Statistics Homework Solutions 144

# Statistics Homework Solutions 144 - 144 7 Y 3 05 sY 1 5761...

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144 CHAPTER 6 3. X 1 93, s X 0 84562, n X 7, Y 3 05, s Y 1 5761, n Y 12. The number of degrees of freedom is ν 0 84562 2 7 1 5761 2 12 2 0 84562 2 7 2 7 1 1 5761 2 12 2 12 1 16, rounded down to the nearest integer. t 16 1 93 3 05 0 0 84562 2 7 1 5761 2 12 2 014. The null and alternate hypotheses are H 0 : μ X μ Y 0 versus H 1 : μ X μ Y 0. Since the alternate hypothesis is of the form μ X μ Y , the P -value is the sum of the areas to the right of t 2 014 and to the left of t 2 014. From the t table, 0 05 P 0 10. A computer package gives P 0 061. The null hypothesis is suspect. 5. X 2 1062, s X 0 029065, n X 5, Y 2 0995, s Y 0 033055, n Y 5. The number of degrees of freedom is ν 0 029065 2 5 0 033055 2 5 2 0 029065 2 5 2 5 1 0 033055 2 5 2 5 1 7, rounded down to the nearest integer. t 7 2 1062 2 0995 0 0 029065 2 5 0 033055 2 5 0 3444. The null and alternate hypotheses are H 0 : μ X μ Y 0 versus H 1 : μ X μ Y 0. Since the alternate hypothesis is of the form μ X μ Y , the P -value is the sum of the areas to the right of t 0 3444 and to the left of t 0 3444. From the
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