## Question 945:

1

1. Q945_1_q945.xls

To construct the 95% confidence interval we use the following steps:

1. Create the proportion : 200/1000 = .20
2. Multiply .20 * 1-.20 = .16
3. Divide .16/1000 = .00016 and take the square root = .012649. This is the standard error (SE).
4. The margin of error is equal to the standard error times the critical value from the normal distribution for a 95% level of confidence (1.96). The margin of error is 1.96*.012649 = .024792.
5. The 95% confidence interval is the proportion plus and minus the margin of error: .175 to .224 or expressed as a percentage 17.5% to 22.5%

The sample size for a margin of error is found by working backwards from the Confidence Interval formula and solving for n.  The sample size formula is n = z2*s2 / d2

z is 1.96 for the 95% level of confidence

s is the standard deviation, which is based on the estimated proportion, which is usually set to .5 for estimating sample sizes (at the highest variability)

d is the desired margin of error. The current question is worded with 61% margin of error, which doesn't make sense. I'm assuming it might be 6.1% and work that.

We get n = 1.962*.52 / .0612 = 258.093.  Rounding that up gets 259.

So we would need a sample of 259 to have a margin of error of 6.1%.  You can change the margin of error using the attached excel file.