Question

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. In a clinical test with 1600 subjects, 800 showed improvement from the treatment. Find the margin of error for the 99% confidence interval used to estimate the population proportion. a. 0.0322 b. 0.0184 c. 0.0245 d. 0.0282 ON

Answer 1

Solution :

Given that,

n = 1600

x = 800

Point estimate = sample proportion = $\hat p$ = x / n = 0.5

1 - $\hat p$ = 0.5

At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 2.576 * ($\sqrt$((0.5 * 0.5) / 1600)

= 0.0322

Option a is correct.