04-BS-2, December 2017, Q6 | Nivaar Exam Prep

Question 6 (20 marks: (A)(a) 5, (b) 5; (B)(a) 5, (b) 5)

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.

Given data
Quantity	Value
n	1,600
Number happy (x)	1,000
Sample mean salary	USD 160,000
Sample std. dev. salary	USD 24,000

Find. (A)(a) Test $H_0: p=0.64$. (A)(b) Sample size for error 0.01 at 99% confidence. (B)(a) Test $H_0: \mu=155{,}000$. (B)(b) Sample size for error 400 at 99% confidence.

Approach. Use the large-sample $Z$-test for a proportion and for a mean, then invert the margin-of-error formula ($E=z\sqrt{pq/n}$ or $E=z\sigma/\sqrt{n}$) to solve for the required $n$, rounding up to the next integer.

6(A)(a) Test H₀: p = 0.64. $\hat{p}=1000/1600=0.625$: $$Z = \frac{\hat{p}-p_0}{\sqrt{p_0q_0/n}} = \frac{0.625-0.64}{\sqrt{0.64(0.36)/1600}} = \frac{-0.015}{0.012} = -1.25.$$ Since $z_{0.025}=1.96$ and $|-1.25|<1.96$, we fail to reject H₀ — the proportion is not significantly different from 0.64.
6(A)(b) Sample size for error 0.01, 99% confidence. Using $p_0=0.64$ as the planning value and $z_{0.005}=2.576$: $$n = \frac{z_{0.005}^2\,p_0q_0}{E^2} = \frac{(2.576)^2(0.64)(0.36)}{(0.01)^2} = 8850.7 \Rightarrow \boxed{n=8{,}851}.$$
6(B)(a) Test H₀: μ = 155,000. $$Z = \frac{160{,}000-155{,}000}{24{,}000/\sqrt{1600}} = \frac{5{,}000}{600} = 8.33.$$ Since $|8.33|\gg 1.96$, we reject H₀ at α=0.05 — with this large a sample, even a modest USD 5,000 gap is highly statistically significant: the true mean salary is significantly different from (higher than) USD 155,000.
6(B)(b) Sample size for error USD 400, 99% confidence. Using $s=24{,}000$ as the planning σ and $z_{0.005}=2.576$: $$n = \left(\frac{z_{0.005}\,\sigma}{E}\right)^2 = \left(\frac{2.576(24{,}000)}{400}\right)^2 = 23{,}885.6 \Rightarrow \boxed{n=23{,}886}.$$

Question 6 — final results
Part	Quantity	Value
(A)(a)	Z stat vs Z-crit	−1.25, \|Z\|<1.96 → fail to reject
(A)(b)	Required n	8,851
(B)(a)	Z stat vs Z-crit	8.33, \|Z\|>1.96 → reject H₀
(B)(b)	Required n	23,886

Topic: Large-sample hypothesis testing and sample-size determination for a proportion and a mean.

Key relations: Proportion test $Z=(\hat{p}-p_0)/\sqrt{p_0q_0/n}$; mean test $Z=(\bar{x}-\mu_0)/(\sigma/\sqrt{n})$ (or $s/\sqrt{n}$ for large $n$); sample size $n=z_{\alpha/2}^2 pq/E^2$ or $n=(z_{\alpha/2}\sigma/E)^2$, always rounded UP.

Why this works: For n this large the sampling distribution of $\hat{p}$ and $\bar{X}$ is Normal by the CLT, so a $Z$ test is appropriate rather than a $t$ test; inverting the margin-of-error expression for $n$ answers "how big a sample do I need" directly.

Common pitfall: Rounding sample-size answers DOWN instead of up (which would under-cover the target confidence level); confusing statistical significance (guaranteed at large n even for small practical differences) with practical importance.

Source: Walpole, Myers, Myers & Ye, Ch. 9 (Estimation, sample size) & Ch. 10 (Hypothesis Testing, large-sample tests).

Question 6 of 8

Question 6 (20 marks: (A)(a) 5, (b) 5; (B)(a) 5, (b) 5)