# Interpret a LinearModelFit ANOVATable?

Posted 1 month ago
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 Having evaluated the code below, data = {{0, 2}, {1, 0}, {2, 1}, {3, 8}, {4, 8}, {5, 6}, {6, 7}} lm = LinearModelFit[data, {x, x^2}, x] lm["ANOVATable"] lm["ParameterTable"] Show[ListPlot[data], Plot[lm[x], {x, 0, 10}], Frame -> True] what is the meaning of the ANOVATable p-values list (what statistics and hypotesis is it related)? Which table (ANOVATable or ParameterTable) p-values should I use for regression model coefficients significiance inference? Where can I get detailed information (in Mathematica help it is very poor )?  Attachments: Answer
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Posted 1 month ago
 It is just an illustrative example, not real data to be analyzed. My main question - what do ANOVATable F-statistics and p-values mean for regression case. Answer
Posted 1 month ago
 Your response variable consists of integers. Are these counts? rounded measurements? made-up data? If the responses are counts, you might want to consider Poisson regression (using GeneralizedLinearModelFit). A little background on the data might be helpful because performing statistical methods requires more than just the numbers Answer
Posted 1 month ago
 Thanks for the reply. Could you, please, tell me, what is the equation for calculation two F-staistics presenting in regression ANOVATable above? (I know how F-statistics is calculated for ordinary variance analysis and how it can be interpreted, but I know nothing about it in regression case). Answer
Posted 1 month ago
 One can conclude from your ANOVA table that none of the variables{1,x,x^2) has a decisive influence on Y (see https://en.wikipedia.org/wiki/Analysis_of_variance). Your parameter table show that the variance of the estimators of the coefficients is excessive (see https://en.wikipedia.org/wiki/Student%27s_t-test). In addition, you should have a look on the R^2: lm["RSquared"] lm["AdjustedRSquared"] giving respectively 0.608234 and 0.412351. Rather poor, because the model is too complex. A degree one model is better: lm = LinearModelFit[data, {x}, x] lm["ANOVATable"] lm["ParameterTable"] lm["RSquared"] lm["AdjustedRSquared"] Show[ListPlot[data], Plot[lm[x], {x, 0, 10}], Frame -> True] But the Pvalue of the intercept is very large, while the adjusted R^2 is just slightly better... Thus, it's better to choose a simpler one: m = LinearModelFit[data, {x}, x, IncludeConstantBasis -> False] lm["ANOVATable"] lm["ParameterTable"] lm["RSquared"] lm["AdjustedRSquared"] Show[ListPlot[data], Plot[lm[x], {x, 0, 10}], Frame -> True] giving 0.827217 as adjusted R^2, for Y=1.43 x. Answer