Chi sqaure & confidence interval

sstudent
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Lets say you have 5 trials, and 5 output let's say {5, 43, 60, 30 , 4}...so how would u get the chi sqaure from here & the confidence interval. i haven't work on prob &statc in years, thanks a lot for help folks
 
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I'm supposing you made up this problem... for your random sample of size n=5 is {5, 43, 60, 30 , 4},these are your observed cells/frequencies, but you do not have have your expected cells/frequencies.
 
Basically u can do the confidence for 99%
 
I don't understand what you're trying to do.

For a Chi-Squared test, you need a null hypothesis and alternative hypothesis.
The test statistical value is \chi^2=\sum \frac{(O-E)^2}{E}
where O is a shorthand notation for your observed cells and E is the shorthand notation for expected cell/frequency.

If you are trying to find the confidence interval for the true variance \sigma^2, then the formula for that is

P\left( C_{1}< \frac{(n-1)s^2}{\sigma^2} < C_{2} \right) = 1- \alpha = .99 since 100(1-alpha)% =99%

where C_{1}=\frac{(n-1)s^2}{\chi_{\alpha/2,n-1}} is the lower limit
and
C_2 = \frac{(n-1)s^2}{\chi_{1-\alpha/2,n-1}} is your upper limit.
 
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lets say u consider #5 as the thresh#, and for the numbers> 5, u tryto find the confiendnce interval for 99.99% or greater. thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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