Question 6 of 8: Drill/Driver Kit Lifetime and a Recycling Satisfaction Survey

Question 6: Drill/Driver Kit Lifetime and a Recycling Satisfaction Survey (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. 6(A): n=1,600, mean useful life $\bar L$=1,380 h, s=150 h. 6(B): n=2,500, x=1,800 satisfied.

Find. 6(A)(a) test H₀: μ=1,400 at α=0.05. (b) 95% approximate CI for σ using the given formula, and test H₀: σ=130. 6(B)(a) test H₀: p=0.75 at α=0.05. (b) required n for error 0.01, 99% confidence.

Approach. With n=1,600 large, the mean test uses a z-statistic (s substituted for σ); the approximate σ interval uses the large-sample formula the question itself supplies; the proportion test uses the standard large-sample z for a proportion, and the sample-size question inverts the margin-of-error formula.

1. 6(A)(a) Test H₀: μ=1,400 vs H₁: μ≠1,400, α=0.05. With n large, $$z=\dfrac{\bar L-\mu_0}{s/\sqrt n}=\dfrac{1380-1400}{150/\sqrt{1600}}=\dfrac{-20}{3.75}=-5.333$$ The two-sided critical value is $z_{0.025}=1.96$. Since $|z|=5.333\gt1.96$, $$\boxed{\text{Reject } H_0\text{: the mean useful life is significantly different from }1{,}400\text{ h}}$$
2. 6(A)(b)(i) 95% CI for σ using the given large-sample formula. With $z_{0.025}=1.96$, $n=1600$, $\sqrt{2n}=56.57$: $$\dfrac{z_{\alpha/2}}{\sqrt{2n}}=\dfrac{1.96}{56.57}=0.03465$$ $$\dfrac{150}{1+0.03465}\lt\sigma\lt\dfrac{150}{1-0.03465}$$ $$\boxed{144.96\text{ h}\lt\sigma\lt155.38\text{ h}}$$
3. 6(A)(b)(ii) Test H₀: σ=130 at α=0.05, using the same result. The value 130 lies OUTSIDE the 95% interval just found (144.96, 155.38), since $130\lt144.96$. A hypothesized value falling outside its own $(1-\alpha)100\%$ CI is rejected at level α — equivalently, standardizing s itself (which for large n is approximately normal with mean σ and standard deviation $\sigma/\sqrt{2n}$) gives $z=\dfrac{(150-130)\sqrt{2(1600)}}{130}=8.70\gg1.96$. $$\boxed{\text{Reject } H_0\text{: the true standard deviation is significantly different from }130\text{ h}}$$
4. 6(B)(a) Test H₀: p=0.75 vs H₁: p≠0.75, α=0.05. $\hat p=1800/2500=0.72$. $$z=\dfrac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}=\dfrac{0.72-0.75}{\sqrt{0.75(0.25)/2500}}=\dfrac{-0.03}{0.00866}=-3.464$$ Since $|z|=3.464\gt1.96$, $$\boxed{\text{Reject } H_0\text{: the satisfaction proportion is significantly different from }0.75}$$
5. 6(B)(b) Required sample size for error 0.01, 99% confidence. Using the best available estimate $\hat p=0.72$ and $z_{0.005}=2.576$: $$n=\dfrac{z_{0.005}^2\,\hat p(1-\hat p)}{E^2}=\dfrac{(2.576)^2(0.72)(0.28)}{(0.01)^2}=13{,}377.7$$ Rounding up to ensure the error bound holds, $$\boxed{n=13{,}378}$$