hw06-sol - Homework 6 Solutions 9-101 Estimated mean =...

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Homework 6 Solutions 9-101 Estimated mean = 49.6741 use Poisson distribution with =49.674 All expected frequencies are greater than 3. The degrees of freedom are k p 1 = 26 1 1 = 24 a) 1) The variable of interest is the form of the distribution for the number of cars passing through the intersection. 2) H 0 : The form of the distribution is Poisson 3) H 1 : The form of the distribution is not Poisson 4) = 0.05 5) The test statistic is 0 2 2 1 OE E ii i i k 6) Reject H 0 if  o 2 00524 2 36 42  ., . 7) Estimated mean = 49.6741 0 2 769 57 . 8) Since 769.57 >>> 36.42, reject H 0 . We can conclude that the distribution is not Poisson at = 0.05. b) P-value = 0 (found using Minitab) 9-107 1. The variable of interest is failures of an electronic component. 2. H 0 : Type of failure is independent of mounting position. 3. H 1 : Type of failure is not independent of mounting position. 4. = 0.01 5. The test statistic is:     r i c j ij ij ij E E O 11 2 2 0 6. The critical value is 344 . 11 2 3 , 01 . 0 7. The calculated test statistic is 71 . 10 2 0 8. 2 3 , 01 . 0 2 0 , do not reject H 0 and conclude that the evidence is not sufficient to claim that the type of failure is not independent of the mounting position at = 0.01. P -value = 0.013 10-6 a) 1) The parameter of interest is the difference in mean burning rate, 12 2) H 0 :  0  or 3) H 1 : 0  or 4) = 0.05 5) The test statistic is z xx nn 0 12 0 1 2 1 2 2 2  ()  6) Reject H 0 if z 0 < z /2 = 1.96 or z 0 > z /2 = 1.96 7) x 1 18 x 2 24 1 3 2 3 n 1 = 20 n 2 = 20 32 . 6 20 ) 3 ( 20 ) 3 ( ) 24 18 ( 2 2 0 z 8) Because 6.32 < 1.96 reject the null hypothesis and conclude the mean burning rates differ significantly at = 0.05.
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P -value = 0 ) 1 1 ( 2 )) 32 . 6 ( 1 ( 2 b) xx z nn 12 2 1 2 1 2 2 2 12 12 2 1 2 1 2 2 2
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This note was uploaded on 11/16/2010 for the course ISYE 2028 taught by Professor Shim during the Fall '07 term at Georgia Tech.

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hw06-sol - Homework 6 Solutions 9-101 Estimated mean =...

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