M09_BERE8380_12_OM_C09.QD 2/21/11 3:44 PM Page 1 9.6 The Power of a Tet 9.6 The Power of a Tet 1 Section 9.1 defined Type I and Type II error and their aociated rik. Recall that a repreent the probability that you reject the null hypothei when it i true and hould not be rejected, and b repreent the probability that you do not reject the null hypothei when it i fale and hould be rejected. The power of the tet, 1 - b, i the probability that you correctly reject a fale null hypothei. Thi probability depend on how different the actual population parameter i from the value being hypotheized (under H 0 ), the value of a ued, and the ample ize. If there i a large difference between the population parameter and the hypotheized value, the power of the tet will be much greater than if the difference between the population parameter and the hypotheized value i mall. Selecting a larger value of a make it eaier to reject H 0 and therefore increae the power of a tet. Increaing the ample ize increae the preciion in the etimate and therefore increae the ability to detect difference in the parameter and increae the power of a tet. The power of a tatitical tet can be illutrated by uing the Oxford Cereal Company cenario. The filling proce i ubject to periodic inpection from a repreentative of the conumer affair office. The repreentative job i to detect the poible hort weighting of boxe, which mean that cereal boxe having le than the pecified 368 gram are old. Thu, the repreentative i intereted in determining whether there i evidence that the cereal boxe have a mean weight that i le than 368 gram. The null and alternative hypothee are a follow: H 0 : m Ú 368 1filling proce i working properly2 H 1 : m 6 368 1filling proce i not working properly2 The repreentative i willing to accept the company claim that the tandard deviation,, equal gram. Therefore, you can ue the tet. Uing Equation (9.1) on page 330, with L (the lower critical value) ubtituted for, you can find the value of that enable you to reject the null hypothei: = L - m a>2 = L - m Becaue you have a one-tail tet with a level of ignificance of 0.05, the value of to -1.645 (ee Figure 9.16). The ample ize n = 25. Therefore, L = 368 + 1-1.6452 12 The deciion rule for thi one-tail tet i L = m + a>2 Reject H 0 if 6 363.065; otherwie, do not reject H 0. = 368-4.935 = 363.065 a>2 i equal FIGURE 9.16 Determining the lower critical value for a one-tail tet for a population mean at the 0.05 level of ignificance.05.95 L μ = 368 Region of Rejection L = 1.645 0 Region of Nonrejection
M09_BERE8380_12_OM_C09.QD 2/21/11 3:44 PM Page 2 2 CHAPTER 9 Fundamental of Hypothei Teting The deciion rule tate that if in a random ample of 25 boxe, the ample mean i le than 363.065 gram, you reject the null hypothei, and the repreentative conclude that the proce i not working properly. The power of the tet meaure the probability of concluding that the proce i not working properly for differing value of the true population mean. What i the power of the tet if the actual population mean i 360 gram? To determine the chance of rejecting the null hypothei when the population mean i 360 gram, you need to determine the area under the normal curve below L = 363.065 gram. Uing Equation (9.1), with the population mean m = 360, STAT = - m = 363.065-360 = 1.02 From Table E.2, there i an 84.61% chance that the value i le than +1.02. Thi i the power of the tet where m i the actual population mean (ee Figure 9.17). The probability 1b2 that you will not reject the null hypothei 1m = 3682 i 1-0.8461 = 0.39. Thu, the probability of committing a Type II error i.39%. FIGURE 9.17 Determining the power of the tet and the probability of a Type II error when m = 360 gram Power =.8461.39 μ = 360 L = 363.065 0 +1.02 Now that you have determined the power of the tet if the population mean were equal to 360, you can calculate the power for any other value of m. For example, what i the power of the tet if the population mean i 352 gram? Auming the ame tandard deviation, ample ize, and level of ignificance, the deciion rule i Reject H 0 if 6 363.065 otherwie, do not reject H 0. Once again, becaue you are teting a hypothei for a mean, from Equation (9.1), STAT = - m If the population mean hift down to 352 gram (ee Figure 9.18), then STAT = 363.065-352 = 3.69
