Results 1 to 4 of 4

Math Help - Please help...

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    21

    Unhappy Please help...

    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)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by math beginner View Post
    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 \bar{X} and S^2 be the sample mean and variance respectively.

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

    Therefore:

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


    \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:

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

    \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, \alpha = 0.05, ~ t_{0.025, \, df=4} = 2.776. Therefore .....

    b. n = 8, \alpha = 0.1, ~ t_{0.05, \, df=7} = \, .... Therefore .....
    Last edited by mr fantastic; July 11th 2008 at 05:57 AM. Reason: Added a < that was missing.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    21

    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by math beginner View Post
    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 S^2 is a random variable and therefore L^2 is a random variable.

    E(L^2) is a number.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum