Size sample required to construct a 95% C.I. with margin of error

[goodz]

N = 8
Mean = 2.10
stDev = 0.537

What size sample is required to construct a 95% confidence interval with a margin of error of 0.3?
try n = 12, t(.975) with 11 deg. of freedom = 2.2001
so n = ( (.537* 2.2001) / (.3))^2 = 15.5

try n = 14, i get n = 14.95
n = 15, n = 14.74
Why is this a trial and error?
What I don't get is what are we trying to get n to equal?

the answer: n must be between 14 and 15 so, use n = 15.

 

[statdad]

First question: what is $N = 8$ in your question? Second question: are you trying to look for the sample size to use to generate a CI for a mean? If so, have you not encountered the formula

$$n = \left(\frac{z \times \sigma}{E}\right)^2$$

(E is the margin of error)? The general rule is to use this and round the result to the next highest integer. If you use the method you suggest (which I haven't seen in any text)
* you are correct, you get a game of guessing
* it isn't obvious to me that it would ever converge - that is, if you repeated the same step with 15, then with
the result of that calculation, and so on, I am not sure you would ever zero in on a single value.

The benefit of the formula I presented is that it does give an idea of an appropriate sample size, IF, your guess at the size of the standard deviation is in the correct ballpark.