Question

Construct the 99% interval estimate for the ratio of the population variances using the following results from two independently drawn samples from normally distributed populations. (Round "F" value and final answers to 2 decimal places. You may find it useful to reference the appropriate table: chi-square table or F table) points Sample 1: 1 = 165, sî = 25.7, and n1 = 11 Sample 2: 7 2 = 161.7, să = 22.1, and m2 = 10 eBook Hint Confidence interval Ask Print References

Answer 1

Given that n1 =11 and n2 = 10

degree of freedom (df1) = n1 -1 = 11-1 = 10

degree of freedom (df2) = n2 -1 = 10-1 = 9

alpha = 1 -C = 1-0.99 = 0.01

F critical for left side = =F.INV.RT(alpha,df1,df2) =F.INV.RT(0.01,10,9) = 5.26

F critical for right side = =F.INV.RT(alpha,df2,df1) =F.INV.RT(0.01,9,10) = 4.94

s1^2 = 25.7 and s2^2 = 22.1

Confidence interval calculation

$\frac{s^2_1}{s_2^2}*\frac{1}{F_{left}} < \frac{\sigma_1^2}{\sigma_2^2}<\frac{s^2_1}{s_2^2}*F_{right} \\ . \\ \rightarrow \frac{25.7}{22.1}*\frac{1}{5.26} <\frac{\sigma_1^2}{\sigma_2^2}< \frac{25.7}{22.1}*4.94 \\ . \\ \rightarrow 0.22 < \frac{\sigma_1^2}{\sigma_2^2}< 5.74$