F values come out as Infinity using the anovan function

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ki o kim
ki o kim 2023-10-23
评论： Jeff Miller 2023-10-24
13 design variables, 3 level , 2 strength
Using OA table..
avovan(Y,S)
S data
y data
result
What is a problem??
what it is mean?

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采纳的回答

Jeff Miller
Jeff Miller 2023-10-23
The underlying problem is that the d.f. for error is 0, which is responsible for the 0 MS error.
In essence, there are not enough data to assess error with such a complicated model. You can get F's if you drop some of the design variables and look at simpler models. Maybe drop X3 and X11 since they don't account for much anyway (i.e., they have small MS's)?
  2 个评论
ki o kim
ki o kim 2023-10-24
Thank you for the reply.
If it is currently difficult to obtain sufficient data, I would like to ask whether it is possible to determine the contribution rate to performance based on the proportion of each variable to Sum Sq.
Jeff Miller
Jeff Miller 2023-10-24
Sure, you can convert each of the sums of squares into a proportion of the total sum of square if you want. That's a reasonable thing to do, but there are two points that you might want to keep in mind:
  1. The proportions you get will reflect the current sample only, and you have no way to estimate whether you would get a similar set of proportions if you were to repeat the experiment. So, these proportions would only be guesses, and you have no way to judge how accurately they reflect any real effects.
  2. The sums of squares are simply transformations of the means, so you might find it more intuitive to look at the means themselves rather than these transformations of them. For example, the SS for X1 is just a summary description of the means for the three levels of X1. Why not just look directly at those means instead of looking at this harder-to-understand SS?

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更多回答（1 个）

Sam Chak
Sam Chak 2023-10-23
Ran in:
Hi @ki o kim
In this example, you'll notice that the F-statistic is calculated as the ratio of the mean squared errors: (7.875/0.13889), (4.5/0.13889), and (0.04167/0.13889). However, in your case, the mean squared error is 0. If you haven't encountered division by zero before, performing such a computation will result in either positive or negative infinity.
load popcorn
popcorn
popcorn = 6×3
5.5000 4.5000 3.5000 5.5000 4.5000 4.0000 6.0000 4.0000 3.0000 6.5000 5.0000 4.0000 7.0000 5.5000 5.0000 7.0000 5.0000 4.5000
[p, tbl] = anova2(popcorn, 3);

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