# Tag Info

0

The adjusted $p$-value is not dependent upon the significance level, just like the ordinary $p$-value. You just compare these values to the significance level. With ordinary $p$-values, you compare them to .05 or .01, or whatever your significance level (simple Type I error rate) is. The choice of significance level does not change the $p$-value. The same ...

0

I have no clue what you are trying to do in R. But I don't think your problem lies there. Thing is, the p value tells you how likely your specific datapoints would have been, if the Hypothesis you are trying to disprove were true.

2

It appears that you and they disagree by a factor of $\sqrt{2}$. If you look at the table rows with $k=2$, they have the familiar 1.96. R gives > round(qtukey(.95,5:2,Inf),2) [1] 3.86 3.63 3.31 2.77 > round(qtukey(.95,5:2,Inf)/sqrt(2),2) [1] 2.73 2.57 2.34 1.96 where the 3.858 matches what you found in the reference table, and the second row matches ...

Top 50 recent answers are included