Sample size can be used for different things. For a given statistical power, to generate a given error rate, to generalize to a population. You need to know what sample size is needed for. My guess is to get a specific error rate, sometimes called the margin of error of the polls
I was tasked with determining a "statistical sample size" for a population of 6070. The purpose is as a spot check with binomial results (is or is not); it is not an acceptance sampling; and I was given no other information (ie. margin of error). There is no baseline or historical data about the population. The only information I have to go on is the size of the population. Is there a statistical method of determining an appropriate sample size when nothing else is known except the size of the population?
Yes, and the population size has nothing to do with the answer.
We have sample size, expected successes, actual successes, and probability of that difference.
let n = 120, long run Psuccess = PX =.3, expected successes = .3 * 120 = 36 = X
actual success = 40 = x
40/36 = 1.111
Now to an online binomial calculator, example is statrek. Plug in .3, 120 and 40 and get an array of probabilities, one of which is
Cumulative Prob X>x = .1843.
The probability of the long run success probability being > the sample success value is 18.43%.
And we can double that, trust me, to 37%, and say this:
"A binomial distribution with Ps = .3 will have samples of 120 success number between 32 and 40 = +/-11.1%, 1-.37 = 63 percent of the time.
So, take a sample of maybe 30, get a ballpark guess at PX-I guessed .3
Put the guess PX, an n and an x into the machine.
Change n until you're happy with PX>x
You've solved for n.
Fifteen minutes with an online binomial calculator and you'll become an expert.