Question

1. In a random sample of 100 registered voters, 20 say they plan to vote for Candidate A.

1. Determine a 95% confidence interval for the proportion of all the registered voters who will vote for Candidate A.

2. You are interested in knowing support for candidate by gender to provide strategic advice to candidate B. Suppose your guess based on previous knowledge is that female support for candidate B is around 20 percent, and male support for candidate B is around 50 percent. Suppose you conduct the male and females surveys separately, what should be the sample size approximately for each of the two surveys. For both surveys, you want a margin of error of ±0.03 and a 95% confidence level.

Answer 1

a)

Please refer standard normal distribution table for finding z value for 95% significant level.

Z=1.96

Given

$\overline{p}=20/100=0.20$

Sample size=n=100

Standard error of proportion is given by

$\widehat{\sigma _{\overline{p}}}=\sqrt{\frac{\overline{p}*(1-\overline{p})}{n}}=\sqrt{\frac{0.2*0.8}{100}}=0.04$

Lower bound of confidence interval is given by

$\overline{p}-z*\widehat{\sigma} _{\overline{p}}=0.20-1.96*0.04=0.1216$

Upper bound of confidence interval is given by

$\overline{p}+z*\widehat{\sigma} _{\overline{p}}=0.20+1.96*0.04=0.2784$

b)

We know z=1.96 at 95% significance level. (part a)

i)

For female

$\overline{p}=0.20$

Margin of error is given by

$z\widehat{\sigma _{\overline{p}}}=z\sqrt{\frac{\overline{p}*(1-\overline{p})}{n}}=1.96\sqrt{\frac{0.2*0.8}{n}}$

We are given margin of error as 0.03. So,

$1.96*\sqrt{\frac{0.2*0.8}{n}}=0.03$

$\frac{3.8416*0.2*0.8}{n}=0.0009$

$n=\frac{3.8416*0.2*0.8}{0.0009}=682.95\approx 682$

ii)

For male

$\overline{p}=0.50$

Margin of error is given by

$z\widehat{\sigma _{\overline{p}}}=z\sqrt{\frac{\overline{p}*(1-\overline{p})}{n}}=1.96\sqrt{\frac{0.5*0.5}{n}}$

We are given margin of error as 0.03. So,

$1.96*\sqrt{\frac{0.5*0.5}{n}}= 0.03$

$\frac{3.8416*0.5*0.5}{n}=0.0009$

$n=\frac{3.8416*0.5*0.5}{0.0009}=1067.11\approx 1067$