Question

Assume that a given population is normally distributed with the population mean meu and variance sigma² and that the sample given is chosen sufficiently large such that (x-bar-meu/sigma/√n)~meu(0,1). Show that the Comfidence interval at alpla=0.01 probability is theta1<=meu<=theta2 where theta1=x-bar-2.58(sigma/√n) is the lower C.I and theta2=x-bar+2.58(sigma/√n) is the upper C.I

Answer 1

A $100(1-\alpha)$ confidence interval

$$P(\hat{\theta_L}<\theta<\hat{\theta_R})=1-\alpha$$

Writing $z_{\alpha/2}$ for the z-value above which we find an area of $\alpha/2$ under the normal curve, we can see

$$P(-z_{\alpha/2}<Z<z_{\alpha/2})=1-\alpha$$

where

$$Z={\bar{x}-\mu \over \sigma/\sqrt{n}} \sim N(0, 1)$$

Hence

$$P(-z_{\alpha/2}<{\bar{x}-\mu \over \sigma/\sqrt{n}}<z_{\alpha/2})=1-\alpha$$

$$P(\bar{x}-z_{\alpha/2}{\sigma \over \sqrt{n}}<\mu<\bar{x}+z_{\alpha/2}{\sigma \over \sqrt{n}})=1-\alpha$$

The z-value leaving an area of 0.005 to the right, and therefore an area of 0.995 to the left, is $z_{0.005}=2.58$

Therefore confidence interval at $\alpha=0.01$

$$CI=(\bar{x}-2.58{\sigma \over \sqrt{n}}, \bar{x}+2.58{\sigma \over \sqrt{n}})$$