Question

הסט Determine the margin of error for a confidence interval to estimate the population proportion for the following confidence levels with a sample proportion equal to 0.35 and n 120 Nama 90% c.99% Click the icon to view a portion of the Cumulative Probabilities for the Standard Normal Distribution table Due: Curre Atter Late The margin of enor for a confidence interval to o und to three decimal places as needed) imate the population proportion for the confidence levels

Answer 1

Solution :

Given that,

n = 120

Point estimate = sample proportion = $\hat p$ = 0.35

1 - $\hat p$ = 1-0.35 = 0.65

A) At 90% confidence level

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

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

Z$\alpha$/2 = Z_{0.05 = 1.645}

Z$\alpha$/2 = 1.645   

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

= 1.645 * ($\sqrt$(0.35*(0.65) /120 )

= 0.072

B)

At 95% confidence level

$\alpha$ = 1-0.95% =1-0.95 =0.05

$\alpha$/2 =0.05/ 2= 0.025

Z$\alpha$/2 = Z_{0.025 = 1.960}

Z_$\alpha$/2 = 1.960

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

= 1.960 * ($\sqrt$(0.35*(0.65) /120 )

= 0.085

C)

At 99% confidence level

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

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

Z$\alpha$/2 = Z_{0.005 = 2.576}

Z$\alpha$/2 = 2.576

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

= 2.576 * ($\sqrt$(0.35*(0.65) /120 )

= 0.112