M09_BERE8380_12_OM_C09.QD 2/21/11 3:44 PM Page 3 9.6 The Power of a Tet 3 FIGURE 9.18 Determining the power of the tet and the probability of a Type II error when m = 352 gram Power =.99989 =.00011 μ = 352 L = 363.065 0 +3.69 From Table E.2, there i a 99.989% chance that the value i le than + 3.69. Thi i the power of the tet when the population mean i 352. The probability 1b2 that you will not reject the null hypothei 1m = 3682 i 1-0.99989 = 0.00011. Thu, the probability of committing a Type II error i only 0.011%. In the preceding two example, the power of the tet i high, and the chance of committing a Type II error i low. In the next example, you compute the power of the tet when the population mean i equal to 367 gram a value that i very cloe to the hypotheized mean of 368 gram. Once again, from Equation (9.1), STAT = - m If the population mean i equal to 367 gram (ee Figure 9.19), then STAT = 363.065-367 = -1.31 FIGURE 9.19 Determining the power of the tet and the probability of a Type II error when m = 367 gram Power =.0951 L = 363.065 =.9049 μ = 367 1.31 0 From Table E.2, the probability le than = -1.31 i 0.0951 (or 9.51%). Becaue the rejection region i in the lower tail of the ditribution, the power of the tet i 9.51%, and the chance of making a Type II error i 90.49%. Figure 9.20 illutrate the power of the tet for variou poible value of m (including the three value examined). Thi graph i called a power curve.
M09_BERE8380_12_OM_C09.QD 2/21/11 3:44 PM Page 4 4 CHAPTER 9 Fundamental of Hypothei Teting FIGURE 9.20 Power curve of the cereal-box-filling proce for H 1 : m 6 368 gram Power 1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00.99961.9964.9783.9545.99989.99874.9909.9131.8461 352 353 354.7549.6406.5080.3783.2578.1635.0951.0500 355 356 357 358 359 360 361 362 363 364 365 366 367 368 Poible Value for μ (gram) 1 For ituation involving one-tail tet in which the actual mean, m 1, exceed the hypotheized mean, the convere would be true. The larger the actual mean, m 1, compared with the hypotheized mean, the greater i the power. For two-tail tet, the greater the ditance between the actual mean, m 1, and the hypotheized mean, the greater the power of the tet. From Figure 9.20, you can ee that the power of thi one-tail tet increae harply (and approache 100%) a the population mean take on value farther below the hypotheized mean of 368 gram. Clearly, for thi one-tail tet, the maller the actual mean m, the greater the power to detect thi difference. 1 For value of m cloe to 368 gram, the power i mall becaue the tet cannot effectively detect mall difference between the actual population mean and the hypotheized value of 368 gram. When the population mean approache 368 gram, the power of the tet approache a, the level of ignificance (which i 0.05 in thi example). Figure 9.21 ummarize the computation for the three cae. You can ee the dratic change in the power of the tet for different value of the actual population mean by reviewing the different panel of Figure 9.20. From Panel A and B you can ee that when the population mean doe not greatly differ from 368 gram, the chance of rejecting the null hypothei, baed on the deciion rule involved, i not large. However, when the population mean hift ubtantially below the hypotheized 368 gram, the power of the tet greatly increae, approaching it maximum value of 1 (or 100%). In the above dicuion, a one-tail tet with a = 0.05 and n = 25 wa ued. The type of tatitical tet (one-tail v. two-tail), the level of ignificance, and the ample ize all affect the power. Three baic concluion regarding the power of the tet are ummarized below: 1. A one-tail tet i more powerful than a two-tail tet. 2. An increae in the level of ignificance 1a2 reult in an increae in power. A decreae in a reult in a decreae in power. 3. An increae in the ample ize, n, reult in an increae in power. A decreae in the ample ize, n, reult in a decreae in power.
