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# sol7 - Homework 7 due Mar 21 8-61(9-62 8-68(9-69 9-4(10-4...

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Homework 7 due Mar. 21 8-61 (9-62) 8-68 (9-69) 9-4 (10-4) 9-20 (10-22) 8-61. Mean = np = 6(0.25) = 1.5 Value 0 1 2 3 4 Observed 4 21 10 13 2 Expected 8.8989 17.7979 14.8315 6.5918 1.6479 The expected frequency for value 4 is less than 3. Combine this cell with value 3: Value 0 1 2 3-4 Observed 4 21 10 15 Expected 8.8989 17.7979 14.8315 8.2397 The degrees of freedom are k - p - 1 = 4 - 0 - 1 = 3 a) 1) The variable of interest is the form of the distribution for the random variable X. 2) H 0 : The form of the distribution is binomial with n = 6 and p = 0.25 3) H 1 : The form of the distribution is not binomial with n = 6 and p = 0.25 4) α = 0.05 5) The test statistic is ( 29 χ 0 2 2 1 = - = O E E i i i i k 6) Reject H 0 if χ χ o 2 0 05 3 2 7 81 = . , . 7) ( 29 ( 29 χ 0 2 2 2 4 88989 88989 15 8 2397 8 2397 = - + + - = . . . . 10.39 8) Since 10.39 > 7.81 reject H 0 . We can conclude that the distribution is not binomial with n = 6 and p = 0.25 at α = 0.05. b) P-value = 0.0155 (found using Minitab) 8-68. 1. The variable of interest is failures of an electronic component. 2. H 0 : Type of failure is independent of mounting position.

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sol7 - Homework 7 due Mar 21 8-61(9-62 8-68(9-69 9-4(10-4...

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