Solved – Partial F ratio from ANOVA table

anovamultiple regressionregression

In multiple regression, if you have just an ANOVA table, and nothing else, no specific data, how can you do a partial F test on X1, given X2 is already in the model?

So, you have the ANOVA table:

source        df   SS    MS     F
-----------------------------------
regression    2    1.44  0.72   9.72
error         3    0.22  0.07
total         5    1.66
-----------------------------------

All values are filled in. With only this information, how can you do the partial F test where:

F = MSR(X1|X2) / MSE(X1, X2)
MSR(X1|X2) = SSR(X1, X2) – SS(X2) = 1.44 – ????
MSE (X1, X2) = MSE = 0.07

SSR(X1, X2) can be obtained from the table (SS regression)
MSE(X1, X2) can also be obtained from the table (just MSE)
but I cannot get SS(X2) from the table, as far as I know……

As far as I know, you need specific X and Y values to do this. Any other way from just the table?

Best Answer

As @PatrickCoulombe hints, you can't conduct a partial F-test with (only) the information in that ANOVA table. Let's assume you want to conduct the partial F-test for X1, where your full model includes both X1 and X2. In that case, you would need the ANOVA table for the full model, and the ANOVA table for the reduced model, which would only include X2. The reason you can't find the number to put in place of the "????" is because that number isn't listed in the ANOVA table you have access to--you need the reduced model ANOVA table as well.

The equations you list in the question aren't quite right. Your equation for the F ratio is right, and MSE is right, but your equation for MSR is actually for SS(X1|X2). Having calculated that, you get the MS(X1|X2) by dividing by the appropriate degrees of freedom, which is the degrees of freedom for those regressors that were dropped / you are testing (in your case, I'm guessing the df for X1=1). You calculate the F by dividing the two MSs, as you list; the realized F can be assessed against the theoretical distribution for F with numerator degrees of freedom equal to the df for dropped predictors, and denominator df equal to the df(residual) in the full model.

For a fuller understanding of this topic, it might help you to read my answers here: Testing for moderation with continuous vs categorical moderators, and possibly here: how to interpret type I (sequential) ANOVA and MANOVA.

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Regression – Interpreting F-values in ANOVA Table for Regression Analysis

For age, $F=1.0812=\frac{MS_\text{age}}{MS_\text{Residuals}}$. The same is true of the other $F$ values replacing age with weight or protein. This partials out the variance in carb that is related to the other two factors not being tested directly by the $F$ test in question, whereas using $MS_\text{Residuals}$ from a single-factor GLM in the denominator would not. Thus the hypothesis test is of whether the residual variance in carb that is not explained by your other factors can be explained by your given factor. More specifically, the $p$ value represents the probability that this residual variance would relate to your given factor at least as much as it does in your sample if you were to sample again randomly from a population in which the null hypothesis of no relationship between those residuals and your factor is literally true. As for why 16 and not 18, remember that controlling for these other factors costs you degrees of freedom: one apiece.

To elaborate in response to your edit/comment, another way of looking at your $F=3.5821$ is as $F=\frac{MS_\text{weight}}{MS_\text{Residuals}}$ for a general linear model with only one factor (weight). With that one-factor GLM's ${MS_\text{Residuals}}$ instead of a three-factor GLM's ${MS_\text{Residuals}}$ as the denominator (because you're not controlling for anything, hence what would've been the residuals are the observations $-\ \mu_\text{carb}$ instead, $\mu_\text{carb}$ being the intercept of the null model), you haven't partialed out any of the variance that weight can't explain but age or protein can, so the effect of weight appears less clear by itself.

When you control for the effects of age and protein, you reduce the amount of variance that still needs explaining in your model. This makes the predictive job a little easier for weight, because it no longer has to contend with the independent effects of age or protein in explaining carb. In a post hoc / retrospective sense, you can look back and say, "Well, no wonder weight didn't predict these observations as well by itself; age and protein vary in my sample too, and their independent effects were mucking things up for poor ol' weight!" Of course, these results are even better in an epistemic, hypothesis-testing sense if you expected in advance that this would happen, and chose multiple regression to examine hypotheses of independent effects you also specified in advance.

Solved – Contingency Table from ANOVA

The short answer, no, at least to my knowledge.

I think you are confused about a few things here including what a contingency table is. A contingency table is nothing more than a table that displays frequency distributions. When you refer to TP,FP,TN,&FN you are referring to a hat you describe is a special contingency table known as a confusion matrix. A confusion matrix is (in binary situations) a 2x2 matrix containing the predicted group memberships. For example:

   A  B      A   B
A  5  2    A TP  FP
B  3  8    B FN  TN

This is typically used to evaluate the performance of a predictive algorithm. In this case there are 5 true positives, 2 false positives, 3 false negatives, and 8 true negatives. You can ultimately use this table to derive many other statistics (e.g. Accuracy, Kappa, Sensitivity, etc).

The generic contingency table is simply a representation of the distribution of groups in a population. For example (copying from wikipedia):

         Right-handed  Left-handed  Total
Males    43            9            52
Females  44            4            48
Total    87            13           100

Now, this is, more-or-less, raw data that you can use to test a hypothesis. Assuming the data fits some assumptions, some tests include chi-squared, fisher's, and the G-test among others.

This is different from ANOVA (which in this case would be two-way). ANOVA uses some quantitative value as opposed to just counts. Your output is intended to be used to interpret how the quantitative variable is impacted by the factor.

Best Answer

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