Suposing $\sigma^2$ unkown, find the minimum value for n that gurantees with a probability of 90% that the 95% confidence interval for $\mu$ is of a length no more than $\frac{\sigma}{4}$
Ive been trying to figure this one out using the method to find confidence intervals for sample means, but i get stuck on how to implement the 90% probability
i used that when $\sigma^2$ is uknown the confidence interval is given by: $P(\overline{X}-t^{1-\alpha/2}_{n-1}\frac{S}{\sqrt{n}}<\mu<\overline{X}+t^{1-\alpha/2}_{n-1}\frac{S}{\sqrt{n}})=1-\alpha$
and from there i get that the inequality for the length of the interval will be given by:
$2t^{0.975}_{n-1}\frac{S}{\sqrt{n}}<\frac{\sigma}{4}$ (using $\alpha =.05$)
but i dont know how to proceed from here because i need to achieve: $P(2t^{0.975}_{n-1}\frac{S}{\sqrt{n}}<\frac{\sigma}{4})=.9$
Hint: According to CLT for sample variance, if the 4th central moment of iid function exists, then sample variance converges in distribution to a normal distribution. By applying Delta method, we can get sample standard deviation in distribution to a normal distribution as well.