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STAT 350: Lecture 20

The SCENIC data set, continued

See Lecture 18 for plots of the data and Lecture 19 for our first analysis.

We have found that STAY, CULTURE and CHEST are significant and that we must retain one of the three variables BED, NURSES and CENSUS which measure size of the hospital. Picking the variable of the three which produces the largest multiple tex2html_wrap_inline21 we go with NURSES. Now we look at the question of adding further variables to that 4 covariate model.

 
> anova(fit.n,fit.full)
Analysis of Variance Table
Response: Risk
 Model   Resid. Df    RSS   Test Df SumSq   F Value    Pr(F) 
 FULL       108    95.63982 
 REDUCED    104    95.63982    4    2.9895 0.8127053 0.5198417
This suggests we need not consider adding further variables.

However, we should examine diagnostics and consider the question of how variables are likely to influence RISK.

Suggestion: Transform other variables.

Define NURSE.RATIO = NURSES/CENSUS. Idea: large values indicate more intensive nursing care.

Define CROWDING = CENSUS/BEDS. Idea: large values indicate a crowded hospital.

Add these variables to the model.

> Nurse.Ratio <- scenic$Nurse/scenic$Census
> sc.ext <- data.frame(scenic, Nurse.Ratio)
> Crowding <- scenic$Census/scenic$Beds
> sc.ext <- data.frame(sc.ext, Crowding)
> fit.l20 <- lm(Risk ~ Stay + Culture + Chest +
    Nurses + Crowding + Nurse.Ratio, data = sc.ext)
> summary(fit.l20)
Residuals:
    Min      1Q  Median     3Q   Max
 -2.036 -0.6102 0.01268 0.3956 2.798
Coefficients:
              Value Std. Error t value Pr(>|t|)
(Intercept) -1.2762  0.8753    -1.4581  0.1478
       Stay  0.2196  0.0594     3.6983  0.0003
    Culture  0.0424  0.0099     4.2740  0.0000
      Chest  0.0093  0.0055     1.7040  0.0913
     Nurses  0.0014  0.0007     1.9627  0.0523
   Crowding  1.4296  0.9455     1.5121  0.1335
Nurse.Ratio  0.8238  0.3298     2.4979  0.0140

Residual standard error: 0.9359 on 106 df
Multiple R-Squared: 0.5389
F-statistic: 20.65 on 6 and 106 df,
the p-value is 6.661e-16

Correlation of Coefficients:
            (Intercept)    Stay Culture   Chest  Nurses Crowding
       Stay -0.3314
    Culture  0.1738     -0.1725
      Chest -0.1170     -0.3422 -0.3010
     Nurses  0.3162     -0.2737 -0.0803  0.1608
   Crowding -0.7108     -0.2136 -0.0321 -0.0605 -0.3032
Nurse.Ratio -0.6321      0.2561 -0.1365 -0.2548 -0.3056  0.3849

Conclusion: NURSE.RATIO is a useful predictor.

Can we discard CHEST, CROWDING? NURSES marginal but seems reasonable to keep this variable since we are keeping NURSE.RATIO.

fit.l20.t <- lm(Risk ~ Stay + Culture + Nurse.Ratio 
      + Nurses, data = sc.ext)
> summary(fit.l20.t)
Residuals:
    Min      1Q  Median     3Q   Max 
 -2.214 -0.6387 0.06483 0.5021 2.655

Coefficients:
              Value Std. Error t value Pr(>|t|) 
(Intercept) -0.0831  0.6092    -0.1365  0.8917 
       Stay  0.2767  0.0549     5.0417  0.0000 
    Culture  0.0482  0.0096     5.0311  0.0000 
Nurse.Ratio  0.7695  0.2994     2.5701  0.0115 
     Nurses  0.0016  0.0007     2.2607  0.0258 

Residual standard error: 0.9511 on 108 df
Multiple R-Squared: 0.5149 
F-statistic: 28.66 on 4 and 108 df, 
         the p-value is 3.331e-16 

Correlation of Coefficients:
            (Intercept)    Stay Culture Nurse.Ratio 
       Stay -0.8669                                
    Culture  0.1569     -0.3317                    
Nurse.Ratio -0.6468      0.3148 -0.2287            
     Nurses  0.1916     -0.3356 -0.0521 -0.1851    
> anova(fit.l20,fit.l20.t)
Analysis of Variance Table    Response: Risk
Model   Res df   ESS   test df   SS    F  P value
FULL      106   92.852
REDUCED   108   97.689   2      4.84 2.76 0.068

Conclusion: Can discard CHEST, CROWDING but not NURSES.

Remaining Issues


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Richard Lockhart
Fri Feb 28 10:08:55 PST 1997