- #1
Rasalhague
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Well, just as I thought I'd got the hang of this...
Koosis: Statistics: A Self-Teaching Guide, 4th ed., §§ 6.29-43.
The "degreeses of freedom" are 3 and 36. This critical value, 4.38, is found by looking up the score for 1% in the table at the back of the book, or in Excel with F.INV.RT(0.01,3,36) = 4.38.
But this is a two-tailed test, right? So why do we not divide the desired significance level by 2 before inputting it into this function, as was explained by Koosis in §§ 6.19-22, and on this page? Thus: F.INV.RT(0.01/2,3,36) = 5.06. This latter method makes sense because, in a two-tailed test a result could achieve significance by being far enough from the mean of the F distribution on either side, so extreme results on one side of the mean will only constitute half of that 1%.
Koosis concludes that the result, which is 4.42, is significant, since it's greater than 4.38. But I'd have concluded that it's not significance, since it's less than 5.06.
A researcher believes that the color of a toy will affect how long children will play with it [...] her null hypothesis is that [itex]\mu_1 = \mu_2 = \mu_3 = \mu_4[/itex]. The alternative is that not all the means are equal; that is, that the color of the toy does make a difference.
[...]
For a 1% significance level, the critical region is F > or = 4.38.
Koosis: Statistics: A Self-Teaching Guide, 4th ed., §§ 6.29-43.
The "degreeses of freedom" are 3 and 36. This critical value, 4.38, is found by looking up the score for 1% in the table at the back of the book, or in Excel with F.INV.RT(0.01,3,36) = 4.38.
But this is a two-tailed test, right? So why do we not divide the desired significance level by 2 before inputting it into this function, as was explained by Koosis in §§ 6.19-22, and on this page? Thus: F.INV.RT(0.01/2,3,36) = 5.06. This latter method makes sense because, in a two-tailed test a result could achieve significance by being far enough from the mean of the F distribution on either side, so extreme results on one side of the mean will only constitute half of that 1%.
Koosis concludes that the result, which is 4.42, is significant, since it's greater than 4.38. But I'd have concluded that it's not significance, since it's less than 5.06.
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