Question

Eric wants to estimate the percentage of elementary school children who have a social media account. He surveys 450 elementary school children and finds that 280 have a social media account. Identify the values needed to calculate a confidence interval at the 99% confidence level. Then find the confidence interval.

z0.10	z0.05	z0.025	z0.01	z0.005
1.282	1.645	1.960	2.326	2.576

Use the table of common z-scores above.

• Round the final answer to three decimal places.

$p'=\ $ sigma sub p prime is equal to $\sigma_{p'}\ =\ $ z sub alpha over 2 is equal to $z_{\frac{\alpha}{2}}\ =\ $ (, )

Answer 1

Sample proportion = 280 / 450 = 0.622

99% confidence interval for p is

- Z/2 * sqrt [ ( 1 - ) / n ] < p < + Z/2 * sqrt [ ( 1 - ) / n ]

0.622 - 2.576 * sqrt [ 0.622 * (1 - 0.622) / 450] < p < 0.622 + 2.576 * sqrt [ 0.622 * (1 - 0.622) / 450]

0.563 < p < 0.681

99% CI is ( 0.563 , 0.681 )