Statistics Homework Solutions 157

Statistics Homework Solutions 157 - 157 SECTION 6.12...

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SECTION 6.12 157 Section 6.11 1. ν 1 7, ν 2 20. From the F table, the upper 5% point is 2.51. 3. (a) The upper 1% point of the F 5 7 distribution is 7.46. Therefore the P -value is 0.01. (b) The P -value for a two-tailed test is twice the value for the corresponding one-tailed test. Therefore P 0 02. 5. The sample variance of the breaking strengths for composite A is σ 2 1 202 7175. The sample size is n 1 9. The sample variance of the breaking strengths for composite B is σ 2 2 194 3829. The sample size is n 2 14. The null and alternate hypotheses are H 0 : σ 2 1 σ 2 2 1 versus H 1 : σ 2 1 σ 2 2 1. The test statistic is F σ 2 1 σ 2 2 1 0429. The numbers of degrees of freedom are 8 and 13. Since this is a two-tailed test, the P -value is twice the area to the right of 1.0429 under the F 8 13 probability density function. From the F table, P 0 2. A computer package gives P 0 91. We cannot conclude that the variance of the breaking strength varies between the composites. Section 6.12
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