• Jul 10th 2008, 04:36 AM
math beginner
Suppose that X1,....,Xn form a random sample from a normal distribution for which both the mean m and variance s^2 are unknown.
Let the random variable L denote the length of the shortest confidence interval for m that can be constructed from the observed values in the sample.
Find the value of E(L^2) for the following values of the sample size n and the confidence coefficient y:
a. n = 5, y = 0.95 (Ans: 6.16 s^2)
b. n = 8, y = 0.90 (Ans:2.05 s^2)
• Jul 11th 2008, 01:37 AM
mr fantastic
Quote:

Originally Posted by math beginner
Suppose that X1,....,Xn form a random sample from a normal distribution for which both the mean m and variance s^2 are unknown.
Let the random variable L denote the length of the shortest confidence interval for m that can be constructed from the observed values in the sample.
Find the value of E(L^2) for the following values of the sample size n and the confidence coefficient y:
a. n = 5, y = 0.95 (Ans: 6.16 s^2)
b. n = 8, y = 0.90 (Ans:2.05 s^2)

Let the random variables $\displaystyle \bar{X}$ and $\displaystyle S^2$ be the sample mean and variance respectively.

Theorem (there's a proof somewhere in these forums): The random variable $\displaystyle \sqrt{n} \, \left( \frac{\bar{X} - \mu}{S}\right)$ follows the t-distribution with n-1 degrees of freedom.

Therefore:

$\displaystyle \Pr \left( -t_{\alpha/2} < \sqrt{n} \, \left( \frac{\bar{X} - m}{S} \right) < t_{\alpha/2} \right) = 1 - \alpha$

$\displaystyle \Rightarrow \Pr \left( \bar{X} - \frac{t_{\alpha /2} \, S}{\sqrt{n}} < m < \bar{X} + \frac{t_{\alpha/2} \, S}{\sqrt{n}} \right) = 1 - \alpha$.

Therefore:

$\displaystyle L = \frac{2 t_{\alpha/2}}{\sqrt{n}} \, S \Rightarrow L^2 = \frac{4 \left(t_{\alpha/2}\right)^2}{n} \, S^2$

$\displaystyle \Rightarrow E(L^2) = \frac{4 \left(t_{\alpha/2}\right)^2}{n} \, E(S^2) = \frac{4 \left(t_{\alpha/2}\right)^2}{n} s^2$.

a. n = 5, $\displaystyle \alpha = 0.05, ~ t_{0.025, \, df=4} = 2.776$. Therefore .....

b. n = 8, $\displaystyle \alpha = 0.1, ~ t_{0.05, \, df=7} = \, ....$ Therefore .....
• Jul 11th 2008, 08:45 PM
math beginner
Thank you!
Thank you very much.

I have a slight confusion in that I am not sure what difference it makes when the question asks for E(L^2) vs when the question asks for L^2.

Why is it that the question asks for E(L^2) rather than just L^2?
• Jul 11th 2008, 08:57 PM
mr fantastic
Quote:

Originally Posted by math beginner
Thank you very much.

I have a slight confusion in that I am not sure what difference it makes when the question asks for E(L^2) vs when the question asks for L^2.

Why is it that the question asks for E(L^2) rather than just L^2?

Because $\displaystyle S^2$ is a random variable and therefore $\displaystyle L^2$ is a random variable.

$\displaystyle E(L^2)$ is a number.