M09_BERE8380_12_OM_C09.QD 2/21/11 3:44 PM Page 5 9.6 The Power of a Tet 5 FIGURE 9.21 Determining tatitical power for varying value of the population mean Panel A Given: α =.05, σ =, n = 25 One-tail tet μ = 368 (null hypothei i true) Region of Rejection Region of Nonrejection L = 363.065 L = 368 (1.645) = 363.065 25 Deciion rule: Reject H 0 if < 363.065; otherwie, do not reject α =.050 1 α =.95 Panel B Given: α =.05, σ =, n = 25 One-tail tet H 0 : μ = 368 368 μ = 367 (true mean hift to 367 gram) μ 363.065 367 STAT = = = 1.31 σ 3 n Power =.0951 Power =.0951 =.9049 Panel C Given: α =.05, σ =, n = 25 One-tail tet H 0 : μ = 368 367 μ = 360 (true mean hift to 360 gram) μ 363.065 360 STAT = = = +1.02 σ 3 n Power =.8461 Power =.8461 =.39 Panel D Given: α =.05, σ =, n = 25 One-tail tet H 0 : μ = 368 360 μ = 352 (true mean hift to 352 gram) μ 363.065 352 STAT = = = +3.69 σ 3 n Power =.99989 Power =.99989 352 Region of Rejection =.00011 L = 363.065 Region of Nonrejection
M09_BERE8380_12_OM_C09.QD 2/21/11 3:44 PM Page 6 6 CHAPTER 9 Fundamental of Hypothei Teting Problem for Section 9.6 APPLYING THE CONCEPTS 9.80 A coin-operated oft-drink machine i deigned to dicharge at leat 7 ounce of beverage per cup, with a tandard deviation of 0.2 ounce. If you elect a random ample of 16 cup and you are willing to have an a = 0.05 rik of committing a Type I error, compute the power of the tet and the probability of a Type II error 1b2 if the population mean amount dipened i actually a. 6.9 ounce per cup. b. 6.8 ounce per cup. 9.81 Refer to Problem 9.78. If you are willing to have an a = 0.01 rik of committing a Type I error, compute the power of the tet and the probability of a Type II error 1b2 if the population mean amount dipened i actually a. 6.9 ounce per cup. b. 6.8 ounce per cup. c. Compare the reult in (a) and (b) of thi problem and in Problem 9.78. What concluion can you reach? 9.82 Refer to Problem 9.78. If you elect a random ample of 25 cup and are willing to have an a = 0.05 rik of committing a Type I error, compute the power of the tet and the probability of a Type II error 1b2 if the population mean amount dipened i actually a. 6.9 ounce per cup. b. 6.8 ounce per cup. c. Compare the reult in (a) and (b) of thi problem and in Problem 9.78. What concluion can you reach? 9.83 A tire manufacturer produce tire that have a mean life of at leat 25,000 mile when the production proce i working properly. Baed on pat experience, the tandard deviation of the tire i 3,500 mile. The operation manager top the production proce if there i evidence that the mean tire life i below 25,000 mile. If you elect a random ample of 100 tire (to be ubjected to detructive teting) and you are willing to have an a = 0.05 rik of committing a Type I error, compute the power of the tet and the probability of a Type II error 1b2 if the population mean life i actually a. 24,000 mile. 9.84 Refer to Problem 9.81. If you are willing to have an a = 0.01 rik of committing a Type I error, compute the power of the tet and the probability of a Type II error 1b2 if the population mean life i actually a. 24,000 mile. c. Compare the reult in (a) and (b) of thi problem and (a) and (b) in Problem 9.81. What concluion can you reach? 9.85 Refer to Problem 9.81. If you elect a random ample of 25 tire and are willing to have an a = 0.05 rik of committing a Type I error, compute the power of the tet and the probability of a Type II error 1b2 if the population mean life i actually a. 24,000 mile. c. Compare the reult in (a) and (b) of thi problem and (a) and (b) in Problem 9.81. What concluion can you reach? 9.86 Refer to Problem 9.81. If the operation manager top the proce when there i evidence that the mean life i different from 25,000 mile (either le than or greater than) and a random ample of 100 tire i elected, along with a level of ignificance of a = 0.05, compute the power of the tet and the probability of a Type II error 1b2 if the population mean life i actually a. 24,000 mile. c. Compare the reult in (a) and (b) of thi problem and (a) and (b) in Problem 9.81. What concluion can you reach?