Bernhard Kabelka - Mathematical Statistics: "Basic Theory of Estimation

Confidence Intervals

$[A(X_1,\,...,\,X_n),B(X_1,\,...,\,X_n)]$ is called a confidence interval with coverage probability if for all

$\P _{theta}(A <= theta <= B) >= gamma$

If $X_1,\,X_2,\,...$ are i.i.d. normal random variables with mean and variance , then

$\overlineX_n$ has a normal distribution with mean and variance $\frac{sigma^2n$ .
$\frac{n-1}{sigma^2*S_n^2$ has -distribution with degrees of freedom.
$\overlineX_n$ and are independent.
$\frac{\overlineX_n- µ{\sqrt{S_n^2/n}$ has a t-distribution with degrees of freedom.

With the help of this theorem one can obtain confidence intervals for the normal distribution:

confidence interval for ( known):
$[ \overlineX_n- z_{\frac{1+gamma}2......2} * \sqrt{\frac{sigma^2n ]$
confidence interval for ( unknown):
$[ \overlineX_n- t_{n-1,\frac{1+gamma......ma}2 * \sqrt{\frac{S_n^2n ]$
confidence interval for ( known):
$[ \frac{sum_{i=1}^n (X_1-µ)^2{Chi-Quadrat_{n,\f......n (X_1-µ)^2{Chi-Quadrat_{n,\frac{1-gamma}2 ]$
confidence interval for ( unknown):
$[ \frac{(n-1) * S_n^2{Chi-Quadrat_{......suremath{S_n^2{Chi-Quadrat_{n,\frac{1-gamma}2 ]$

When looking at proportions, that is to say

$\P (X=1) = theta, \P (X=0) = 1 - theta {for a} \; theta in (0,1),$

one can use the fact that $\overlineX_n$ has an approximate normal distribution with mean and variance $\frac{theta(1-theta)}n$ . This leads to an approximate confidence interval

$[ \overlineX_n- z_{\frac{1+gamma}2......{1+gamma}2 * \sqrt{\frac{theta(1-theta)}n ]$

but, unfortunately, we do not know the exact value of . So we could replace it by its estimator $\overlineX_n$ , or solve the equation

$\overlineX_n\pm z_{\frac{1+gamma}2 * \sqrt{\frac{theta(1-theta)}n = theta$

with respect to and use the two solutions as the limits of our confidence interval.