Question 6 of 8

Question 6

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) $n=1{,}095$ trucks, $x=365$ failures. (B) $n=900$ bottles, $\bar x=3{,}990$ mL, $s=120$ mL.

Find. A(a) test $H_0:p=0.30$; A(b) $n$ for $E=0.01$ at 99% confidence; B(a) test $H_0:\mu=4{,}000$; B(b) $n$ for $E=2$ mL at 99% confidence.

Approach. Both tests use the large-sample $Z$ statistic (proportion and mean respectively); both sample-size questions invert the margin-of-error formula $E=z_{\alpha/2}\cdot(\text{se})$ for $n$.

1. A(a) Test $H_0:p=0.30$ vs $H_1:p\ne0.30$. $\hat p=365/1{,}095=0.3333$: $$z=\frac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}=\frac{0.3333-0.30}{\sqrt{0.30(0.70)/1{,}095}}=\boxed{2.407},\qquad z_{0.025}=1.960.$$ Since $|z|=2.407>1.960$, we reject $H_0$ — the true failure proportion is significantly different from 0.30.
2. A(b) Sample size, $E=0.01$, 99% confidence. Using $\hat p=0.3333$ as the planning estimate and $z_{0.005}=2.576$: $$n=\frac{z_{0.005}^2\,\hat p(1-\hat p)}{E^2}=\frac{(2.576)^2(0.3333)(0.6667)}{(0.01)^2}=14{,}744.2\ \Rightarrow\ \boxed{n=14{,}745}.$$
3. B(a) Test $H_0:\mu=4{,}000$ vs $H_1:\mu\ne4{,}000$. $$z=\frac{\bar x-\mu_0}{s/\sqrt n}=\frac{3{,}990-4{,}000}{120/\sqrt{900}}=\frac{-10}{4}=\boxed{-2.50},\qquad z_{0.025}=1.960.$$ Since $|z|=2.50>1.960$, we reject $H_0$ — the mean fill volume is significantly different from (below) the claimed 4,000 mL, so the data do not support the producer's claim.
4. B(b) Sample size, $E=2$ mL, 99% confidence. Treating $s=120$ mL as the population estimate, $z_{0.005}=2.576$: $$n=\left(\frac{z_{0.005}\,s}{E}\right)^2=\left(\frac{2.576(120)}{2}\right)^2=23{,}885.6\ \Rightarrow\ \boxed{n=23{,}886}.$$