Question 6 of 8: Large-Sample Tests — Tile Weight and Road Satisfaction
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, May 2013 — 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.
Check: the printed sample-size line for Question 7 reads "thirteen bulbs were randomly selected...one result had to be discarded" immediately above a table showing nA=13, nB=12. Read as: Make A's sample stayed at 13 (no discard), Make B's original 13 dropped to 12 after the discard — the table's printed nA=13, nB=12 are used as given.
Given. Part A: n=625, x̄=1,300.0 g, s=5.0 g. Part B: n=2,500, x=1,200 dissatisfied.
Find. (A)(a) test H₀: μ=1,300.5 at α=0.05; (A)(b) approximate 95% CI for σ and test H₀: σ=6.0; (B)(a) test H₀: p=0.5 at α=0.05; (B)(b) required sample size for a 99% CI on p with error 0.01.
Approach. With n = 625 and n = 2,500 both large, use the Normal (z) approximation throughout — for the mean (known-form s as an estimate of σ), for the CI-based test on σ using the supplied large-sample formula, for the proportion test, and for the minimum-n formula $n=z_{\alpha/2}^2 pq/E^2$.
6A(a) z-test for the mean, H₀: μ=1300.5 vs H₁: μ≠1300.5.
$$z=\dfrac{\bar x-\mu_0}{s/\sqrt n}=\dfrac{1300.0-1300.5}{5.0/\sqrt{625}}=\dfrac{-0.5}{0.20}=\boxed{-2.50}.$$
Critical value $z_{0.025}=1.96$. Since $|z|=2.50>1.96$, reject H₀: the mean weight is significantly different from 1,300.5 g.
6A(b) Large-sample CI for σ, then test H₀: σ=6.0. With $z_{0.025}=1.96$ and $s/\sqrt{2n}=5.0/\sqrt{1250}=0.1414$, the correction factor is $z_{0.025}\cdot s/\sqrt{2n}=1.96(0.1414)/5.0\times5.0=0.05544$ (i.e. $z_{\alpha/2}/\sqrt{2n}=0.05544$, applied directly to s):
$$\dfrac{s}{1+0.05544}<\sigma<\dfrac{s}{1-0.05544}\ \Rightarrow\ \boxed{4.737<\sigma<5.293\text{ g}}.$$
Since the hypothesized $\sigma_0=6.0$ g lies outside this 95% interval, reject H₀: the true standard deviation is significantly different from 6.0 g at α=0.05.
6B(a) z-test for a proportion, H₀: p=0.5 vs H₁: p≠0.5. $\hat p=1200/2500=0.48$.
$$z=\dfrac{\hat p-p_0}{\sqrt{p_0q_0/n}}=\dfrac{0.48-0.50}{\sqrt{(0.5)(0.5)/2500}}=\dfrac{-0.02}{0.01}=\boxed{-2.00}.$$
Critical value $z_{0.025}=1.96$. Since $|z|=2.00>1.96$, reject H₀: the proportion dissatisfied is significantly different from 0.5.
6B(b) Minimum sample size for a 99% CI on p with error 0.01. With no reliable prior estimate of the satisfied proportion offered for design purposes, use the conservative $p=q=0.5$ (maximizes $pq$, guarantees the error bound):
$$n=\dfrac{z_{0.005}^2\,pq}{E^2}=\dfrac{(2.576)^2(0.5)(0.5)}{(0.01)^2}=16{,}587.2\ \Rightarrow\ \boxed{n=16{,}588}\ (\text{round up}).$$
(If the survey's own estimate of satisfied $\hat q=1-0.48=0.52$ is used instead of the conservative 0.5, $n=(2.576)^2(0.52)(0.48)/(0.01)^2\approx16{,}561$ — essentially the same order, since $\hat p\hat q$ is close to its 0.25 maximum.)
