confidence interval ~ minimum sample size with a maximum width

My task is: minimum sample size, which guarantees that the width of the resulting confidence interval is a maximum of W = 0.06. Is my way right?

$$\ X1, ... , Xn ~ (iid) N(μ,σ²), σ=9, W= 0.06, α=0.10$$

$$\ W = (z1-α/2) * \left(\frac{σ}{\sqrt n}\right)$$ $$\ ...$$ $$\ N >= \left(\frac{(z1-α/2) * σ}{W}\right)^2$$

$$\ N >= \left(\frac{1.645 * 9}{0.06}\right)^2$$

$$\ N >= 60885.5625 <> N>= 60886$$

There is also asked why we couldn't do this without knowing σ. I think it is because we would have to use a t-distribution to find the correct quantils (z1)? And without σ we'd have to guess and that would be hard without knowing anything about any variables.

It depends on the assumed distribution.

If the distribution is normal with (quite well) known sigma, then you get a good approximation with

n = ((z*sigma)/w)²

where z is the normal quantile for the given confidence level and w is the desired half-width of the confidence interval.

Note, the W parameter is the half-width (it is unclear for your cited confidence interval whether the W = 0.06 is the associated +/- half-width).

Also, per a source:

When the number of degrees of freedom is large, then the t ‐distribution, of course, converges to the normal distribution. Note that if we knew the population standard deviation, even if the sample size was small, we would not need to use the t‐distribution.

So, don't worry about the t-distribution especially for large n.

I would also recommend reading this article 'A Method for Selecting the Size of the Initial Sample in Stein's Two Sample Procedure', relating to a double sampling approach.

• Thanks for your help! You are right my w=0.06 is not the half width, so I used n=((2*z*sigma/w)², could have also used w=0.03. – jfnotk May 28 '20 at 18:54