The following log-likelihoods are from multinomial logistic…
The following log-likelihoods are from multinomial logistic regression models fit to a job satisfaction dataset (Response: High/Med/Low; predictors: Sector and Region): Minimal model (js.min.out): log Lik. = −1980.874 (df = 2) Main model (js.out): log Lik. = −1943.667 (df = 6) Saturated model(js.sat.out): log Lik. = −1940.553 (df = 8) Use the log-likelihood values to compute the deviance statistic to test: H₀: The minimal model is appropriate H_a: The main model (js.out) is better than the minimal model This deviance has 4 degrees of freedom. Carry out the test and state your conclusion.
Read DetailsThe following table displays job satisfaction responses (Hig…
The following table displays job satisfaction responses (High, Medium, Low), employment sector (Private/Public), and region (Urban/Rural) collected from a recent workforce survey. Response Sector Region Count1 High Private Urban 1982 Med Private Urban 1423 Low Private Urban 874 High Private Rural 1155 Med Private Rural 1386 Low Private Rural 1727 High Public Urban 1638 Med Public Urban 1559 Low Public Urban 10910 High Public Rural 9411 Med Public Rural 14812 Low Public Rural 187 A statistician used R to build the following multinomial logistic regression model: log(πMed / πLow) = β0,Med + β1,Med x1 + β2,Med x2log(πHigh / πLow) = β0,High + β1,High x1 + β2,High x2 where:x1 = 0 if Sector = Private, x1=1 if Sector = Publicx2 = 0 if Region = Rural, x2=1 if Region = Urban R OUTPUT (FULL MODEL) > js.out summary(js.out) Call:multinom(formula = Response ~ Sector + Region, data = js, weights = Count) Coefficients: (Intercept) Sector.L Region.LMed -0.103421 0.06724812 0.3158204High -0.317653 -0.12038946 0.6421537 Std. Errors: (Intercept) Sector.L Region.LMed 0.06184320 0.08731045 0.08754612High 0.06401887 0.08952033 0.08913748 Residual Deviance: 3887.334AIC: 3899.334 > logLik(js.out)’log Lik.’ -1943.667 (df=6) Then the statistician built the following minimal model: log(πMed / πLow) = β0,Medlog(πHigh / πLow) = β0,High > js.min.out summary(js.min.out) Call:multinom(formula = Response ~ 1, data = js, weights = Count) Coefficients: (Intercept)Med 0.1062438High -0.0834271 Std. Errors: (Intercept)Med 0.06017652High 0.06123894 Residual Deviance: 3961.748AIC: 3965.748 > logLik(js.min.out)’log Lik.’ -1980.874 (df=2) Finally, the statistician built the following saturated model: > js.sat.out summary(js.sat.out) Call:multinom(formula = Response ~ Sector * Region, data = js, weights = Count) Coefficients: (Intercept) Sector.L Region.L Sector.L:Region.LMed -0.097814 0.05841736 0.3074592 -0.1823457High -0.311042 -0.13250481 0.6318274 -0.3140826 Std. Errors: (Intercept) Sector.L Region.L Sector.L:Region.LMed 0.06207431 0.08779203 0.08779203 0.1243618High 0.06431005 0.09014227 0.09014227 0.1276641 Residual Deviance: 3881.106AIC: 3897.106 > logLik(js.sat.out)’log Lik.’ -1940.553 (df=8) (a) [5 POINTS]Compute the deviance test: H0: Minimal model is appropriateHa: Full model is better Note this deviance should have 4 d.f. Carry out this test, show all work, and clearly state yourconclusion. (b) [5 POINTS]Compute the deviance test: H0: Full model is appropriateHa: Saturated model is better Note this deviance should have 2 d.f. Carry out this test, show all work, and clearly state yourconclusion. (c) [5 POINTS]Using the full model, compute: π̂Low, π̂Med, π̂High for an Urban Private-sector worker.
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