Sample Size Problem, help Please?

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.

An analyst selected 100 docs as the proper sample size to account for a certainty level of 90%. How many documents should he select to account for a certainty level of 99.7% (which is certainty factor of 3)

I've been mulling over this for a while and for some retarded reason I can't come up with results. Can i get an answer and explanation please. Thanks. ~Flip

Answer 1

Maybe because there is a bunch of extraneous data there.

The answer:
0.997 x 9000 = 8973 pages

which makes sense because if you want a 100% certainty, you would only be able to use the entire number of pages as a sample. Any smaller size sample would introduce uncertainty.

Answer 2

ow?

Population variance is 400hr^2, so the standard deviation is 20 hours.

The formula: n = (Z*std_dev/e)^2
where Z is from your 90% level of certainty,
std_dev is 20 from above

We have two parts. 1st find e using "An analyst selected 100 docs...)" His L.O.S was 90%, so Z=1.645.
Solving 100 = (1.645*20/e)^2 gives us e=3.280.

for L.O.S = 99.7, Z=2.75.
n=(2.75*20/3.20)^2 = 295.41
So we want to round up (always round up when determining sample sizes) and get 296 docs for a 99.7% certainty level.

Answer 3

Based on what you have said about using 100 documents for 90% certainty leve,

in simple words you can ask that if 90% is 100 then what is 99.7%

= (99.7*100)/90
= 110.778

so round it up and read 111 documents

Answer 4

Not all monkeys are red, so find it yourself.