Introduction
This notes $\textbf{lists}$ the classical $(1-\alpha)100\%$ confidence intervals for the parameters involving in the Normal Distribution. Only one method, Inverting Acceptance region of a test is used in this note
Suggested Reading: [CABE] Casella, G., & Berger, R. L. (2002). Statistical inference (Vol. 2). Pacific Grove, CA: Duxbury; specifically, Chapters 8 and 9 of CABE
Keywords:
- Estimator
- Estimate
- Likelihood function
- Maximum Likelihood Estimates (MLE)
- Test of hypotheses
- Acceptance Region
- Tail Area Probabilities
- Normal, t, F, and Chi square Distributions
Objectives
- Method of finding CI
- Normal Models
Let $X_1,X_2,\cdots, X_n$ be iid random sample from normal distribution with mean $\theta$ and variance ${\sigma}^2$
Case 1 : Single mean, where ${\sigma}^2$ is known
$\theta \leq \theta_0$ Vs $\theta > \theta_0$
Acceptance region is
$\overline {x} – \theta_0 \leq z_\alpha \frac{\sigma} {\sqrt n}$
Corresponding interval is
$[\overline {x}-z_\alpha \frac{\sigma}{\sqrt n},\infty)$
$\theta \geq \theta_0$ Vs $\theta < \theta_0$
Acceptance region is
$\overline{x} – \theta_0 > z_{1-\alpha} \frac{\sigma} {\sqrt n}$ or
$\overline {x} – \theta_0 > -z_\alpha \frac{\sigma} {\sqrt n}$
$\Rightarrow \overline {x} + z_\alpha \frac{\sigma} {\sqrt n} > \theta_0$
Interval is
$(-\infty,\overline {x} + z_\alpha \frac{\sigma} {\sqrt n})$
$\theta = \theta_0$ Vs $\theta \neq \theta_0$
Acceptance Region
$|\overline {x} – \theta_0| \leq z_{\alpha/2} \frac{\sigma} {\sqrt n}$
$\Rightarrow – z_{\alpha / 2} \frac{\sigma}{\sqrt n } \leq \overline{x} – \theta_0 \leq z_{\alpha /2}\frac{\sigma}{\sqrt n }$
$\Rightarrow$ Interval is
$(\overline {x} – z_{\alpha / 2} \frac{\sigma}{\sqrt n } , \overline {x} + z_{\alpha / 2} \frac{\sigma}{\sqrt n })$
If ${\sigma^2}$ is unknown, Replace $z$ by $t_{n-1}$ and $\sigma^2$ by the sample variances $S^2$
Case 2 : Independent Two Sample ($\sigma_1^2$,$\sigma_2^2$ Known)
$\theta_1 – \theta_2 \leq \delta$ Vs $\theta_1 – \theta_2 > \delta$
Acceptance Region
$\overline {x} – \overline {y} – \delta \leq z_\alpha \sqrt {\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}$
Confidence interval is
$[\overline {x}-\overline {y} -z_\alpha \sqrt {\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}~, ~~\infty)$
$\theta_1 – \theta_2 \geq \delta$ Vs $\theta_1 – \theta_2 < \delta$
Acceptance Region
$\overline {x} – \overline {y} – \delta \geq z_{1-\alpha} \sqrt {\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}$
$= – z_\alpha \sqrt {\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}$
Confidence interval is
$( -\infty~ ,~~ \overline {x}-\overline {y} + z_\alpha \sqrt {\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}]$
$\theta_1 – \theta_2 = \delta$ Vs $\theta_1 – \theta_2 \neq \delta$
Confidence interval is
$(\overline {x}-\overline {y} – z_{\alpha/2} \sqrt {\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}},~~\overline {x}-\overline {y} + z_{\alpha/2} \sqrt {\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}})$
if $\sigma_1^2 , \sigma_2^2$ are unknown Replace $z$ by $t_{m+n-2}$ and $SE = S_p \sqrt{\frac{1}{m}+\frac{1}{n}}, ~~ S_p^2 = \frac{(m-1)S_1^2 + (n-1)S_2^2}{m+n-2}$
Case 3 : Bivariate Samples
use Case 1 to have intervals with $W_i = x_i – y_i$
$S_W =\frac{\sum (W_i – W)^2}{n-1}$
$\overline {W}=\overline {x}-\overline {y}$
$[ \overline {W} – z_\alpha \frac{S_W}{\sqrt n} , \infty)$
$(-\infty , \overline {W} + z_\alpha \frac{S_W}{\sqrt n}]$
$(\overline {W} – z_\alpha \frac{S_W}{\sqrt n} , \overline {W} + z_\alpha \frac{S_W}{\sqrt n})$
Case 4 : Single Variance in Normal$(\mu, \sigma^2)$
1. $\mu$ is known
$X_1,X_2,\cdots,X_n \sim \text{Normal}~(\mu,\theta) ~~~~~\theta = \sigma^2$ and $\mu$ is Known
$\theta \leq \theta_0$ Vs $\theta > \theta_0$
Acceptance Region is
$\frac {\sum(x_i-\mu)^2}{\theta_0} < \chi^2_{\alpha,n}$
$\Rightarrow$ Interval is
$[\frac{\sum(x_i-\mu)^2 }{\chi^2_{\alpha,n}}, \infty )$ for $(\theta=\sigma^2)$
Or,
$[\sqrt{\frac{\sum(x_i-\mu)^2}{\chi^2_{\alpha,n}}},\infty)$ for $\theta = \sigma$
$\theta \geq \theta_0$ Vs $\theta < \theta_0$
Acceptance Region
$\frac{\sum(x_i-\mu)^2}{\theta_0} > \chi^2_{1-\alpha,n}$
Interval is $\Big(0, \frac{\sum(x_i-\mu)^2}{\chi^2_{(1-\alpha),n}}\Big]$
$\theta = \theta_0 ~~ Vs~~ \theta \neq \theta_0$
Interval is
$\Big(\frac{\sum(x_i-\mu)^2}{\chi^2_{n,\alpha/2}},\frac{\sum(x_i-\mu)^2}{\chi^2_{(1-\alpha /2 ),n}}\Big)$
2. $\mu$ is unknown
Replace $\mu$ by $\overline{x}$ and the degrees of freedom is $n – 1$
Case 5 : Equality of variances in Normal$(\mu_1, \sigma_1^2)$ and Normal $(\mu_2, \sigma_2^2)$
1. $\mu_1~\&~\mu_2$ are known
$X_1,\cdots,X_n \sim \text {Normal}~(\mu_1,\sigma_1^2)$ and $Y_1,\cdots,Y_n \sim \text {Normal}~(\mu_2,\sigma_2^2)$
TEST 1
$\Big[\frac{1}{F_{\alpha,m,n}} \frac{\sum(x_i-\mu_1)^2}{\sum(y_j-\mu_2)^2} \frac{n}{m},\infty)$
TEST 2
$(0, \frac{1}{F_{1-\alpha,m,n}} \frac{\sum(x_i-\mu_1)^2}{\sum(y_j-\mu_2)^2} \frac{n}{m}\Big]$
TEST 3
$\Big(\frac{1}{F_{\alpha/2,m-1,n-1}} \frac{\sum(x_i-\mu_1)^2}{\sum(y_j-\mu_2)^2} \frac{n}{m},\frac{1}{F_{1-\alpha/2,m-1,n-1}} \frac{\sum(x_i-\mu_1)^2}{\sum(y_j-\mu_2)^2} \frac{n}{m}\Big)$
2. $\mu_1~\&~\mu_2$ are Unknown
Replace the test statistic by $\frac{S^2_{X}}{S^2_{Y}}$