Question 6 of 8

Question 6 (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. (A) $n=1{,}600$, $\bar L=251{,}000$ km, $s=30{,}000$ km. (B) $n=2{,}200$, $x=1{,}950$ satisfied.

Find. (A)(a) test $H_0:\mu=250{,}000$; (A)(b) 95% CI for $\sigma$ (and $\sigma^2$), and test $H_0:\sigma=25{,}000$; (B)(a) test $H_0:p=0.90$; (B)(b) required $n$ for error 0.01 at 99% confidence.

Approach. Large-$n$ tests on the mean and proportion use the $z$-statistic; the standard-deviation CI uses the given large-sample approximation formula directly; the required-sample-size formula inverts the margin-of-error expression for a proportion using the best available point estimate of $p$.

1. (A)(a) Test H₀: μ = 250,000 km (α = 0.05, z-test, large n). $z=\dfrac{\bar L-\mu_0}{s/\sqrt n}=\dfrac{251{,}000-250{,}000}{30{,}000/\sqrt{1{,}600}}=\dfrac{1{,}000}{750}=\boxed{1.333}$. Critical $z_{0.025}=1.960$. Since $|z|=1.333<1.960$, $\boxed{\text{fail to reject }H_0}$ — mean lifetime is NOT significantly different from 250,000 km.
2. (A)(b)(i) 95% CI for σ using the given large-sample formula. $z_{0.025}=1.960$, $\sqrt{2n}=\sqrt{3{,}200}=56.57$, so $z_{0.025}/\sqrt{2n}=0.03465$. $\sigma_{\text{low}}=\dfrac{30{,}000}{1.03465}=28{,}995$ km, $\sigma_{\text{high}}=\dfrac{30{,}000}{0.96535}=31{,}077$ km, i.e. $\boxed{28{,}995\text{ km}<\sigma<31{,}077\text{ km}}$. Squaring: $\boxed{8.407\times10^8\text{ km}^2<\sigma^2<9.658\times10^8\text{ km}^2}$.
3. (A)(b)(ii) Test H₀: σ = 25,000 km at α = 0.05 using the same CI. The hypothesized value 25,000 km lies below the entire 95% CI (28,995–31,077 km), so $\boxed{\text{reject }H_0}$ — the true standard deviation IS significantly different from 25,000 km.
4. (B)(a) Test H₀: p = 0.90 (α = 0.05, z-test on a proportion). $\hat p=\dfrac{1{,}950}{2{,}200}=0.8864$. $z=\dfrac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}=\dfrac{0.8864-0.90}{\sqrt{0.90(0.10)/2{,}200}}=\dfrac{-0.01364}{0.006396}=\boxed{-2.132}$. Since $|z|=2.132>1.960$, $\boxed{\text{reject }H_0}$ — the satisfaction proportion IS significantly different from 0.90.
5. (B)(b) Required sample size (error 0.01, 99% confidence). $n=\dfrac{z_{0.005}^2\,\hat p(1-\hat p)}{e^2}$, using $\hat p=0.8864$ as the best prior estimate (from part (a)): $n=\dfrac{(2.576)^2(0.8864)(0.1136)}{(0.01)^2}=\dfrac{6.6306(0.1007)}{0.0001}=\dfrac{0.6679}{0.0001}=6{,}679$, rounded up to the next integer $\boxed{n=6{,}683}$ (the small discrepancy from a hand-rounded 6,679 is resolved by carrying full precision on $z_{0.005}=2.5758$: $n=6{,}682.9\to6{,}683$).