Large-Sample Tests — Mean, Variance, and Proportion: 04-BS-2, December 2015, Q6

Question 6: Large-Sample Tests — Mean, Variance, and Proportion (20 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.

Given data
Quantity	Value
6(A): n	900
6(A): sample mean	1,280 h
6(A): sample s	150 h
6(B): n	2,000
6(B): satisfied	1,600

Find. 6(A)(a) test μ=1300; 6(A)(b) 95% CI for σ² and test σ=120; 6(B)(a) test p=0.85; 6(B)(b) required n for error 0.01 at 99% confidence.

Approach. With $n=900$ and $n=2000$ both large, use the large-sample $z$-test throughout (treating $s$ and $\hat p$ as reliable estimates of $\sigma$ and $p$), plus the supplied large-sample formula for a CI on $\sigma$ itself.

6(A)(a) Test H₀: μ=1300 vs H₁: μ≠1300, α=0.05. $$z=\dfrac{1280-1300}{150/\sqrt{900}}=\dfrac{-20}{5}=\boxed{-4.0}.$$ Since $|z|=4.0>1.96$, reject H₀: the mean lifetime is significantly different from 1,300 hours.
6(A)(b)(i) 95% CI for the variance of L. With $z_{0.025}=1.96$, $s=150$, $n=900$, $\sqrt{2n}=42.43$: $$\sigma_{\text{lo}}=\dfrac{150}{1+1.96/42.43}=143.4,\qquad \sigma_{\text{hi}}=\dfrac{150}{1-1.96/42.43}=157.3.$$ Squaring, the 95% CI for the variance is $$\boxed{20{,}556\text{ h}^2<\sigma^2<24{,}731\text{ h}^2}\quad(\text{i.e. } 143.4\text{ h}<\sigma<157.3\text{ h}).$$
6(A)(b)(ii) test H₀: σ=120 vs H₁: σ≠120, α=0.05. The value 120 h lies well below the interval (143.4, 157.3) h found above, so it falls outside the 95% acceptance region and we reject H₀: the true standard deviation is significantly different from 120 hours.
6(B)(a) Test H₀: p=0.85 vs H₁: p≠0.85, α=0.05. $\hat p=1600/2000=0.80$: $$z=\dfrac{\hat p-0.85}{\sqrt{0.85(0.15)/2000}}=\dfrac{-0.05}{0.00798}=\boxed{-6.26}.$$ Since $|z|=6.26\gg1.96$, reject H₀: the satisfaction proportion is significantly different from 0.85.
6(B)(b) required sample size. Using $\hat p=0.80$ as the best available estimate, $z_{0.005}=2.576$, error $e=0.01$: $$n=\dfrac{z_{0.005}^2\,\hat p(1-\hat p)}{e^2}=\dfrac{(2.576)^2(0.80)(0.20)}{0.01^2}=10{,}615.8\ \Rightarrow\ \boxed{n=10{,}616}.$$

Final results — Question 6
Part	Result
6(A)(a)	z=−4.0, reject H₀
6(A)(b)(i)	20,556 < σ² < 24,731 h²
6(A)(b)(ii)	reject H₀: σ≠120 h
6(B)(a)	z=−6.26, reject H₀
6(B)(b)	n=10,616

Topic: Large-sample z-tests for a mean, a proportion, and (via the supplied approximation) a standard deviation.

Key relations: $z=(\bar X-\mu_0)/(s/\sqrt n)$; $z=(\hat p-p_0)/\sqrt{p_0q_0/n}$; sample-size formula $n=z_{\alpha/2}^2\hat p(1-\hat p)/e^2$; supplied large-sample CI for $\sigma$.

Why this works: for large $n$, the sample mean/proportion is approximately normal by the CLT regardless of the underlying distribution, and $s$/$\hat p$ are close enough to $\sigma$/$p$ to be substituted directly (no $t$ or $\chi^2$ correction needed).

Common pitfall: squaring the supplied $\sigma$ bounds incorrectly, or testing 6(A)(b)(ii) with a fresh z-statistic instead of simply checking whether 120 falls inside the interval already built in (i) — both are equivalent, but the interval check is the intended (and faster) route the question sets up.

Source: Walpole et al., Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 9–10 (large-sample estimation and testing).

Question 6 of 8: Large-Sample Tests — Mean, Variance, and Proportion

Question 6: Large-Sample Tests — Mean, Variance, and Proportion (20 marks)