Question

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

Answer 1

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

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{(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{(0.45(1-0.45))/(875)}\\\\SE=0.03297`