Answer to Question #220115 in Statistics and Probability for ahmed

Question

4. Refer to the previous problem. How large a sample would be required to estimate the population proportion to within .05 with 95 percent confidence (.30 is the best available estimate of p): (a) If the finite population correction can be ignored? (b) If the finite population correction is not ignored and N 1500

Answer 1

The margin of error for constructing a 100(1-α) % confidence interval for population proportion when the finite population correction can be ignored is

$E = Z_{α/2}\times \sqrt{\frac{\hat{p}(1- \hat{p})}{n}}$

$\hat{p}$ = sample proportion

n = sample size

α = 1 – conference level

When the finite population correction is not ignored, then the margin of error is

$E = Z_{α/2}\times \sqrt{\frac{\hat{p}(1- \hat{p})}{n}} \times \sqrt{\frac{N-n}{N-1}}$

N= the population size

$α=1-0.95=0.05 \\ Z_{α/2}=Z_{0.05/2}=Z_{0.025}=1.96 \\ E=0.05 \\ \hat{p}=0.30$

(a)

$E = Z_{α/2}\times \sqrt{\frac{\hat{p}(1- \hat{p})}{n}} \\ 0.05 = 1.96 \times \sqrt{\frac{0.30(1- 0.30)}{n}} \\ n = \frac{(1.96)^2 \times 0.30 \times 0.70}{(0.05)^2} \\ n=322.69 ≈323$

The required sample size is n=323.

(b) N=1500

$E = Z_{α/2}\times \sqrt{\frac{\hat{p}(1- \hat{p})}{n}} \times \sqrt{\frac{N-n}{N-1}} \\ 0.05 = 1.96\times \sqrt{\frac{0.30(1- 0.30)}{n}} \times \sqrt{\frac{1500-n}{1500-1}} \\ (0.05)^2 = (1.96)^2 \times \frac{0.30 \times 0.70}{n} \times \frac{1500-n}{1499} \\ \frac{1500-n}{n}= \frac{(0.05)^2 \times 1499}{(1.96)^2 \times 0.30 \times 0.70} \\ \frac{1500-n}{n}= 4.64526 \\ \frac{1500}{n}-1= 4.64526 \\ n = \frac{1500}{4.64526+1} \\ n = 265.71 ≈266$

The required sample size is n=266.