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# Lec24 - IOE/Stat IOE/Stat 265 Fall 2009 Lecture#24...

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IOE/Stat 265, Fall 2009 Lecture #24: CH 14: Goodness of Fit and Contingency Analysis (Independence) 1 C t i l D t A l i Categorical Data Analysis 14.1 Goodness-of-Fit 2 G d f i (C i h ) 14.2 Goodness-of-Fit (Composite Hypotheses) 14.3 Two-Way Contingency Tables 2 Testing Goodness of Fit How do we decide whether our distribution assumptions are correct? Chi-Square Distribution useful to compare Chi Square Distribution useful to compare "observed frequencies" (or observed probabilities) with "expected frequencies" d di t ib ti d under any distribution assumed. 3 Example 1 : Printed Circuit B d Boards Defects assumed to follow a Poisson Distribution Number of Defects Observed Frequency Poisson Distribution. A random sample of 60 boards. 0 32 1 15 Is the Poisson assumption reasonable? 2 9 3 4 4

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Solution What would you expect if the underlying assumption (Poisson) was true? Sample Mean (0 32 1 15 2 9 3 4)/ 75 60 X Under a Poisson assumption then we would estimate .75 = + + + = 75 . 0 = λ ± 5 We would expect a distribution like this: ( ) 0 0.75 0.75 ( 0) 0.472 e P x = = = ( ) 1 0.75 0! 0 75 e 0.75 ( 1) 0.354 1! P x = = = ( ) 2 0.75 0.75 ( 2) 0.133 2! e P x = = = 6 etc. If this were true we would expect E 0 = ( 60 )(0.472) = 28.32 E = ( 60 )(0 354) = 21 24 E 1 = ( )(0.354) = 21.24 E 2 = ( 60 )(0.133) = 7.98 7 No. Defects Observed Frequency Expected Frequency 0 32 28.32 1 15 21.24 2 9 7.98 3 4 2 46 8 2.46
As long as we don't have too many cells with < 5 expected observations we can assume a normal distribution for the number of observations in each cell (0 1) i i O E N ~ (0,1) i E ( ) 2 2 (1) ~ i i O E E χ i ( ) 2 2 ~ k i i O E χ ( ) 1 k i i E = Two Minor Problems Two Minor Problems … 1) We had to estimate "p" parameters to get the expected frequencies .... lose one degree of expected frequencies lose one degree of freedom for each estimate 2) We know that the probabilities have to add up to 1 .... lose one degree of freedom for this constraint ( ) 2 2 k i i O E 10 ( 1) 1 ~ k p i i E χ = For the Example Problem: ( ) 2 k O E 2 (0.05, 1 2) 1 4.46 5.99 i i k p i i E χ = = = < = MINITAB can be used … but it’s not automatic 11 Example 2 - Machine Breakdowns Historically breakdowns follow the pattern shown Cause Frequency Electrical 0.2 Mechanical 0.5 O h 0 3 12

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Lec24 - IOE/Stat IOE/Stat 265 Fall 2009 Lecture#24...

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