# Thread: confidence interval chi square probability search

1. ## confidence interval chi square probability search

1000 samples each of size n=121, where observations Xi ~ iid N ( m, sigma^2), for each sample we calculate Confidence Interval: I = [ 0.75 s^2, s^2 + s^2/4 ] where s^2=(1/(n-1))SIGMA(Xi - X bar)^2

how many samples include the true value of sigma^2, ie 1000 * prob.

we know that (n-1) s^2/sigma^2 ~ Chi-sqrd with (n-1) degrees of freedom

so i started:

P (0.75s^2 < sigma^2 < s^2 + s^2/4) = P (0.75 < sigma^2/s^2 < 5/4) =
=P (0.75/(n-1) < sigma^2/(s^2(n-1)) < 1.25/(n-1)) =
=P (1.25/(n-1) < (n-1)s^2/sigma^2 < 0.75/(n-1)) =
=P (1.25/120 < Chi-sqrd with (n-1) df < 0.75/120) =
= chi n-1 df (0.75/120) - chi n-1 df (1.25/120) =
= chi n-1 df (0.00625) - chi n-1 df (0.0104167) ..... =
=1 -alpha

and now im stuck, i am not sure if i am allowed to deduct them, and whether the numbers represent the quantiles of chi-squared, and if yes, i have a complications to find them in the table, was also thinking to get each separaetely the p value for v=120 and then deduct but dont like it....

how do i finish it now to get the probability...

thank you, if any

nicky

2. It would be a lot easier if you used TeX.

$P(3S^2/4<\sigma^2<5S^2/4)=P(3 (120)S^2/(4\sigma^2)<120< 5(120)S^2/(4\sigma^2))$

$= P(3 C/4<120<5C/4)$ where $C\sim\chi^2_{120}$

$= P(96