Question

Let X_1,X_2,......,X_n be independently and identically distributed b(1, p) random 

variables. Obtain a confidence internal for p using Chebychev’s inequality.

Answer 1

 $Chebyshev's$ inequality is given as

$p(|x-\mu|\lt k\sigma )\geqslant 1-(\dfrac{1}{k^2})$

where

$x$ is the point estimate of the parameter whose confidence interval needs to be determined.

$\sigma$ is the standard deviation of that parameter.

We are required to determine the confidence interval for the parameter $p$ of the binomial distribution.

Given the random variables $X_1,X_2,X_3,...,X_n$ , we need to determine the point estimate for $p$.

To do so, let us define the random variable $Y=\sum ^n_{i=1}X_i=X_1+X_2+....+X_n$ ,

Let $\hat{p}$ be the point estimate for $p$ then $\hat{p}=\sum^n_{i=1}X_i/n=Y/n$ .

The standard deviation$\ sd\ (p)$ for $p$ is given as,

$sd(p)=\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

 

In $Chebyshev's$ inequality stated above we find the value for $k$ using the 95% confidence.

The right hand side of the inequality is equated with 95% in order to find $k$ as below,

$1-(1/k^2)=0.95\implies0.05=1/k^2\implies k^2=20\implies k=4.5(1dp)$

A 95% confidence interval for $p$ is given as

$|\hat{p}-p|\lt k*sd(p)$

This can be re-written as

$\hat{p}-k*sd(p)\lt p\lt\hat{p}+k*sd(p)............(i)$

Replacing for $k$ and $sd(p)$ in $equation (i)$ we have $\hat{p}-4.5*\sqrt{\hat{p}(1-\hat{p})/n}\lt p\lt \hat{p}+4.5*\sqrt{\hat{p}(1-\hat{p})/n}$ .

Therefore the 95% confidence interval for the parameter $p$ is given as

$C.I=(\hat{p}-4.5*\sqrt{\hat{p}(1-\hat{p})/n },\space \hat{p}+4.5*\sqrt{\hat{p}(1-\hat{p})/n })$