Construct the confidence interval for the ratio of the population variances given the following sample statistics. Round your answers to four decimal places. n1=12 , n2=22, s12=92.101, s22=51.453, 95% level of confidence
Answer:
Given,
n1 = 12
n2 = 22
s12 = 92.101
s22 = 51.453
degree of freedom = n1 - 1
= 12 -1
df1 = 11
df2 = n2 - 1
= 22 - 1
= 20
(s1^2/s2^2) / F(alpha/2 , 11 , 20) < / < (s1^2/s2^2) / F(1 - alpha/2 , 11 , 20)
= 92.101/51.453 / 2.7209 < / < 92.101/51.453 / 0.3100
= 0.6579 < / < 5.7742
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