Verified answer

95% confidence interval for the population proportion is

sample proportion +/- z-score for 95% confidence*standard error of proportion

Sample proportion = p = 9% = 0.09

Sample size = n = 256

z-score for 95% confidence = 1.96

Standard error of proportion = sqrt (p*(1-p)/n)

= sqrt (0.09*0.91/256)

= 0.0179

The confidence interval is

0.09 +/- 1.96*0.0179

0.09 +/- 0.0351

lower limit is 0.09 - 0.0351 = 0.0549 or 5.49%

upper limit is 0.09 + 0.0351 = 0.1251 or 12.51%

The 95% confidence interval for the actual % of people who voted illegally in the country is

(5.49%, 12.51%)

Since n=256 is large, you use normal approximation to binomial.

Then ...

The 95% confidence interval for the true proportion P based on a sample estimate p=0.09 is given by

p ± z(α/2)*√[p(1-p)/n], where z(α/2) = z(0.025) = 1.960, n = 256.

0.09±0.035, interval about (0.05, 0.13)

or (5%, 13%).

You have to be 95 percent within the range of people that actually voted illegally and the ones who mathematically voted illegally. So if out of 256 there is 9% than out of the countries amount of people, that's ~ 23.04 people. So. If there's 25,600,000 people we move the decimal to... 2,304,000 .. Annd it's a classroom statistic so don't expect it to be within reason.. -.- but the math there is accurate as an example. When u get the real number just multiply it by .09.. Then calculate a what 5% of that is.. There's the 95% accuracy range from your calculated #