Question 510:

Asked on November 13, 2008

Tags: Sample Size , Confidence Level

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 +/- 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 servicing 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.
2. 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.
3. So the SEM is .0255 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) = .0255
4.  Squaring both sides = .21/n  = .000651
5. Isolating n gets us .000651n = .21 or n = .21/.000651 = 322.6
6. Since we cannot ask part of a person, we round up to the nearest person to get 323.
7. We can confirm our results using the confidence interval around a proportion calculator, the proportion was .3*323 = 97. Entering 97 and 323 gets us in-fact the desired margin of error of .05.
8. So we'd need to sample 323 customers to have a margin of error of +/- 5%, which gives us a 95% confidence interval of between 25% and 35%.