Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (loser66):

On this, at 3:24, he talks about program inverse chi-square distribution. Does anyone know about it? Please post the code or give me the link. Thanks in advance https://www.youtube.com/watch?v=nFieXUy9Rxs

OpenStudy (loser66):

@myininaya

OpenStudy (zarkon):

try solve(\(\chi^2\)cdf(0,\(x\),n)-(1-\(\alpha\)),\(x\),n) where \(n\) is the degrees of freedom and \(\alpha\) is the number such that \[P(X\ge \chi^2_{n})=\alpha\]

OpenStudy (loser66):

Suppose I need find \(P(a< X_4^2<b)=0.90\), how to hit the keys on my ti83?

OpenStudy (zarkon):

you trying to find the confidence interval?

OpenStudy (loser66):

That is the last step and I can find it by hand.

OpenStudy (zarkon):

in general a and b are not unique

OpenStudy (loser66):

Yes, I know. Suppose we choose \(P( X_4^2 <a) =0.05\) then how to find a?

OpenStudy (zarkon):

\[\text{solve}(\chi^2cdf(0,x,4)-.05,x,4)\]

OpenStudy (zarkon):

then \[\text{solve}(\chi^2cdf(0,x,4)-.95),x,4)\] for b

OpenStudy (loser66):

It does not work. Do you mean \(X^2cdf(0,x,4) - X^2 cdf(0.05,x,4)\)?

OpenStudy (zarkon):

works on mine

OpenStudy (zarkon):

ah...jas

OpenStudy (zarkon):

\[\text{solve}(\chi^2cdf(0,x,4)-.05,x,4)\] \[\text{solve}(\chi^2cdf(0,x,4)-.95,x,4)\]

OpenStudy (zarkon):

had an extra parenthesis

OpenStudy (loser66):

solve (.....) I don't have that word on my calculator. :)

OpenStudy (zarkon):

you do... look in the catalog

OpenStudy (zarkon):

2nd 0

OpenStudy (loser66):

YYYYYYYYYYYes. Thanks a ton....

OpenStudy (zarkon):

vid

OpenStudy (zarkon):

happy you got it

OpenStudy (loser66):

woah... You should teach your students this method. It save a lot of lives. :)

OpenStudy (zarkon):

i wrote a program to find the CI on the 83/84 and I share it with my students

OpenStudy (loser66):

Best Professor ever. :)

OpenStudy (zarkon):

sure ;)

OpenStudy (loser66):

I am greedy :) but I would like to have that program too. :) Please.

OpenStudy (zarkon):

all you need to do : have inputs \(s^2\),d.f.(D) and Confidence level (C) then compute \[\chi^2_{L}=\text{solve}(\chi^2cdf(0,x,D)-(1-C)/2,x,D)\] \[\chi^2_{R}=\text{solve}(\chi^2cdf(0,x,D)-(1+C)/2,x,D)\] C.I for \(\sigma^2\)is given by \[\left[\frac{D s^2}{\chi^2_{R}},\frac{D s^2}{\chi^2_{L}}\right]\]

OpenStudy (loser66):

I need a concrete example to understand it. Example 8.7 text book, how to use this code ?

OpenStudy (zarkon):

example 8.7 doesn't use a chi square it uses a z-interval...that is built in to your ti 83

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!