Question

Given n = 875 and pˆ = 0.45, find the margin of error E that corresponds to a 95% confidence level.

Answer 1

Answer:

The margin of error E that corresponds to a 95% confidence level is: $E=0.03297$

Step-by-step explanation:

The propocion is $p = 0.45$

The sample size $n = 875$

95% confidence level

Level of significance = 1-level of confidence

Level of significance = 1-0.95 = 0.05

The formula for the error is:

$E = Z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}$

The value for $Z_{0.05/2} =Z_{0.0250}= 1.96$ according to the normal table

So:

$E = 1.96\sqrt{\frac{0.45(1-0.45)}{875}}\\\\SE=0.03297$