370 LOGLINEAR MODELS FOR CONTINGENCY TABLES Table 9.18 Software Output (Based on

Question

370 LOGLINEAR MODELS FOR CONTINGENCY TABLES Table 9.18 Software Output (Based on SAS) for Fitting a Loglinear Model to Table9.17 Criteria For Assessing Goodness of Fit Criterion Deviance Pearson Chi-Square 7 Value 12.3687 12.1996 DF Analysis of Parameter Estimates Standard LR 958 Confidence Wald Chi- DF Estimate 0.3219 0.4237 Limits Parameter EI SN SN TF SN JP TF JP Square Error 0.1360 0.1520 e n1 0.5886 5.60 0.7242 7.77 n 1.2202 0.1451.5075 0.9382 70.69 f j 0.5585 0.1350-0.8242 -0.2948 17.12 0.0553 0.1278 9.3 Refer to the previous exercise. Table 9.18 shows the fit of the model that assumes conditional independence between E/I and T/F and between E/I and J/P but has the other pairwise associations. a. Compare this to the fit of the model containing all the pairwise associations, which has deviance 10.16 with df= 5, what do you conclude? b. Show how to use the limits reported to construct a 95% profile likelihood con- fidence interval for the conditional odds ratio between the S/N and J/P scales. Interpret. c. SAS (PROC GENMOD) reports maximized log-likelihood values of 3475.19 for the mutual independence model, 3538.05 for the homogeneous association model, and 3539.58 for the model containing all the three-factor interaction terms. Write the loglinear model for each case, and show that the numbers of model parameters are 5, 11, and 15, so residual df 11, 5, and 1 d. According to AIC, which model in (c) seems best? Why?

Explanation / Answer

a. Critical value for Chi-square statististic with df=5, is 11.017(from table critical values of chi- sqaured with df =5 crossing at 0.95 column value) which is greater than the deviance 10.16. So we have the sufficient evidence to support the claim that the model is a good fitting one. We fail to reject the null hypothesis. We conclude that the model is a good-fitting one.

b. Wald statistic is the ratio of the regression coefficient to the standard error of the regression coefficient and follows approximately normal distribution. The Wald statistic is 70.69 for the SN*JP pair. This value 70.69 is greater than the critical value of z=+1.96 for the data being normally distributed at the significance level of 95%. So it means that these two SN and JP independent variables contribute to the model strongly on the good -fitting model aspect in the presence of one another. The LR 95% confidence limits -1.5076 to -0.9382 is some where in between +1.96 to -1.96, not in the critical region.

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