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Feb 16, 2018 at 21:39 comment added amoeba Hey @caracal, nice to see such an old answer suddenly accepted! :-) Cheers.
Feb 16, 2018 at 19:34 vote accept caracal
Jan 9, 2015 at 11:08 answer added amoeba timeline score: 5
Jan 9, 2015 at 11:01 history edited amoeba CC BY-SA 3.0
made the title more precise + syntax highlighting
Nov 12, 2011 at 3:56 answer added CDX timeline score: 0
Nov 11, 2011 at 23:39 answer added gung - Reinstate Monica timeline score: 2
Nov 7, 2011 at 13:34 comment added caracal @MikeWierzbicki You're right in that if the full model contains only IV1 (1st example), then the two expressions for the denominator are identical. However, when the full model contains additional effects, the denominator for testing $A$ changes even though the model comparison (~ 1 vs. ~ IV1 + 1 for type 1 SS) does not. In the 3 examples, the mean square for $A$ does not change (same model comparison in all cases), but the mean square error does. I'm interested in what justifies the changing error term when the actual comparison stays the same.
Nov 5, 2011 at 20:55 comment added user5594 I may just be confused with your question, but for testing the effect of $A$ with Type 1 SS, the denominator is the one you have in your second expression. The F-value in the output from running anova(lm(DV ~ IV1)) is calculated via your second expression. That is, if you ran anova(lm(DV ~ 1)) and anova(lm(DV ~ IV1)) and plugged the corresponding values into your second expression, you get $F=0.9342$. Let me know if I'm completely missing your concern.
Oct 20, 2011 at 19:06 history tweeted twitter.com/#!/StackStats/status/127098280491483136
Oct 20, 2011 at 18:00 history edited caracal CC BY-SA 3.0
deleted 150 characters in body
Oct 20, 2011 at 14:23 history edited user88 CC BY-SA 3.0
edited title
Oct 20, 2011 at 13:38 history asked caracal CC BY-SA 3.0