Graduation- chi square test statistic

Hi, the test stat given for the chi-square test is not quite what I'm familiar with, it's given in terms of "standardised deviations". From CT3 and my general knowledge of stats I know the test stat for the chi squared test to be (O-E)^2/E summed up over the different data points. Can I use this?

Thanks

 

same there...
let A: Actual deaths
E:Expected Death
so the standardised deviations are \(\frac{A_i-E_i}{\sqrt(E_i)} \)....
Chi-square is sum of standard-deviations-squared.(Actual and Observed are same thing)

 

Hi thanks but dint think this answers my question and I'm well aware of actual/observed .... I deal with that stuff everyday at work!

My questions is that I thought the sum of the formula I gave in my original post gives a chi squared distribution. The formula in my original post is not quite the standardised deviance.

Maybe someone can tell me what I'm missing here?

Edit: oh youre saying they're the same- well just squares it. Ok I can see that using the formula you've used. but I thought for the standardised deviance you divide by the standard deviation of the actual number of claims/deaths/whatever

The whole question is a pretty mute point to be honest- it's not difficult ... but just interested to understand why the difference.

Thanks

 

I have this question too!

 

I'm not entirely sure what the question is!

The underlying (approximate) normal distributions we're using for Dx (the number of deaths at age x) have expected value = variance, so the square root of the expected value in the denominator of zx is also the standard deviation. This means that zx ~ N(0,1) and the sum of the zx's squared will follow a chi-squared distribution.

See pages 26 and 27 of Chapter 12 for the distributions and forms of zx.

 

But, that is true when we are approximating a poisson model. Using a binomial model leads to a different variance. Yes, we can still approximate variance here as expected value, but for that we require q(x) to be approximately zero.

 

I agree. The assumption that qx is small underlies our approximation using the chi-square distribution for the binomial model.