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(9-20)
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EXAMPLE 9-3
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Consider the rocket propellant problem of Example 9-2. Suppose that the analyst wishes to design the test so that if the true mean burning rate differs from 50 centimeters per second by as much as 1 centimeter per second, the test will detect this (i.e., reject H0: 50) with a high probability, say 0.90. Now, we note that 2, 51 50 1, 0.05, and 0.10. z0.10 1.28, the sample size required to detect this Since z 2 z0.025 1.96 and z departure from H0: 50 is found by Equation 9-19 as 1z z 22
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The approximation is good here, since 1 z 1 5.202 0, which is small relative to .
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Using Operating Characteristic Curves When performing sample size or type II error calculations, it is sometimes more convenient to use the operating characteristic curves in Appendix Charts VIa and VIb. These curves plot as calculated from Equation 9-17 against a parameter d for various sample sizes n. Curves are provided for both 0.05 and 0.01. The parameter d is de ned as 0 0 0 0
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(9-21)
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CHAPTER 9 TESTS OF HYPOTHESES FOR A SINGLE SAMPLE
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so one set of operating characteristic curves can be used for all problems regardless of the values of 0 and . From examining the operating characteristic curves or Equation 9-17 and Fig. 9-7, we note that 1. The further the true value of the mean is from 0, the smaller the probability of type II error for a given n and . That is, we see that for a speci ed sample size and , large differences in the mean are easier to detect than small ones. 2. For a given and , the probability of type II error decreases as n increases. That is, to detect a speci ed difference in the mean, we may make the test more powerful by increasing the sample size.
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EXAMPLE 9-4
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Consider the propellant problem in Example 9-2. Suppose that the analyst is concerned about the probability of type II error if the true mean burning rate is 51 centimeters per second. We may use the operating characteristic curves to nd . Note that 51 50 1, n 25, 2, and 0.05. Then using Equation 9-21 gives 0
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and from Appendix Chart VIa, with n 25, we nd that 0.30. That is, if the true mean burning rate is 51 centimeters per second, there is approximately a 30% chance that this will not be detected by the test with n 25. EXAMPLE 9-5 Once again, consider the propellant problem in Example 9-2. Suppose that the analyst would like to design the test so that if the true mean burning rate differs from 50 centimeters per second by as much as 1 centimeter per second, the test will detect this (i.e., reject H0: 50) with a high probability, say, 0.90. This is exactly the same requirement as in Example 9-3, where we used Equation 9-19 to nd the required sample size to be n 42. The operating characteristic curves can also be used to nd the sample size for this test. Since 1 2, 0.05, and d 0 0.10, we nd from Appendix Chart VIa that the 00 required sample size is approximately n 40. This closely agrees with the sample size calculated from Equation 9-19. In general, the operating characteristic curves involve three parameters: , d, and n. Given any two of these parameters, the value of the third can be determined. There are two typical applications of these curves: 1. For a given n and d, nd (as illustrated in Example 9-3). This kind of problem is often encountered when the analyst is concerned about the sensitivity of an experiment already performed, or when sample size is restricted by economic or other factors. For a given and d, nd n. This was illustrated in Example 9-4. This kind of problem is usually encountered when the analyst has the opportunity to select the sample size at the outset of the experiment.
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Operating characteristic curves are given in Appendix Charts VIc and VId for the onesided alternatives. If the alternative hypothesis is either H1: 0 or H1: 0, the abscissa scale on these charts is d 0
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(9-22)
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