Using continuity correction for normal approximation or not?

Question

Below is a question on a recent actuarial exam, Exam 3L of the CAS. I didn't know whether or not to use the continuity correction when using the normal approximation to do hypothesis testing involving a bernoulli trial. The answer is vastly different (based on the answer choices given in the exam) depending on whether or not you use it. The problem does not specify to use it or to not use it, though some previous problems have specified one or the other, whereas others have not specified either.

You are given the following:

• Accidents happen during a work day at a probability of p when a machine is operated.
• The null hypothesis $H_0$ is that the probability of an accident is 0.05; the alternative hypothesis $H_1$ is that the probability is less than 0.05.
• If less than 20 accidents are observed in 365 work days, then reject the null hypothesis.

Using the normal approximation, calculate the probability of Type II error using the value 0.03 as the true probability of an accident occurring.

Any help would be greatly appreciated. Just to be clear, I am hoping to better understand when it should be used and when it should not be used, in general, in addition to learning the best way to do this problem.

Answer 1

I can't speak to what the people setting the exam might do; sometimes the actuarial choices on statistical matters baffle me.

I can only speak to what I see as the statistics issues.

Given 20 is pretty much near the middle of the null distribution, which itself is reasonably well approximated by a normal, the continuity correction will greatly improve the accuracy of probability calculations there. So if you were trying to compute the type I error rate, it's quite useful.

(These are ridiculous type I error rates, by the way; the mean, median and mode of the null are included in the rejection region! A more sensible critical value would be somewhere around 13 or likely even less; best places to put the critical value depends on the relative cost of the two types of error)

However, while the continuity correction works well for calculating the Type I error rate, for the considered alternative (p = 0.03), the critical value is way in the tail and then the continuity correction often unhelpful; I'd have leaned toward avoiding it. (And since the alternative is what the question is about... that's where it matters)

But I'd be unsurprised if the actuaries have not covered such details in the course - that the continuity correction works very well when you have exact symmetry and more generally works well when you're toward the mean of the binomial, and often does badly when you are in the far tail of asymmetric distributions (p far from 0.5), though it depends on which direction you're looking. I don't know if you're supposed to consider this issue in this way.

It turns out that the results in this case are:

The exact Type II error rate is 0.0079, with continuity correction it's 0.0044, and without it's 0.0027 (assuming I got it right the second time around).

It looks like my inclination to avoid it in this case was of no benefit, though neither approximation is very good.

Answer 2

This is a "find the mistakes" question.
When you see "accident", think Poisson, not Bernoulli, process.
The process average "p" needs a rate of occurrence: .05/workday. It is not a "Probability".
The H0 and H1 as stated, call for a Type1 Risk. Let's make it 5%.
And the question becomes: What is the largest number of accidents in the past workyear that would have you reject H0?
Using the Normal Approximation, the answer: 10
There follows •If less than 20 accidents are observed in 365 work days, then reject the null hypothesis. "20" is a mistake; it should be "11"; but, in any case, less than "18".
The test question is then modified.