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

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IOE/Stat 265, Fall 2009 cture #24: Lecture #24: CH 14: Goodness of Fit and ontingency Analysis (Independence) Contingency Analysis (Independence) 1 Categorical Data Analysis ± 14.1 Goodness-of-Fit ± 14.2 Goodness-of-Fit (Composite Hypotheses) ± 14.3 Two-Way Contingency Tables 2 esting Goodness of Fit Testing Goodness of Fit ± How do we decide whether our distribution ssumptions are correct? assumptions are correct? hi- quare Distribution useful to compare ± Chi Square Distribution useful to compare "observed frequencies" (or observed probabilities) with "expected frequencies" under any distribution assumed. 3 Example 1 : Printed Circuit Boards ± Defects assumed to follow a oisson Distribution Number of Defects Observed Frequency Poisson Distribution. ± A random sample of 60 boards. 03 2 ± Is the Poisson assumption reasonable? 1 15 2 9 34 4
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olution Solution ± What would you expect if the underlying assumption oisson) was true? (Poisson) was true? ± Sample Mean ± Under a Poisson assumption then we would estimate (0 32 1 15 2 9 3 4)/ .75 60 X =⋅ + +⋅+⋅ = 75 . 0 = λ ± 5 We would expect a distribution e this: like this: () 0 0.75 0.75 0 ) 0 . 4 7 2 e == = 1 0.75 (0 0! 75 Px ( ) 0.75 ( 1) 0.354 1! = 2 0.75 0.75 ( 2) 0.133 ! = 6 2! etc. If this were true we would expect E = ( 60 )(0.472) = 28.32 0 ( 0 0 354) = 21 24 E 1 = ( 60 )(0.354) = 21.24 E 2 = ( 60 )(0.133) = 7.98 7 No. Defects Observed Frequency Expected Frequency qy 0 32 28.32 1 15 21.24 29 7 . 9 8 46 8 3 4 2.46
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± 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) ii i OE N E ( ) 2 2 (1) ~ χ ( ) 2 2 k () 1 ~ = wo Minor Problems Two Minor Problems … 1) We had to estimate "p" parameters to get the xpected 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 ) pp to 1 . ... lose one degree of freedom for this constraint 2 10 ( ) 2 (1 ) 1 ~ kp −− = or the Example Problem: For the Example Problem: 2 ( ) 2 (0.05, 1 2) 1 4.46 5.99 −−= = =< = MINITAB can be used … but it’s not automatic 11 Example 2 - Machine Breakdowns ±
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Lec24 - IOE/Stat IOE/Stat 265, Fall 2009 Lecture #24:...

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