Appropriate ANOVA?

OK.  First, since I don’t know how much you know about regression analysis, excuse me if I start too elementary…   :-)  Second, it looks like you’re using Minitab, which I have, too.  If you’ll email me your Minitab file it’ll be a lot easier working with you – and I don’t mind talking with you over the phone, which would be even better.
 
Just because a predictor variable (an “X” – also called an explanatory variable, an independent variable) accounts for a relatively large proportion of the variation in a response variable ( the “Y”) doesn’t necessarily establish cause and effect between the two – it only means for sure they are correlated (i.e., they tend to “move together”).
 
In order to justify keeping an X in a regression model, it should account for a statistically significant proportion of the total variation in Y.  This will be shown by the p-value of the t statistic associated with the X’s estimated coefficient.  Typically you want the p-value to be less than 0.05 (on a scale of 0 to 1).  This means you’ve chosen an alpha value of 0.05, which is the probability of making a Type I Error (concluding the true value of the coefficient of an X is significantly different from zero, when it really isn’t).  Since you didn’t include all the output, I don’t have the X and X**2 coefficients’ t statistics and p-values.  But often, while the linear term’s (X) coefficient is significant, the quadratic term’s (X**2) is not.  So check that, and if the p-value of the estimated coefficient of the X**2 term is greater than 0.05 it should not be included in the model.
 
R-squared (R**2) is the multiple correlation coefficient and is the amount of the total variation of Y about its mean accounted for by all the terms you’ve included in the current model – 30.8% in your case.  Because R**2 will increase every time you add another term to a model (just due to the math), even if its coefficient is *not* significant, R**2 adjusted attempts to compensate for this by “penalizing” R**2 as the number of terms in the model increases.
 
Yes, with almost 70% of the variation in Y unaccounted for, other factors not (yet) in the model are associated with Y.  “Error” sum of squares (SS) in this case is just the (as yet) unaccounted for variation of Y about its mean (that almost 70%).  I’m not sure what is meant by (included in) “Patient Complexity of Care” – sounds similar to acuity to me.  If your Complexity metric is estimated from a combination of other factors, I’d try including them directly in the regression instead of Complexity.  You could also attempt to distinguish differences in LOS by including a dummy variable (surrogate) for case type (initial diagnosis), age, gender, race or ethnicity, etc. – any factor that might contribute to LOS.  You can use Stepwise Regression (Forward Selection or Backward Elimination) to let the statistical significance of the estimated coefficients “automatically” (sort of) build a model.
 
OK – I’ll leave it there for now and wait for feedback from you as to whether this is too elementary, helpful, or overwhelming…   :-)
 
Wayne
 
*** Wayne G. Fischer, BS, MS, PhD ***
Certified Manager of Quality & Organizational Excellence
Certified Quality Engineer / Certified Quality Auditor
Perioperative Quality Analyst
U of TX MD Anderson Cancer Center
Houston, TX  77030
office = 713.794.4340 / cell = 281.360.7584

  So, it sounds like your mean LOS and the variance are coupled – increase mean LOS and increase variance.  You might try logging the LOS and plotting/regressing that against PCC. If the residual pattern changes from a funnel shape to more like a shotgun blast then this would suggest a model on log LOS might be the better choice and the act of logging has decoupled the mean and the variance.

Robert, wondering:
Given that there are so many factors that contribute to LOS other than the acuteness of the patient (many of which can be controlled/influenced by nurse/doc/other interactions), why would you not have suggested a multiple regression with other measurable variables?  Since there is roughly 70% of variation left in the current model, just wondering – I had suggested that and since I’m positive you are more statistically savvy than I am, I was hoping you could help me understand.  Appreciate the feedback….

Thanks – I just wanted to make sure I wasn’t providing unsound advice.