Solution for Cpt 9 Problems

Solution for Cpt 9 Problems - Solutions for Cpt 9 Problems...

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Solutions for Cpt 9 Problems 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 σ - - - - - = = + + = 1.73 9.2 H 0 : 1 = 2 H 1 : 1 2 Decision rule: If Z < – 2.58 or Z > 2.58, reject H 0 . Test statistic: = 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 Z = ( X 1 X 2 ) – ( 1 2 ) 1 2 n 1 + 2 2 n 2 = (72 – 66) – 0 20 2 40 + 10 2 50 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 - - - - - = = = + + (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 - + = - a + 1 2 3.7296< 12.2704 - <
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9.10 (a) 0 : I II H μ = Mean waiting time of Bank 1 and Bank 2 is the same. 1 : I II H Mean waiting time of Bank 1 and Bank 2 is different. PHStat output: t Test for Differences in Two Means Data Hypothesized Difference 0 Level of Significance 0.05 Population 1 Sample Sample Size 15 Sample Mean 4.286667 Sample Standard Deviation 1.637985 Population 2 Sample Sample Size 15 Sample Mean 7.114667 Sample Standard Deviation 2.082189 Intermediate Calculations Population 1 Sample Degrees of Freedom 14 Population 2 Sample Degrees of Freedom 14 Total Degrees of Freedom 28 Pooled Variance 3.509254
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Solution for Cpt 9 Problems - Solutions for Cpt 9 Problems...

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