chisq.test function in R

[Naitik Shah]

An insurer believes that the distribution of the number of claims on a particular type of policy is binomial with parameters n = 3 and p . A random sample of the number of claims on 153 policies revealed the following results:
Number of Claims | 0 | 1 | 2 | 3 |
Number of Policies | 60 | 75 | 16 | 2 |
(a) Show that the method of moments estimate for p is 0.246.
(b) Carry out a goodness of fit test for the specified binomial model for the number of claims on each policy, ensuring that the expected frequencies are greater than 5.
(c) Use the CDF of a chi squared distribution to find the correct p-value.

Can anyone help me for the parts (b) & (c) for the above question?

R Code for part (a) is as follows:
obs <- c(60,75,16,2)
x <- c(0,1,2,3)
n <- 3
mu <- sum(x*obs)/sum(obs)
n <- 3
p <- mu/n

 

[John Lee]

The built in chisq.test function doesn't work well here as it doesn't correctly calculate the degrees of freedom when you fit a model - so you simply use R as a calculator and do what you would do on a bit of paper.

 

[Naitik Shah]

Hi John,

Thanks for your prompt response.

But would it be possible for you to help me out with the (c) part and how do I go about doing the manual procedure on R?

For the latter, do I just substitute the values in the formula?

Best,
Naitik Shah

 

[John Lee]

Yes, so if you store the observed frequencies in "obs" and the expected frequencies in "exptd" then recall the formula is:
\(\sum \frac{(obs-exptd)^2}{exptd}\) ~ \( \chi^2\)
So if you calculate this in R and store the result in "stat" then recall we reject H0: "model is a good fit" for large values.
So the p-value will just be the probability that the chi-square is greater than stat.