23-Ind-B1 Reliability and Maintainability · May 2015
Question 4 of 10: Proportion of Female Engineering Students — CI and Two-Proportion Tests
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 4 (Section A.4): Proportion of Female Engineering Students — CI and Two-Proportion Tests (5 marks)
Find. (a) 95% CI for $p_{2010}$; (b) test $p_{2010}\ne p_{2000}$; (c) test $p_{2010}\gt 0.35$.
Approach. (a)–(b) use the standard one- and two-proportion large-sample $z$ procedures (pooled proportion for the test in (b)); (c) is a one-sided one-proportion $z$-test against a fixed benchmark of 0.35.
(a) 95% CI for $p_{2010}$. $\hat p_{2010}=421/1290=0.3264$. $$\hat p\pm z_{0.025}\sqrt{\frac{\hat p(1-\hat p)}{n}}=0.3264\pm 1.96\sqrt{\frac{0.3264(0.6736)}{1290}}=\boxed{(0.3008,\ 0.3519)}$$
(b) Two-proportion test, $H_0: p_{2010}=p_{2000}$. $\hat p_{2000}=220/785=0.2803$. Pooled proportion $\bar p=\dfrac{220+421}{785+1290}=0.3089$. $$z=\frac{\hat p_{2010}-\hat p_{2000}}{\sqrt{\bar p(1-\bar p)\left(\frac{1}{n_{2010}}+\frac{1}{n_{2000}}\right)}}=\frac{0.3264-0.2803}{0.0209}=\boxed{2.204}$$ At $\alpha=0.05$ two-tailed ($|z_{crit}|=1.96$, $p=0.028$), reject $H_0$: the female proportion in 2010 is significantly different (higher) than in 2000.
(c) One-sided test, $H_0: p_{2010}\le 0.35$ vs. $H_a: p_{2010}\gt 0.35$. $$z=\frac{0.3264-0.35}{\sqrt{0.35(0.65)/1290}}=\boxed{-1.780}$$ $p=P(Z>-1.780)=0.962$; since $z\lt z_{0.05}=1.645$, fail to reject $H_0$: there is no evidence the 2010 proportion exceeds 35% — consistent with the observed $\hat p_{2010}=0.326$ sitting below 0.35 outright.
