Question

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=19, s12=8.041, s22=4.964, 90% level of confidence

Answer 1

At 90% confidence level, the critical values are

F_L = F(0.95, 11, 18) = 0.3744

F_R = F(0.05, 11, 18) = 2.3742

The 90% confidence interval for $\frac{\sigma_{1}^2}{\sigma_{2}^2}$ is

$\frac{s_{1}^2}{s{_2}^2} * \frac{1}{F_{R}} < \frac{\sigma_{1}^2}{\sigma_{2}^2} < \frac{s_{1}^2}{s{_2}^2} * \frac{1}{F_{L}}$

$= \frac{8.041}{4.964} * \frac{1}{2.3742} < \frac{\sigma_{1}^2}{\sigma_{2}^2} < \frac{8.041}{4.964} * \frac{1}{0.3744}$

$= 0.6823 < \frac{\sigma_{1}^2}{\sigma_{2}^2} < 4.3266$