Hey JB1974.
Can you outline specifically what hypotheses you are trying to test.
Hello,
I need a little help here - 2 questions
Question 1:
I need a formula to determine the sample size required to detect a change in proportion. I use n=Z^{2}pq/h^{2}. (h = half interval width)
However, I cannot replicate ny answers on either minitab or SPCXL.
e.g. Manufacturing components with a historical pass rate of say 50%. What sample size do I need to be 95% confident that the pass rate has improved to 55%.
Question 2:
For the difference between 2 sample proportions, what formula can you use to determine sample sizes for a 5% difference at 95% confidence?
THank you,
j
Chiro,
The testing of the components results in a simple pass/fail. We do not have measurement data
THere are 2 possibilities:
1. Historical yield is 47%. How many samples do I need to test to demonstrate a 5% increase in yield with 95% confidence. Do I need to take "power" into account? It is not in the formula I have shown in my first post.
2. I need to make a change to a process. THerefore I would like to run a number of samples with the original conditions, followed by running a second set of samples with the changed conditions. Is there a way to determine how many samples in each sample set I need to run in order to show at least a 5% improvement in yield at 95% confidence with the new conditions. Again, does power need to be taken into account.
THank you for you interest.
j
In this case if you are using a large sample size your interval will be something like (x_bar-1.96*se,x_bar+1.96*se) where se is the standard error that for a proportion using normal approximation will be SQRT(x_bar(1-x_bar)/n) where n is your sample size, and x_bar is the sample mean (point estimate of the proportion variable).
Now if you are finding a number where-by 1.96*se = 0.05, then you have x_bar which means you can re-arrange to get n in terms of x_bar from the formula for se.
The thing is though that this is a two-tailed test and if you want a one-tailed test then you will need to adjust the calculation (and consider that you can only get proportions between 0 and 1).
If you don't want to use the Normal approximation, then the same idea is used but for the likelihood (which is the binomial distribution).
Are you using a normal approximation or not?