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