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I am given the following table

Source of variation ___ sum of squares __ degrees of freedom __ mean squares __ test statistic __ p

Among or between ____1874_______________ ----______________-----___________------____0.009

Within groups/error_____16580_____________----_______________-----___________

total_________________-----_______________----

There are 5 groups which 25 plants are separated into.

My attempt at filling in the table:

Source of variation ___ sum of squares __ degrees of freedom __ mean squares __ test statistic __ p

Among or between ___1874____________5-1________________1874/4_________ (1874/4)/(16580/20) ____0.009

Within groups/error___16580___________25-5_______________16580/20___________

total_______________1874+1650 ______ 25-1

This division of two divisions belong under test statistic and 0.0009 belongs under p

I opted for showing the equations rather than the answer to show where I got the values from. Please do excuse me for the formatting of the tables.

How did I do? Also with a p-value < 0.05 I can reject the null hypothesis correct?

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closed as off-topic by Michael Chernick, kjetil b halvorsen, mdewey, Peter Flom Apr 14 '17 at 13:28

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As far as I can see your table is correct. Well, nearly SST should be 1874+16580

If you set alpha to 0.05 you can reject the null hypothesis of the ANOVA.

Remark: since this is not a concrete problem but you are rather learning about ANOVA in general, please add the self-study tag to your question.

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