Question

From a population with a variance of 900, a sample of 225 items is selected. at 95% confidence, what is the margin of error for the population mean? round your answer to three decimal places.

Answer 1

The margin of error is 3.92.

To find the margin of error, we use the formula

z*(σ/√n)

We first find the z-score.  
Convert 95% to a decimal:  0.95
Subtract from 1: 1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that the z-score for this is 1.96:
1.96*(σ/√n)

Since the variance is 900, the standard deviation (σ) is √900 = 30:
1.96*(30/√n)

The sample size, n, is 225:
1.96*(30/√225) = 1.96*(30/15) = 1.96*2  = 3.92