# Type 2 Error

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• Published : September 23, 2010

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Calculating the Probability of a Type II Error

To properly interpret the results of a test of hypothesis requires
that you be able to judge the pvalue of the test.

However, to do so also requires that you have an understanding of the relationship between Type I and Type II errors.

Here, we describe how the probability of a Type II error is computed.

A Type II error occurs when a false null hypothesis is not rejected. For example, if a rejection region is as follows:

xbar < 127.06 or xbar > 132.94

and the null hypothesis is false, then the probability of a Type II error is defined as

= P(127.06 < xbar < 132.94 (given that H0 is false)

The condition that the null hypothesis is false only tells us that the mean is not equal to 130. If we want to compute , we need to specify a value for . Suppose that we want to determine the probability of making a Type II error when, in actual fact, = 135, 131, 139, and/or any other value.

A Windmill Example:

The feasibility of constructing a profitable electricityproducing windmill depends on the average velocity of the wind. For a certain type of windmill, the average wind speed would have to exceed 20 mph in order for its construction to be feasible. To test whether or not a particular site is appropriate for this windmill, 50 readings of the wind velocity are taken, and the average is calculated. The test is designed to answer the question, is the site feasible? That is, is there sufficient evidence to conclude that the average wind velocity exceeds 20 mph? We want to test the following hypotheses.

H0: A 20
HA: A > 20

If, when the test is conducted, a Type I error is committed (rejecting the null hypothesis when it is true), we would conclude mistakenly that the average wind velocity exceeds 20 mph. The consequence of this decision is that the windmill would be built on an inappropriate site. Because this error is quite costly, we specify a small value for a, = 0.01.

If a Type II error is committed (not rejecting the null hypothesis when it is false), we would conclude mistakenly that the average wind velocity does not exceed 20 mph. As a result, we would not build the windmill on that site, even though the site is a good one. The cost of this error may not be very large, since, if the site under consideration is judged to be inappropriate, the search for a good site would simply continue.

But suppose that a site where the wind velocity is greater than or equal to 25 mph is extremely profitable. To judge the effectiveness of this test (to determine if our selection of = 0.01 and n = 50 is appropriate), we compute the probability of committing this error. Our task is to calculate when = 25. (Assume that we know that ( = 12 mph.)

Our first task is to set up the rejection region in terms of xbar.

Rejection region: z > z = z0.01 = 2.33 (look up 0.9900 in Table)

So we have z = (xbar-) / (/n) = (xbar-20) / (12/50) > 2.33

Rejection region: xbar > 23.95

Region where H0 is not rejected: xbar < 23.95

Thus: = P(xbar < 23.95 (given that = 25) =

P{[(xbar-) / (/n)] < [(23.95-25) / (12/50)] =

P(z < -0.62) = 0.5 – 0.2324 = 0.2672

23.95| 22| 1.95| 1.697| 1.15| 0.3749| 0.8749|
23.95| 22.5| 1.45| 1.697| 0.85| 0.3023| 0.8023|
23.95| 23| 0.95| 1.697| 0.56| 0.2123| 0.7123|
23.95| 23.5| 0.45| 1.697| 0.27| 0.1064| 0.6064|
23.95| 24| -0.05| 1.697| -0.03| 0.0120| 0.4880|
23.95| 24.5| -0.55| 1.697| -0.32| 0.1255| 0.3745|
23.95| 25| -1.05| 1.697| -0.62| 0.2324| 0.2676|
23.95| 25.5| -1.55| 1.697| -0.91| 0.3186| 0.1814|
23.95| 26| -2.05| 1.697| -1.21| 0.3869| 0.1131|
23.95| 26.5| -2.55| 1.697| -1.50| 0.4332| 0.0668|
23.95| 27| -3.05| 1.697| -1.80| 0.4641| 0.0359|

This is the graph for associated with numbers from 18 to 32.5:

The...