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Statistics Homework Solutions 133

Statistics Homework Solutions 133 - 133 SECTION 6.1 Chapter...

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SECTION 6.1 133 Chapter 6 Section 6.1 1. (a) X 783, s 120, n 73. The null and alternate hypotheses are H 0 : μ 750 versus H 1 : μ 750. z 783 750 120 73 2 35. Since the alternate hypothesis is of the form μ μ 0 , the P -value is the area to the right of z 2 35. Thus P 0 0094. (b) The P -value is 0.0094, so if H 0 is true then the sample is in the most extreme 0.94% of its distribution. 3. (a) X 1 90, s 21 20, n 160. The null and alternate hypotheses are H 0 : μ 0 versus H 1 : μ 0. z 1 90 0 21 20 160 1 13. Since the alternate hypothesis is of the form μ μ 0 , the P -value is the sum of the areas to the right of z 1 13 and to the left of z 1 13. Thus P 0 1292 0 1292 0 2584. (b) The P -value is 0.2584, so if H 0 is true then the sample is in the most extreme 25.84% of its distribution. 5. (a) X 51 2, s 4 0, n 110. The null and alternate hypotheses are H 0 : μ 50 versus H 1 : μ 50. z 51 2 50 4 0 110 3 15. Since the alternate hypothesis is of the form μ μ 0 , the P -value is area to the right of z 3 15. Thus P 0 0008. (b) If the mean tensile strength were 50 psi, the probability of observing a sample mean as large as the value of 51.2
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