Sample Size problem?

Question

A systems analyst needs to summarize 9000 documents. The docs describe end-user (non)usage and (dis)satisfation. Usage is reported as teh number of hours per week; the populationvariance for usage is 400 hours hours. Satisfaction is reported as satisfied or non-satisfied. The analyst wants to avoid sampling errors if possible, and does not want to review all 9000 docs unless required.

1) How many docs should the analyst select to obtain 99.7% certainty for estimating the number of hours of usage if the acceptable error of the estimate is 2 hours?

2) For the sample size in 1) explain 2 specific methods for selecting the desired sample.

3) If the analyst believes (believes, being key) one third of the documents suggests a variation from normal user satisfaction how many documents should the analyst select to estimate the percentage of dissatisfied users at the 99.7% certainty level if the acceptable error is 0.1?

Answer 1

For Question 1, the formula for this is N=(Zc*s/E)^2. You'll need to look up in a table the z-score associated with 99.7% confidence (which is a very non-standard value). That number is Zc. E is your margin of error (here 2 hours). "s" is the standard deviation, which is the square root of 400. You'll plug the numbers into the formula to get the answer.

For Question 2, you'll need to discuss the methods of sample selection (such as systematic samples or methods for selecting random samples) that you learned in your statistics class.

For Question 3, the formula for this is N=P*Q*(Zc/E)^2. Again, you'll need to look up in a table the z-score associated with 99.7% confidence (the same number as for Question 1). That number is Zc. "E" is the acceptable error (0.1). You may want to clarify whether that means 10% or 0.1% (which would actually be .001 as E). P and Q estimated percentages with and without the desired characteristic (here 1/3 or .33333... and 2/3 or .66666...).