Sept '10 Q5 - confidence interval of binomial parameter

Hi,

A random sample of 200 travel insurance policies contains 29 on which the
policyholders made claims in their most recent year of cover.

Calculate a 99% confidence interval for the proportion of policyholders who make claims in a given year of cover.

The answer booklet uses Ch 11 pg 16 of the notes which explains that if n is large then we can use the normal approximation of the Binomial. Thus, the confidence interval is (where p is the estimator of theta = 29/200):

p ± 2.5758 * sqrt( p(1-p) / n)

which gives a CI of (0.0809, 0.2091).

However, pg 9 of Ch 11 discusses the use of the t result when forming a confidence interval for samples with an unknown variance.

If we break the exam question into a series of 200 Bernoulli trials we should be able to obtain the same confidence interval on the mean of the trials, which is the same as the proportion of policyholders that have made claims in the last 12 months. We can take Xbar to be 29/200, and S to be 29/200 * (1 - 29/200). n is of course 200.

I know the Tables doesn't give a t value with 199 degrees of freedom, but an online calculator gives a t value of 2.6007. Using this to obtain a confidence interval gives (0.0802, 0.2098), which is a "looser" interval than the first.

Can anyone explain why I'm not obtaining the same confidence interval? Granted, they are very similar, but not the same. I wouldn't want to use one method in an exam when the other was more appropriate.

Thanks.

 

Guess I've found the answer to my question online.

From what I understand, the approximation of the Binomial to the Normal is the same as using the t distribution with inifinte degrees of freedom. Thus, the t-distribution with 199 degrees of freedom is a more "accurate" 99% CI, but I guess its application required the use of a t value which cannot be obtained using the Tables. Therefore the Normal approximation is used.

It seems a rule of thumb is that if n is large enough that the t value cannot be found using the Tables, then the Normal approximation is suitable.

Does that sound reasonable?

 

Thanks