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An approximate test based on the normal approximation to the binomial will be given. As noted above, this approximate procedure will be valid as long as p is not extremely close to zero or one, and if the sample size is relatively large. Let X be the number of observations in a random sample of size n that belongs to the class associated with p. Then, if the null hypothesis H0: p p0 is true, we have X N[np0, np0(1 p0)], approximately. To test H0: p p0, calculate the test statistic X np0 1np0 11 p0 2
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Note that the standard normal distribution is the reference distribution for this test statistic. Critical regions for the one-sided alternative hypotheses would be constructed in the usual manner. EXAMPLE 9-10 A semiconductor manufacturer produces controllers used in automobile engine applications. The customer requires that the process fallout or fraction defective at a critical manufacturing step not exceed 0.05 and that the manufacturer demonstrate process capability at this level of quality using 0.05. The semiconductor manufacturer takes a random sample of 200 devices and nds that four of them are defective. Can the manufacturer demonstrate process capability for the customer We may solve this problem using the eight-step hypothesis-testing procedure as follows: 1. The parameter of interest is the process fraction defective p. 2. H0: p 0.05 3. H1: p 0.05 This formulation of the problem will allow the manufacturer to make a strong claim about process capability if the null hypothesis H0: p 0.05 is rejected. 4. 0.05 5. The test statistic is (from Equation 9-32) z0 x np0 1np0 11 p0 2
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where x 4, n 200, and p0 0.05. 6. Reject H0: p 0.05 if z0 z0.05 1.645 7. Computations: The test statistic is z0 8. 4 20010.052 120010.05210.952 1.95
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1.95 z0.05 1.645, we reject H0 and conclude that the Conclusions: Since z0 process fraction defective p is less than 0.05. The P-value for this value of the test statistic z0 is P 0.0256, which is less than 0.05. We conclude that the process is capable.
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Another form of the test statistic Z0 in Equation 9-32 is occasionally encountered. Note that if X is the number of observations in a random sample of size n that belongs to a class of interest, then P X n is the sample proportion that belongs to that class. Now divide both numerator and denominator of Z0 in Equation 9-32 by n, giving Z0 or Z0 X n 1p0 11
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This presents the test statistic in terms of the sample proportion instead of the number of items X in the sample that belongs to the class of interest. Statistical software packages usually provide the one sample Z-test for a proportion. The Minitab output for Example 9-10 follows.
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Test and CI for One Proportion Test of p 0.05 vs p 0.05 Sample X N Sample p 95.0% Upper Bound Z-Value P-Value 1 4 200 0.020000 0.036283 1.95 0.026 * NOTE * The normal approximation may be inaccurate for small samples.
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Notice that both the test statistic (and accompanying P-value) and the 95% one-sided upper con dence bound are displayed. The 95% upper con dence bound is 0.036283, which is less than 0.05. This is consistent with rejection of the null hypothesis Ho: p 0.05.
9-5.2 Small-Sample Tests on a Proportion (CD Only) 9-5.3 Type II Error and Choice of Sample Size
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It is possible to obtain closed-form equations for the approximate -error for the tests in Section 9-5.1. Suppose that p is the true value of the population proportion. The approximate -error for the two-sided alternative H1 : p p0 is a p0 p 1p11 z p0 2 n b a p0 p 1p11 z p0 2 n b (9-34)
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