## Question 508:

Asked on November 13, 2008

Tags:
Sample Size ,
Confidence Level ,
Margin of Error

1

## Answer:

No answer provided yet.

- For this question we need to find the sample size needed in order to meet the margin of error of +/- 2% using a 95% confidence level. The critical value associated with this level is 1.96, which we can look up using the percentile to z-score calculator. We're given an expected agreement rate of 30%, which we can use to build our standard deviation. The variance is .3*.7 = .21 and the square root of this is the standard deviation =SQRT(.21) = .458. Now we setup an equation to solve for the unknown sample size given our standard deviation and critical value of z.
- The margin of error is the standard error of the mean SEM * the critical z value of 1.96, so working backwards from a margin of .02, the SEM = .02/1.96= .0102.
- So the SEM is .0102 which is made up of the SD/ SQRT(n). Using the standard deviation we found of .458 and solving for the unknown sample size n we have the equation .458/sqrt(n) = .0102
- Squaring both sides = .2098/n = .000104
- Isolating n gets us .000104n = .2098 or n = .2098/.000104 = 2017.3
- Since we cannot ask part of a person, we round up to the nearest person to get 2018.
- We can confirm our results using the confidence interval around a proportion calculator, the proportion for was .3*2018 = 605. Entering 605 and 2018 gets us inf-act the desired margin of error of .02.
- So we'd need to sample 2018 people to have a margin of error of +/- 2%, which gives us a 95% confidence interval of between 28% and 32%.