Large-Sample Proportion and Mean — Test and Sample-Size Planning: 04-BS-2, December 2013, Q6

Question 6: Large-Sample Proportion and Mean — Test and Sample-Size Planning (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) $n=1{,}800$, $x=1{,}450$ happy $\Rightarrow \hat p=0.8056$. (B) $\bar x=$$275,000, $s=$$45,000, $n=1{,}800$.

Find. (A)(a) test $H_0:p=0.85$; (A)(b) required $n$ for error 0.01 at 99% confidence. (B)(a) test $H_0:\mu=$$260,000; (B)(b) required $n$ for error $1,000 at 99% confidence.

Approach. Large-sample $z$-tests (Normal approximation, $n$ in the thousands) for both hypothesis tests; invert the margin-of-error formula, $E=z_{\alpha/2}\sqrt{p(1-p)/n}$ or $E=z_{\alpha/2}\,\sigma/\sqrt n$, to solve for $n$.

(A)(a) Test $H_0:p=0.85$ vs. $H_1:p\ne0.85$. $\hat p=1{,}450/1{,}800=0.8056$. $$z=\dfrac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}=\dfrac{0.8056-0.85}{\sqrt{(0.85)(0.15)/1{,}800}}=\dfrac{-0.0444}{0.008416}=\boxed{-5.28}$$ Since $|-5.28|\gg z_{0.025}=1.96$: reject $H_0$ — the true happiness proportion is significantly different from (below) 0.85.
(A)(b) Sample size for a proportion, $E=0.01$, 99% confidence. $z_{0.005}=2.576$. Using the best available estimate $\hat p=0.8056$ (Walpole's "estimated $p$" formula): $$n=\left(\dfrac{z_{0.005}}{E}\right)^2\hat p(1-\hat p)=\left(\dfrac{2.576}{0.01}\right)^2(0.8056)(0.1944)=(66{,}358)(0.1566)=\boxed{10{,}393\Rightarrow n=10{,}393}$$ (round up to the next integer; using the conservative $p=0.5$ instead would give the upper-bound design value $n=16{,}588$).
(B)(a) Test $H_0:\mu=$$260,000 vs. $H_1:\mu\ne$$260,000. With $n=1{,}800$ large, use $z=\dfrac{\bar x-\mu_0}{s/\sqrt n}=\dfrac{275{,}000-260{,}000}{45{,}000/\sqrt{1{,}800}}=\dfrac{15{,}000}{1{,}060.7}=\boxed{14.14}$. Since $14.14\gg1.96$: reject $H_0$ — the data do not support the financial officer's claim; the true mean salary is significantly higher than $260,000.
(B)(b) Sample size for a mean, $E=$$1,000, 99% confidence. $$n=\left(\dfrac{(2.576)(45{,}000)}{1{,}000}\right)^2=(115.9)^2=\boxed{13{,}436\Rightarrow n=13{,}436}$$

Final results — Question 6
Part	Result
(A)(a) $z$, verdict	−5.28; reject $H_0:p=0.85$
(A)(b) required $n$	10,393 (using $\hat p$); 16,588 (conservative $p=0.5$)
(B)(a) $z$, verdict	14.14; reject $H_0:\mu=$$260,000
(B)(b) required $n$	13,436

Topic: Large-sample $z$-tests and sample-size determination for a proportion and a mean.

Key relations: $z=\dfrac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}$; $n=(z_{\alpha/2}/E)^2\,p(1-p)$; $z=\dfrac{\bar x-\mu_0}{s/\sqrt n}$; $n=(z_{\alpha/2}\,\sigma/E)^2$.

Why this works: With $n$ in the thousands the CLT makes $\hat p$ and $\bar x$ essentially Normal regardless of the underlying population shape, so the $z$-test applies directly; the sample-size formulas are just the margin-of-error expression solved for $n$, always rounded UP so the achieved error does not exceed the target.

Common pitfall: Using $p=0.5$ (the maximum-variance, most conservative choice) when a genuine prior estimate $\hat p$ is available and the question doesn't call for a worst-case design — the two give very different $n$; also forgetting to round the sample-size result UP to the next whole respondent.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 9 (confidence intervals & sample size) & Ch. 10 (large-sample tests).

Question 6 of 8: Large-Sample Proportion and Mean — Test and Sample-Size Planning

Question 6: Large-Sample Proportion and Mean — Test and Sample-Size Planning (20 marks)