# finding best confidence interval for Normal distribution $n(\theta,\theta^2)$

A random sample of size $n$ drawn from a population $X = n(\theta, \theta^2)$ which statistics gives the best confidence interval for $\theta$

a) $Q_1 =\dfrac{\sqrt n(\bar X-\theta)}{\theta}$

b) $Q_2 =\dfrac{(\bar X-\theta)}{\sqrt{S^2/n}}$

c) $Q_3 =\dfrac{(n-1)S^2}{\sigma^2}$

d) $Q_4 =\dfrac{n(\bar X-\theta)^2}{S^2}$

e) $Q_5 = \dfrac{\bar X}{\sqrt{\theta^2/n}}$

My solution - I understand $Q_1$ has $Z(0,1)$ distribution, $Q_2$ has $t_{n-1}$ distribution, $Q_3$ has $\chi_{n-1}^2$ distribution, $Q_4$ has $F_{1,n-1}$ distribution and $Q_5$ has $Z(\theta,1)$ distribution. I understand $Q_5$ is not pivotal, so we can reject it. But between the other statistics which one gives the best confidence interval. I would be really grateful if someone answers me. Thank you.

• You have to do more than know the distribution of the point estimate--specifically, you need to calculate the variance, which for a Wald-type confidence interval (implicitly assumed in this question), gives you a measure of its width. The one with the smallest variance is the "best." – heropup Mar 18 '18 at 17:35