CT3 Q12.18 chi square test

Can some one please explain me the steps of how did we arrive at the expected value of the the observed data in question 12.18 of chapter hypothesis testing?
And how did we further solve it to prove that it isn't a Poisson distribution ?

PDF page 525.

Thank you for your help and response.

 

Hello shdh,
In, this question we are testing the hypothesis that the number of claims, \(X\) follows a Poisson distribution.
We estimated \(\lambda\) using MLEs as:
\[\hat\lambda = 0.1658\]
So,
\[X\sim Poi(0.1658)\]
Now, using this we calculate the Expected frequencies as follows:
When the number of claims is 0, the expected frequency is = \(3420P[X=0] = 3420(e^{-0.1658}) = 2897.48\)
Now, we can calculate the rest of the probabilities in a similar fashion.

We are performing the chi-square goodness of fit test for our null hypothesis,
\(H_{0}: \) The given data conforms to a Poisson distribution.

As the test statistic was greater than the tabulated value, we rejected the null hypothesis and concluded that the given data does not conform to a Poisson distribution.

Hope this helps.
Kaustav.

 

Thank you very much. It has been of great help to me. Thanks once again.