The error can be computed using Equation 9 12 as follows Thus the probability

The error can be computed using equation 9 12 as

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The -error can be computed using Equation 9-12 as follows: Thus, the probability is about 0.7 that the semiconductor manufacturer will fail to conclude that the process is capable if the true process fraction defective is p = 0.03 (3%). That is, the power of the test against this particular alternative is only about 0.3. This appears to be a large -error (or small power), but the difference between p = 0.05 and p = 0.03 is fairly small, and the sample size n = 200 is not particularly large. Calculations are on the board! (an alternative way) 67 . 0 44 . 0 1 200 / 03 . 0 1 03 . 0 200 / 95 . 0 05 . 0 ) 645 . 1 ( 03 . 0 05 . 0 1
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9-5 Tests on a Population Proportion Example 9-11 Suppose that the semiconductor manufacturer was willing to accept a -error as large as 0.10 if the true value of the process fraction defective was p = 0.03. If the manufacturer continues to use a = 0.05, what sample size would be required? The required sample size can be computed from Equation 9-15 as follows: where we have used p = 0.03 in Equation 9-15. Conclusion: Note that n = 832 is a very large sample size. However, we are trying to detect a fairly small deviation from the null value p 0 = 0.05. 832 05 . 0 03 . 0 97 . 0 03 . 0 28 . 1 95 . 0 05 . 0 645 . 1 ~ 2 n
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9-7 Testing for Goodness of Fit The test is based on the chi-square distribution. Assume there is a sample of size n from a population whose probability distribution is unknown. Let O i be the observed frequency in the i th class interval. Let E i be the expected frequency in the i th class interval. The test statistic is (9-16) k i i i i E E O X 1 2 2 0 ) (
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9-7 Testing for Goodness of Fit EXAMPLE 9-12 Printed Circuit Board Defects Poisson Distribution The number of defects in printed circuit boards is hypothesized to follow a Poisson distribution. A random sample of n = 60 printed boards has been collected, and the following number of defects observed. Number of Defects Observed Frequency 0 32 1 15 2 9 3 4
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9-7 Testing for Goodness of Fit Example 9-12 The mean of the assumed Poisson distribution in this example is unknown and must be estimated from the sample data. The estimate of the mean number of defects per board is the sample average, that is, (32·0 + 15·1 + 9·2 + 4·3)/60 = 0.75. From the Poisson distribution with parameter 0.75, we may compute p i , the theoretical, hypothesized probability associated with the i th class interval. Since each class interval corresponds to a particular number of defects, we may find the p i as follows: 041 . 0 1 ) 3 ( 133 . 0 ! 2 75 . 0 ) 2 ( 354 . 0 ! 1 75 . 0 ) 1 ( 472 . 0 ! 0 75 . 0 ) 0 ( 3 2 1 4 2 75 . 0 3 1 75 . 0 2 0 75 . 0 1 p p p X P p e X P p e X P p e X P p
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9-7 Testing for Goodness of Fit Example 9-12 The expected frequencies are computed by multiplying the sample size n = 60 times the probabilities p i . That is, E i = np i . The expected frequencies follow: Number of Defects Probability Expected Frequency 0 0.472 28.32 1 0.354 21.24 2 0.133 7.98 3 (or more) 0.041 2.46
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9-7 Testing for Goodness of Fit Example 9-12 Since the expected frequency in the last cell is less than 3, we combine the last two cells: The chi-square test statistic in Equation 9-16 will have k p 1 = 3 1 1 = 1 degree of freedom, because the mean of the Poisson distribution was estimated from the data.
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