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Chap09_part1 - 368 Chapter 9 Two-Sample Tests with...

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112 Chapter 9: Two-Sample Tests with Numerical Data CHAPTER 9 9.1 1 2 1 2 2 2 2 2 1 2 1 2 ( ) ( ) (72 66) 0 20 10 40 50 X X Z n n μ μ σ σ - - - - - = = + + = 1.73 9.2 H 0 : μ 1 = μ 2 H 1 : μ 1 μ 2 Decision rule: If Z < – 2.58 or Z > 2.58, reject H 0 . Test statistic: Z = ( X 1 X 2 ) – ( μ 1 μ 2 ) σ 1 2 n 1 + σ 2 2 n 2 = (72 – 66) – 0 20 2 40 + 10 2 50 = 1.73 Decision: Since Z calc = 1.73 is between the critical bounds of Z = ± 2.58, do not reject H 0 . There is inadequate evidence to conclude the two population means are different. 9.3 p value = 2(1.0 – 0.9582) = 0.0836 9.4 (a) 2 2 2 2 2 1 1 2 2 1 2 ( 1) ( 1) (7) 4 (14) 5 ( 1) ( 1) 7 14 p n S n S S n n - + - + = = - + - + = 22 1 2 1 2 2 1 2 ( ) ( ) (42 34) 0 3.8959 1 1 1 1 22 8 15 p X X t S n n μ μ - - - - - = = = + + (b) df = ( n 1 – 1) + ( n 2 – 1) = 7 + 14 = 21 (c) Decision rule: df = 21. If t > 2.5177, reject H 0 . (d) Decision: Since t calc = 3.8959 is above the critical bound of t = 2.5177, reject H 0 . There is enough evidence to conclude that the first population mean is larger than the second population mean. (e) We are sampling from two independent normal distributions having equal variances. (f) ( 29 ( 29 2 1 2 1 2 1 1 1 1 42 34 2.0796 22 8 15 p X X t S n n - + = - q + 1 2 3.7296< 12.2704 μ μ - < 112
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113 Chapter 9: Two-Sample Tests with Numerical Data 9.5 (a) H 0 : μ 1 = μ 2 Light bulbs produced by machine 1 have the same average life expectancy as light bulbs produced by machine 2. H 1 : μ 1 μ 2 Light bulbs produced by machine 1 have a different average life expectancy as light bulbs produced by machine 2. Decision rule: If Z < – 1.96 or Z > 1.96, reject H 0 . Test statistic: Z = ( X 1 X 2 ) – ( μ 1 μ 2 ) σ 1 2 n 1 + σ 2 2 n 2 = (375 – 362) – 0 110 2 25 + 125 2 25 = 0.39 Decision: Since Z calc = 0.39 is between the critical bounds of ± 1.96, do not reject H 0 . There is not enough evidence to conclude that light bulbs produced by machine 1 have a different average life expectancy from light bulbs produced by machine 2. (b) p value = 2(1.0 – 0.6517) = 0.6966 The probability of obtaining samples whose means differ by 13 hours or more when the null hypothesis is true is 0.6966. 9.6 (a) H 0 : μ 1 μ 2 where Populations: 1 = new machine, 2 = old machine The average breaking strength of parts produced by the new machine is not greater than the average breaking strength of parts produced by the old machine. H 1 : μ 1 μ 2 The average breaking strength of parts produced by the new machine is greater than the average breaking strength of parts produced by the old machine. Decision rule: If Z > 2.33, reject H 0 . Test statistic: Z = ( X 1 X 2 ) – ( μ 1 μ 2 ) σ 1 2 n 1 + σ 2 2 n 2 = (72 – 65) – 0 9 2 100 + 10 2 100 = 5.20 Decision: Since Z calc = 5.20 is above the critical bound of 2.33, reject H 0 . There is enough evidence to conclude that the average breaking strength of parts produced by the new machine is greater than the average breaking strength of parts produced by the old machine. (b) p value = virtually zero. The probability of obtaining samples whose means differ by 7 or more units of strength when the null hypothesis is true is virtually zero.
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