Question

8.40

stion 4 (6 pt) (Ex. 8.40 on page 409 is modified): Suppose that random variable Y is an observation from a normal distribution with unknown mean u and variance l Find and verify a pivotal quantity that you can use to derive confidence limits for the mean u. Find a 95% lower confidence limit for. a. b. 8.40 Suppose that the random variable Yis an observation from a normal distribution with unknown mean μ and variance 1 . Find a a 95% confidence interval for μ. b 95% upper confidence limit for μ. c 95% lower confidence limit for " .^.m" "'.

Answer 1

Let's consider a pivotal quantity,

$Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

where

Z ~N(0,1)

Let $-Z_{\frac{\alpha}{2}}, Z_{\frac{\alpha}{2}}$ be the lower and upper quartile of Z

Therefore

$P[-Z_{\frac{\alpha}{2}}\leq Z\leq Z_{\frac{\alpha}{2}}]=1-\alpha$

Vn

ㄑㄧ

Vn Vn

Vn Vn

$P[\bar{x}+Z_{\frac{\alpha}{2}}*\frac{\sigma}{\sqrt{n}}\geq \mu\geq \bar{x}-Z_{\frac{\alpha}{2}}*\frac{\sigma}{\sqrt{n}}]=1-\alpha$

Vn

Therefore (1-$\alpha$)*100% confidence interval for the population mean is

Vn

Therefore 95% confidence interval for the population mean is

$P[\bar{x}-Z_{\frac{0.05}{2}}*\frac{\sigma}{\sqrt{n}}\leq \mu\leq \bar{x}+Z_{\frac{0.05}{2}}*\frac{\sigma}{\sqrt{n}}]=1-0.05$

$P[\bar{x}-Z_{0.025}*\frac{\sigma}{\sqrt{n}}\leq \mu\leq \bar{x}+Z_{0.025}*\frac{\sigma}{\sqrt{n}}]=0.95$

When $Y\sim(\mu, 1)$

The 95% lower confidence limit is obtained by

$\bar{x}-Z_{0.025}*\frac{\sigma}{\sqrt{n}}$

ie $\bar{x}-1.96*\frac{1}{\sqrt{n}}$

$\bar{x}-\frac{1.96}{\sqrt{n}}$