Statistics Homework Solutions 143

Statistics Homework Solutions 143 - 143 SECTION 6.7 0...

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SECTION 6.7 143 15. (a) 101 153 0 660131. (b) 90 0 544444 49. (c) X 1 101, n 1 153, p 1 101 153 0 660131, X 2 49, n 2 90, p 2 49 90 0 544444, p 101 49 153 90 0 617284. z 0 660131 0 544444 0 617284 1 0 617284 1 153 1 90 1 79. (d) Since the alternate hypothesis is of the form p X p Y 0, the P -value is the sum of the areas to the right of z 1 79 and to the left of z 1 79. Thus P 0 0367 0 0367 0 0734. Section 6.7 1. (a) X 3 05, s X 0 34157, n X 4, Y 1 8, s Y 0 90921, n Y 4. The number of degrees of freedom is ν 0 34157 2 4 0 90921 2 4 2 0 34157 2 4 2 4 1 0 90921 2 4 2 4 1 3, rounded down to the nearest integer. t 3 3 05 1 8 0 0 34157 2 4 0 90921 2 4 2 574. 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 area to the right of t 2 574. From the t table, 0 025 P 0 050. A computer package gives P 0 041. We can conclude that the mean strength of crayons made with dye B is greater than that made with dye A.
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