Question 6 of 8: Large-Sample Tests — Tire Weight and Water-Quality Proportion

Question 6: Large-Sample Tests — Tire Weight and Water-Quality 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. (A) $n=1{,}300$, $\bar t=45.00$ kg, $s=0.50$ kg. (B) $n=2{,}500$, $x=1{,}600$ satisfied.

Find. (A)(a) test $H_0:\mu=44.5$. (b) 95% CI for $\sigma^2$ via the supplied large-sample formula, and test $H_0:\sigma=0.4$. (B)(a) test $H_0:p=0.7$. (b) required $n$ for error 0.01 at 99% confidence.

Approach. With $n$ this large, use $z$-tests throughout (sample $s$/$\hat p$ standing in for the population values); for the variance interval, substitute directly into the given large-sample formula.

1. (A)(a) Test $H_0:\mu=44.5$ vs. $H_1:\mu\ne44.5$, $\alpha=0.05$. $$z=\dfrac{\bar t-\mu_0}{s/\sqrt n}=\dfrac{45.00-44.5}{0.50/\sqrt{1{,}300}}=36.06$$ Critical value $z_{0.025}=1.96$. Since $36.06\gg1.96$, reject $H_0$. $$\boxed{\text{Reject } H_0:\ \text{mean tire weight is significantly different from }44.5\text{ kg}}$$ (With $n=1{,}300$, even a tiny 0.5 kg gap is many standard errors wide — statistical significance here reflects the huge sample, not necessarily practical importance.)
2. (A)(b)(i) 95% CI for the variance, via the supplied formula. With $z_{\alpha/2}=z_{0.025}=1.96$ and $\sqrt{2n}=\sqrt{2(1{,}300)}=50.99$: $$\dfrac{s}{1+\dfrac{z_{\alpha/2}}{\sqrt{2n}}}\lt\sigma\lt\dfrac{s}{1-\dfrac{z_{\alpha/2}}{\sqrt{2n}}}\ \Rightarrow\ \dfrac{0.50}{1+\frac{1.96}{50.99}}\lt\sigma\lt\dfrac{0.50}{1-\frac{1.96}{50.99}}$$ $$\dfrac{0.50}{1.03844}\lt\sigma\lt\dfrac{0.50}{0.96156}$$ $$\boxed{0.4815\ \text{kg}\lt\sigma\lt 0.5200\ \text{kg}\ \Rightarrow\ 0.2318\ \text{kg}^2\lt\sigma^2\lt 0.2704\ \text{kg}^2}$$
3. (A)(b)(ii) Test $H_0:\sigma=0.4$ via the CI. The hypothesized value 0.4 kg lies below the entire 95% CI (0.4815, 0.5200) for $\sigma$, so it is not a plausible value at $\alpha=0.05$. $$\boxed{\text{Reject } H_0:\ \sigma\text{ is significantly different from }0.4\text{ kg}}$$
4. (B)(a) Test $H_0:p=0.7$ vs. $H_1:p\ne0.7$, $\alpha=0.05$. $\hat p=\dfrac{1{,}600}{2{,}500}=0.64$. $$z=\dfrac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}=\dfrac{0.64-0.7}{\sqrt{0.7(0.3)/2{,}500}}=-6.55$$ Since $|-6.55|\gt1.96$, reject $H_0$. $$\boxed{\text{Reject } H_0:\ p\text{ is significantly different from }0.7}$$
5. (B)(b) Required sample size for error 0.01 at 99% confidence. Using $\hat p=0.64$ as the planning estimate and $z_{0.005}=2.576$: $$n=\dfrac{z_{0.005}^2\,\hat p(1-\hat p)}{E^2}=\dfrac{2.576^2(0.64)(0.36)}{0.01^2}=15{,}286.8$$ Rounding up to guarantee the error bound: $$\boxed{n=15{,}287}$$