Sample Size Estimation with unknown variance

Question

After spending a lot of time I'm still nowhere.

What I have:

Let d denote the desired margin of error so that the interval for µ is of width 2d. Let (1 − α) denote the probability associated with the margin of error. Then $(\frac{z_{α/2}σ}{d})^2$

Using the estimator $X∼N(θ, θ(1 − θ)/n)$

Do I simply plug this into $(\frac{z_{α/2}σ}{d})^2$ ?

2nd Question. What happens to n if there is no idea about the value of $θ$?

Answer 1

1. Yes, if you mean what I think you mean. Your work was a little hard for me to follow. This is how I would do it: $$P(|\theta-\hat\theta|<d|) = P(|X-\hat X|< nd) = P\left(\frac{|X-\hat X|}{\sigma_x} < \frac{nd}{\sigma_x}\right) < z_{\alpha/2}$$ where $n\theta = \hat X$ and $n\hat\theta = X$.

$$z_{\alpha/2} = \frac{nd}{\sigma_x}$$ Now $\sigma_x^2 \approx n\theta(1-\theta)$, so $$z_{\alpha/2} = \frac{\sqrt{n_{min}}d}{\sqrt{\theta(1-\theta)}}$$ $$\sqrt{n_{min}} = \sqrt{\theta(1-\theta)}\frac{z_{\alpha/2}}{d}$$ $$n \geq n_{min} = \theta(1-\theta)\left(\frac{z_{\alpha/2}}{d}\right)^2$$

2. You have to do something to get information about the value of $\theta$. Take a preliminary sample, estimate it based on preexisting data, or make a guess.