(a) On the F-table, column for 5 and row for 8 will give the area on the right 0.05 with F0.05:5,8 = 3.69
(d) F0.95:8,5 = 1/ F0.05:5,8 = 1/3.69 =0.271
(e) The percentile is the area on the left of the value and it means the area on the right of the value is 0.01. On the F-table, look at column for 10 and row for 12 with the area on the right 0.01.
P( F F0.01:10,12 )= 0.99 then F0.01:10,12 = 4.30
(h) P(0.177 F 4.74) = P(F0.99:10,5 F F0.05:10,5) =1-(0.01+0.05)=0.94 where F0.99:10,5 = 1/F0.01:5,10 = 1/5.64=0.177
Example 8: A study was performed to determine whether men and women differ in their repeatability in assembling components on printed circuit boards. Two samples of 26 men and 21 women were selected and each subject assembled the units. The two sample standard deviations of assembly time were smen=0.98 min and swomen=1.02 min. Is there evidence to support the claim that men and women differ in repeatability for this assembly task? Use the significance level 0.02 and state any necessary assumptions about underlying distribution of the data.
=0.10, m=26, n=21
Test statistics: 
Decision: Reject H0 if F F1-/2;m-1,n-1 = F0.99;25,20 =1/2.70=0.37 or F F/2;m-1,n-1= F0.01;25,20 =2.84. Since 0.37 < F=0.9231 <2.84, fail to reject H0.
Conclusion: men and women do not differ in repeatability for this assembly task
Example 9 (Exercise 9.62):
=0.10, m=48, n=45
Test statistics: 
Decision: Reject H0 if F F1-/2;m-1,n-1 = F0.95;47,44 =1/1.6336=0.61214 or F F/2;m-1,n-1= F0.05;47,44 =1.6412 (The F-table on your book is not very detailed, you can use both degrees of freedom as 40 to look at the table. I gave you the real values using the computer programs.)
Since 1.2219 is not larger than 1.6412 or smaller than 0.61214, fail to reject H0.
Conclusion: the variances are the same.
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