Answer to Question #205736 in Statistics and Probability for zakkir

Question #205736

Let X1,X2,......,Xbe independently and identically distributed b(1, p) random 

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


1
Expert's answer
2022-01-11T15:07:08-0500

 "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 })"

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