Statistics Homework Solutions 138

# Statistics Homework Solutions 138 - 138 CHAPTER 6 1 60...

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138 CHAPTER 6 (c) Yes, the P -value is small, so we conclude that μ 35. 5. (a) X 61 3, s 5 2, n 12. There are 12 1 11 degrees of freedom. The null and alternate hypotheses are H 0 : μ 60 versus H 1 : μ 60. t 61 3 60 5 2 12 0 866. Since the alternate hypothesis is of the form μ μ 0 , the P -value is the area to the right of t 0 866. From the t table, 0 10 P 0 25. A computer package gives P 0 202. We cannot conclude that the mean concentration is greater than 60 mg/L. (b) X 61 3, s 5 2, n 12. There are 12 1 11 degrees of freedom. The null and alternate hypotheses are H 0 : μ 65 versus H 1 : μ 65. t 61 3 65 5 2 12 2 465. Since the alternate hypothesis is of the form μ μ 0 , the P -value is the area to the left of t 2 465. From the t table, 0 01 P 0 025. A computer package gives P 0 0157. We conclude that the mean concentration is less than 65 mg/L. 7. (a) 3.8 4 4.2 (b) Yes, the sample contains no outliers. X 4 032857, s 0 061244, n 7. There are 7 1 6 degrees of freedom.
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