Q&A Bank Part 4 Solution 4.5 page 5

[Dikshay Ramnani]

After the observed test statistic is calculated in this solution as 3.730

The 5% critical values for F(4,9) distribution should be 1/F(9,4) = 1/5.999=0.1667 and F(4,9)=3.633

And since 3.730 does not lie between 0.1667 and 3.633, we have sufficient evidence at the 5% level to reject H0.

This type of same thing is used in CT3 paper solution of 28th Oct 2014 solution Q.12 part (iii)

But the solution of this is given as:
The 5% critical values for an F4,9 distribution are 0.1123 and 4.718. Since 3.633 lies
between these, we have insufficient evidence at the 5% level to reject H0 . Therefore
we conclude that the populations are both normal.

Thanks in advance

 

[Hemant Rupani]

F(4,9)=4.718 at 2.5%(as it is two side test)