Question 6 of 8: Airline Luggage & Municipal-Services Survey
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, May 2014 — 04-BS-2 Probability and Statistics (2 hours, closed book except one hand-written information sheet and an approved calculator; any 5 of the 8 questions constitute a complete paper — all 8 are answered below as a full study resource).
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed. (sampling distributions, estimation, hypothesis testing, correlation/regression); standard Normal, t, χ² and F tables as supplied with the exam.
Find. (A)(a) test H0: μ≤5.5 vs H1: μ>5.5; (A)(b) test H0: σ=1.2 via the given large-sample CI formula; (B)(a) test H0: p=0.65 vs H1: p≠0.65; (B)(b) required n for a 99% CI with margin of error 0.01.
Approach. With n≥1,800 in both parts, the sample s and p̂ are treated as known for a large-sample z-test/CI (CLT). Part (A)(b) tests σ0 by checking whether it falls inside the supplied approximate CI rather than computing a separate test statistic. Part (B)(b) inverts the margin-of-error formula $E=z_{\alpha/2}\sqrt{p(1-p)/n}$ for n, using the observed p̂ as the best available estimate.
(A)(a) One-sided test of the mean luggage weight. $H_0:\mu=5.5$ vs $H_1:\mu>5.5$. With n=1,800 large, use $\sigma\approx s$:
$$z=\dfrac{\bar w-\mu_0}{s/\sqrt n}=\dfrac{5.850-5.5}{1.350/\sqrt{1800}}=\boxed{11.00}.$$
The one-sided critical value is $z_{0.05}=1.645$. Since $11.00\gg1.645$, reject H0: the mean hand-luggage weight is significantly larger than 5.5 kg.
(A)(b) Testing σ=1.2 via the given large-sample CI. With $z_{0.025}=1.960$, $s=1.350$, $n=1800$: $z_{\alpha/2}/\sqrt{2n}=1.960/\sqrt{3600}=0.03267$, so
$$\dfrac{1.350}{1+0.03267}<\sigma<\dfrac{1.350}{1-0.03267}\ \Rightarrow\ \boxed{1.307<\sigma<1.396\ \text{kg}}.$$
Since $\sigma_0=1.2$ lies outside this 95% interval, reject H0: the true standard deviation of hand-luggage weight is significantly different from (larger than) 1.2 kg.
(B)(a) Test of the dissatisfaction proportion. $\hat p=1960/2800=0.70$. $H_0:p=0.65$ vs $H_1:p\ne0.65$, using $p_0$ in the standard error under $H_0$:
$$z=\dfrac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}=\dfrac{0.70-0.65}{\sqrt{0.65(0.35)/2800}}=\boxed{5.547}.$$
The two-sided critical value is $z_{0.025}=1.960$. Since $5.547>1.960$, reject H0: the dissatisfaction proportion is significantly different from (higher than) 0.65.
(B)(b) Sample size for a 99% CI with E=0.01. Using the observed $\hat p=0.70$ as the best prior estimate of p and $z_{0.005}=2.576$,
$$n=\dfrac{z_{\alpha/2}^2\,\hat p(1-\hat p)}{E^2}=\dfrac{2.576^2(0.70)(0.30)}{0.01^2}=13{,}933.3\ \Rightarrow\ \boxed{n=13{,}934}\ \text{(rounded up)}.$$
(If no prior estimate were available, the conservative $\hat p=0.5$ would instead give $n=16{,}588$ — the sample already in hand makes the smaller, tighter figure the appropriate one here.)
