Question

In a recent poll, 230 people were asked if they liked dogs, and 80% said they did. Find the margin of error of this poll, at the 99% confidence level.

Give your answer to three decimals

Answer 1

Solution :

Given that,

n = 230

Point estimate = sample proportion = $\hat p$ = 0.80

1 - $\hat p$ = 0.20

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 * ($\sqrt$((0.80 * 0.20) / 230)

= 0.068

The margin of error of this poll, at the 99% confidence level is 0.068.