2 Null hypothesis H 2 001 3Alternative hypothesis H1 2 0019 4 Hypothesis Tests

2 null hypothesis h 2 001 3alternative hypothesis h1

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2. Null hypothesis: H 0 : 2 = 0.01 3. Alternative hypothesis: H 1 : 2 > 0.01
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9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Distribution Example 9-8 4. Test statistic: The test statistic is 5. Reject H 0 : Use = 0.05 , and reject H 0 if . 6. Computations: 7. Conclusions: Since , we conclude that there is no strong evidence that the variance of fill volume exceeds 0.01 (fluid ounces) 2 . So there is no strong evidence of a problem with incorrectly filled bottles. 2 0 2 1 s n 14 . 30 2  07 . 29 01 . 0 ) 0153 . 0 ( 19 14 . 30 07 . 29 2 
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9-5 Tests on a Population Proportion 9-5.1 Large-Sample Tests on a Proportion Many engineering decision problems include hypothesis testing about p . An appropriate test statistic is (9-10) 0 0 1 0 : : H p p H p p ) 1 ( 0 0 0 0 p np np X Z
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9-5 Tests on a Population Proportion Another form of the test statistic Z 0 is or n p p p n X Z / ) 1 ( / 0 0 0 0 n p p p P Z / ) 1 ( ˆ 0 0 0 0
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9-5 Tests on a Population Proportion EXAMPLE 9-10 Automobile Engine Controller 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 finds that four of them are defective. Can the manufacturer demonstrate process capability for the customer? We may solve this problem using the seven-step hypothesis-testing procedure as follows: 1. Parameter of Interest: The parameter of interest is the process fraction defective p. 2. Null hypothesis: H 0 : p = 0.05 3. Alternative hypothesis: H 1 : p < 0.05 This formulation of the problem will allow the manufacturer to make a strong claim about process capability if the null hypothesis H 0 : p = 0.05 is rejected.
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9-5 Tests on a Population Proportion Example 9-10 4. The test statistic is (from Equation 9-10) where x = 4, n = 200, and p 0 = 0.05. 5. Reject H 0 if: Reject H 0 : p = 0.05 if the p-value is less than 0.05. 6. Computations: The test statistic is 7. Conclusions: Since z 0 = 1.95, the P- value is ( 1.95) = 0.0256, so we reject H 0 and conclude that the process fraction defective p is less than 0.05. Practical Interpretation: We conclude that the process is capable. ) 1 ( 0 0 0 0 p np np x z 95 . 1 ) 95 . 0 )( 05 . 0 ( 200 ) 05 . 0 ( 200 4 0 z
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9-5 Tests on a Population Proportion 9-5.2 Type II Error and Choice of Sample Size For a two-sided alternative (9-11) If the alternative is p < p 0 (9-12) If the alternative is p > p 0 (9-13) n p p n p p z p p n p p n p p z p p / ) 1 ( / ) 1 ( / ) 1 ( / ) 1 ( 0 0 /2 0 0 0 /2 0 n p p n p p z p p / ) 1 ( / ) 1 ( 1 0 0 0 n p p n p p z p p / ) 1 ( / ) 1 ( 0 0 0
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9-5 Tests on a Population Proportion 9-5.3 Type II Error and Choice of Sample Size For a two-sided alternative (9-14) For a one-sided alternative (9-15) 2 0 0 0 /2 ) 1 ( ) 1 ( p p p p z p p z n 2 0 0 0 ) 1 ( ) 1 ( p p p p z p p z n
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9-5 Tests on a Population Proportion Example 9-11 Automobile Engine Controller Type II Error Consider the semiconductor manufacturer from Example 9-10. Suppose that its process fallout is really p = 0.03. What is the -error for a test of process capability that uses n = 200 and a = 0.05?
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