# Calculating the Probability of a Type II Error

Published: 2021-07-08 10:55:05

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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. 6 < 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. 5 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| 3. 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 probability of not rejecting the null hypothesis when = 25 is 0. 2676 (see above Figure). This means that, when the mean wind velocity is 25 mph, there is a 26. 76% probability of erroneously concluding that the site is not profitable. If this probability is considered too large, we can reduce it by either increasing or increasing n. For example, if we increase a to 0. 10 and leave n = 50, then = 0. 0475. Rejection region: (xbar-20) / (12/50) > 1. 8 xbar > 22. 17 22. 17| 24. 5| -2. 33| 1. 697| -1. 37| 0. 4147| 0. 0853| 22. 17| 25| -2. 83| 1. 697| -1. 67| 0. 4525| 0. 0475| 22. 17| 25. 5| -3. 33| 1. 697| -1. 96| 0. 4750| 0. 0250| 25 With = 0. 10, however, the probability of building on a site that is not profitable is too large. If we let = 0. 01 but increase n to 100, then = 0. 0329. 22. 796| 23| -0. 204| 1. 20| -0. 17| 0. 0675| 0. 4325| 22. 796| 23. 5| -0. 704| 1. 20| -0. 59| 0. 2224| 0. 2776| 22. 796| 24| -1. 204| 1. 20| -1. 00| 0. 3413| 0. 1587| 22. 796| 24. 5| -1. 704| 1. 20| -1. 42| 0. 4222| 0. 0778| 22. 796| 25| -2. 204| 1. 20| -1. 84| 0. 4671| 0. 329| 22. 796| 25. 5| -2. 704| 1. 20| -2. 25| 0. 4878| 0. 0122| 22. 796| 26| -3. 204| 1. 20| -2. 67| 0. 4962| 0. 0038| 22. 796| 26. 5| -3. 704| 1. 20| -3. 09| 0. 4990| 0. 0010| 22. 796| 27| -4. 204| 1. 20| -3. 50| 0. 5000| 0. 0000| Now both and are quite small, but the cost of sampling has increased. Nonetheless, the cost of sampling is small in comparison to the costs of making Type I and Type II errors in this situation. Another way of judging a test is to measure its power -the probability of its leading us to reject the null hypothesis when it is falserather than measuring the probability of a Type II error. Thus, the power of the test is equal to 1 . In the present example, the power of the test with n = 50 and = . 01 is 1 0. 2676 = 0. 7324. When more than one test can be performed in a given situation, we would naturally prefer to use the test that is correct more frequently. If (given the same alternative hypothesis, sample size, and significance level) one test has a higher power than a second test, the first test is said to be more powerful. To determine the appropriate sample size for specified levels of the Type I and Type II Errors, consult the Text. ————————————————- ———————————————— ASSIGNMENT: ————————————————- ————————————————- In the windmill example presented in class, the Beta and Power functions are computed for n = 50. Compute and display the Beta and Power functions for n = 25, n = 75, and n = 125. Display your results in no more than two graphs. ————————————————- ————————————————- Briefly discuss your results. ————————————————-

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