Question

DOH Determine the margin of error for a 90% confidence interval to estimate the population proportion with a sample proportion equal to 0.90 for the following sample sizes. an 100 bin 200 cn=260 Nam Due Click the icon to view a portion of the Qurtulative Probabilities for the Standard Normal Distribution table. Currea. The main forror for a 90% confidence interval to estimate the population proportion with a sample proportion equal to 0.90 and sample siren 100 is (Round to three decimal places as needed) Atten

Answer 1

Solution :

Given that,

a)

n = 100

Point estimate = sample proportion = $\hat p$ = 0.90

1 - $\hat p$ = 0.10

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

Z_$\alpha$/2 = Z _0.05 = 1.645

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

= 1.645 * ($\sqrt$((0.90 *0.10) / 100)

= 0.049

b)

n = 200

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

= 1.645 * ($\sqrt$((0.90 *0.10) / 200)

= 0.035

c)

n = 260

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

= 1.645 * ($\sqrt$((0.90 *0.10) / 260)

= 0.031