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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.

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  • $\begingroup$ please add the self-study tag $\endgroup$ – Glen_b -Reinstate Monica May 6 '13 at 0:16
  • $\begingroup$ Did the answers come out? $\endgroup$ – Glen_b -Reinstate Monica May 9 '13 at 7:41
  • $\begingroup$ @Glen_b Well, "Preliminary answers" have the non-continuity correction version as the answer but we have a few days to email them and tell them why they should accept others. The exam has several mistakes in it, which is frustrating. $\endgroup$ – Graphth May 10 '13 at 21:17
  • $\begingroup$ Well, if you want to, you can easily argue for the correction in this case, since it's often used and in this case gives a better approximation to the exact binomial probability. It would be hard to argue against it! If I was marking it, I'd accept either... and then if I wrote it, I'd have to smack myself for asking a question where it made enough difference to worry about. $\endgroup$ – Glen_b -Reinstate Monica May 11 '13 at 2:44
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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.

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  • $\begingroup$ Are you sure about the 0.0017. I got something like 0.0044 with the continuity correction. If I reject by getting 19 or less than, then I'm talking about the probability of 20 or more to not reject which would be 19.5 or more using the continuity correction, right? Also, yes, I noticed while taking the test that the mean is inside the rejection region. Thanks for this detailed response. This is the sort of thing that is going to help me understand things better. $\endgroup$ – Graphth May 6 '13 at 1:09
  • $\begingroup$ My mistake; I misread the rejection rule as 20 or less when it's 'less than 20'. Two of those figures change (now fixed). The rejection region includes its boundary so it's kind of weird to state it in terms of an open interval (less than 20) instead of a closed one (19 or less). $\endgroup$ – Glen_b -Reinstate Monica May 6 '13 at 2:01
  • $\begingroup$ The CAS is going to put out the answers tomorrow, or so I have heard. So, I can see if they agree with my using the continuity correction. If not, I can easily argue that it's a much better approximation (though still poor as you said), so I can try to get them to accept my answer :) $\endgroup$ – Graphth May 6 '13 at 2:47
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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.

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