Question 551:

Asked on December 12, 2008

Tags: Sample Size , Confidence Intervals , Proportion

1

Answer:

No answer provided yet.1. For this question we need to find the sample size needed in order to meet the margin of error of +/- 10% using a 99% confidence level. The critical value associated with this level is 2.58, which we can look up using the percentile to z-score calculator or a table of normal values.
2. Next lets find our descriptive statistics. We're told 70% of attendees drive,  which we can use to build our standard deviation. The variance is .7*.3 = .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.
3. The margin of error is the standard error of the mean SEM * the critical z value of 2.58, so working backwards from a margin of .10, the SEM = .10/2.58 = .03876.
4. So the SEM is .03876, 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) = .03876
5.  Squaring both sides = .21/n  = .001502
6. Isolating n gets us .001502n= .21 or n = .21/.001502 = 139.7844
7. Since we cannot ask part of a person, we round up to the nearest person to get 140.
8. We can confirm our results using the confidence interval around a proportion calculator, the proportion was .7*140 = 98. Entering 98 and 140 and selecting a 99% confidence level gets us in-fact the desired margin of error of .10.
9. So we'd need to plan a sample of 140 attendees to have a margin of error of +/- 10%, which gives us a 99% confidence interval of between 60% and 80%.