,n_1-1,n_2-1}},\ \dfrac{s_1^2}{s_2^2}F_{\alpha/2,n_2-1,n_1-1}\right)$ for a variance ratio.

Why this works: $(n-1)s^2/\sigma^2$ is chi-square distributed under normality, and the ratio of two independent chi-squares (each scaled by its df) is $F$-distributed — both intervals here are the standard textbook forms, just evaluated at $n=5$ and $n=10$.

Common pitfall: Swapping which sample's df goes in the numerator vs. denominator of $F$ — the CI for the ratio is NOT symmetric in the two $F$ critical values used, since $F_{\alpha/2,n_1-1,n_2-1}\ne F_{\alpha/2,n_2-1,n_1-1}$.

Source: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed.), ch. 8–9 (interval estimation for variance and ratio of variances).

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