chapter 1

Chapter 14: The Analysis of Categorical Data4 n1 3 n22 2 n3[ ] ⋅ [ θ ( 1−θ) ] ⋅ [ θ ( 1−θ) ] ⋅[ ] [ ] 3 + n1+( ) 3 n2+1 2 n3+ n4= θ233 ( 1 −θ) 367, so15. The part of the likelihood involving θ is ( 1−θ)n4( )4 n5n22 n3+ 3 n4+ 4 n5θ −θ⋅ θ = θ 1−θ4ln ( likelihood ) = 233ln θ + 367ln ( 1−θ)ˆ 233= = .3883,600θ and ( 1 ) = . 6117. Differentiating and equating to 0 yields−θ ˆ [note that the exponent on θ is simply the total #of successes (defectives here) in the n = 4(150) = 600 trials.] Substituting this θ′ into theformula for p yields estimated cell probabilities .1400, .3555, .3385, .1433, and .0227.iMultiplication by 150 yields the estimated expected cell counts are 21.00, 53.33, 50.78, 21.50,and 3.41. the last estimated expected cell count is less than 5, so we combine the last twocategories into a single one ( ≥ 3 defectives), yielding estimated counts 21.00, 53.33, 50.78,24.91, observed counts 26, 51, 47, 26, and χ2 =1. 62 . With d.f. = 4 – 1 – 1 = 2, since21.62 < χ 4.605 , the p-value > .10, and we do not reject H o . The data suggests that. 10,2=the stated binomial distribution is plausible.16.( 0)( 6) + ( 1)( 24) + ( 2)( 42) + ... + ( 8)( 6) + ( 9)( 2)1163λ ˆ = x == = 3.88 , so the300300x−3.88 ( 3.88)estimated cell probabilities are computed from pˆ = e.x!x 0 1 2 3 4 5 6 7 ≥ 8np(x) 6.2 24.0 46.6 60.3 58.5 45.4 29.4 16.3 13.3obs 6 24 42 59 62 44 41 14 8This gives χ2 = 7. 789 . To see whether the Poisson model provides a good fit, we need22χ = χ 12.017 . Since 7 .789 < 12. 017 , the Poisson model does provide a. 10,9−1−1.10,7=good fit.ˆ 380=12017. λ = 3. 167 , so−3.167ˆ( 3.167)p = e.x!xx 0 1 2 3 4 5 6 ≥ 7pˆ .0421 .1334 .2113 .2230 .1766 .1119 .0590 .0427nˆ p 5.05 16.00 25.36 26.76 21.19 13.43 7.08 5.12obs 24 16 16 18 15 9 6 16The resulting value of 2 2χ = 103. 98 , and when compared to χ 18. 474that the Poisson model fits very poorly.445. 01,7=, it is obvious