45 We use the following artificial data to illustrate comments in Section 4.5.3
ID: 3322213 • Letter: 4
Question
45 We use the following artificial data to illustrate comments in Section 4.5.3 about grouped versus ungrouped binary data: x Number of trials Number of successes 4 2 4 4 Denote by Mo the model logit[P(Y = 1)- and by Mi the model logit[P(Y- l)] = + fix. Denote the maximized log-likelihood values by Lo for Mo, Li for Mi, and L for the saturated model. Create a data file in two ways, entering the data as (i) ungrouped data: ni 1,1 = I, , 12, and (ii) grouped data: n, 4, i= 1,2,3. a. Fit Mo and Mi for each data file. Report Lo and L1 in each case. Note they are the same for each form of data entry. b. Show that the deviances for Mo and M1 differ for the two forms of data entry Why is this? [Hint: How many parameters are in the saturated model for each data file?] c. Show that the difference between the deviances for Mo and Mi is the same for each form of data entry. Why is this? (Thus, for testing the effect of s, it does not matter how you enter the data, but it does matter if you want to test goodness of fit.)Explanation / Answer
Answer for A:
Un groupped Data:
Responde Profile:
M0 :Intercept Only Model
Equation:
Log(P/1-P) = -1.5025
-2L0= 16.301
M1: X as the linear Predictor
Equation :
Log(P/1-P) = -1.5027+2.0605X
-2L1= 11.028
Groupped Data:
X Trials Success
0 4 1
1 4 2
2 4 4
M0 :Intercept Only Model
Equation:
Log(P/1-P) = -1.5025
-2L0= 16.301
M1: X as the linear Predictor
Equation :
Log(P/1-P) = -1.5027+2.0605X
-2L1= 11.028
Answer fro C:
Note: For Groupped data and ungroupped data the Parameter estimates and model fit statistics are exactly same.
see below fro teh results:
Parameter Estimates:
Odd Ratio Estimates:
Effect Point Estimate 95% Wald Condidence Limits
X 7.850 0.857 71.908
Trials X Y 1 0 1 2 0 0 3 0 0 4 0 0 5 1 1 6 1 1 7 1 0 8 1 0 9 2 1 10 2 1 11 2 1 12 2 1Related Questions
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