## Question 509:

Asked on November 13, 2008

Tags:
Sample Size ,
Confidence Intervals ,
Margin of Error ,
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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 +/- 5% 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 80%, which we can use to build our standard deviation. The variance is .8*.2 = .16 and the square root of this is the standard deviation =SQRT(.16) = .4. 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 .05, the SEM = .05/1.96= .0255.
- So the SEM is .0255 which is made up of the SD/ SQRT(n). Using the standard deviation we found of .4 and solving for the unknown sample size n we have the equation .4/sqrt(n) = .0255
- Squaring both sides = .16/n = .000651
- Isolating n gets us .000651n = .16 or n = .16/.000651 = 245.78
- Since we cannot ask part of a person, we round up to the nearest person to get 246.
- We can confirm our results using the confidence interval around a proportion calculator, the proportion was .8*246 = 197. Entering 197 and 246 gets us in-fact the desired margin of error of .05.
- So we'd need to sample 246 crazy Cubs Fans to have a margin of error of +/- 5%, which gives us a 95% confidence interval of between 75% and 85%.