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Estimation, Evaluation, and Selection of Actuarial Models

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4.2. A REVIEW OF HYPOTHESIS TESTS 63<br />

The traditional test statistic for this problem is<br />

¯x − 1, 200<br />

z =<br />

3, 435/ √ 20 =0.292<br />

<strong>and</strong> the null hypothesis is rejected if z>1.645. Because 0.292 is less than 1.645, the null hypothesis<br />

is not rejected. The data do not support the assertion that the average claim exceeds 1,200. ¤<br />

The test in the previous example was constructed to meet certain objectives. The first objective<br />

is to control what is called the Type I error. It is the error made when the test rejects the null<br />

hypothesis in a situation where it happens to be true. In the example, the null hypothesis can be<br />

true in more than one way. This leads to the most common measure <strong>of</strong> the propensity <strong>of</strong> a test to<br />

make a Type I error.<br />

Definition 4.3 The significance level <strong>of</strong> a hypothesis test is the probability <strong>of</strong> making a Type I<br />

error, given that the null hypothesis is true. If it can be true in more than one way, the level <strong>of</strong><br />

significance is the maximum <strong>of</strong> such probabilities. The significance level is usually denoted by the<br />

letter α.<br />

This is a conservative definition in that it looks at the worst case. It is typically a case that is<br />

on the boundary between the two hypotheses.<br />

Example 4.4 Determine the level <strong>of</strong> significance for the test in the previous example.<br />

Begin by computing the probability <strong>of</strong> making a Type I error when the null hypothesis is true<br />

with µ =1, 200. Then,<br />

Pr(Z >1.645|µ =1, 200) = 0.05.<br />

That is because the assumptions imply that Z has a st<strong>and</strong>ard normal distribution.<br />

Now suppose µ has a value that is below 1,200. Then<br />

µ <br />

¯X − 1, 200<br />

Pr<br />

3, 435/ √ 20 > 1.645 µ <br />

¯X − µ + µ − 1, 200<br />

=Pr<br />

3, 435/ √ > 1.645<br />

20<br />

µ <br />

¯X − µ<br />

=Pr<br />

3.435/ √ µ − 1, 200<br />

> 1.645 −<br />

20 3.435/ √ .<br />

20<br />

Because µ is known to be less than 1,200, the right h<strong>and</strong> side is always greater than 1.645. The<br />

left h<strong>and</strong> side has a st<strong>and</strong>ard normal distribution <strong>and</strong> therefore the probability is less than 0.05.<br />

Therefore the significance level is 0.05. ¤<br />

The significance level is usually set in advance <strong>and</strong> is <strong>of</strong>ten between one <strong>and</strong> ten percent. The<br />

second objective is to keep the Type II error (not rejecting the null hypothesis when the alternative<br />

is true) probability small. Generally, attempts to reduce the probability <strong>of</strong> one type <strong>of</strong> error increase<br />

the probability <strong>of</strong> the other. The best we can do once the significance level has been set is to make<br />

the Type II error as small as possible, though there is no assurance that the probability will be a<br />

small number. The best test is one that meets the following requirement.

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