Confidence Interval for ratio of variances
Let $X_1, X_2, \cdots , X_{n_1}$ be a random sample of size $n_1$ from $N(\mu_1, \sigma_1^2)$ and $Y_1, Y_2, \cdots , Y_{n_2}$ be a random sample of size $n_2$ from $N(\mu_2, \sigma_2^2)$. Moreover, $X$ and $Y$ are independently distributed.
$100(1-\alpha)$% confidence interval estimate of ratio of variances is
$$ \begin{aligned} \bigg(\frac{s_1^2}{s_2^2}\frac{1}{F_{(\alpha/2,n_1-1,n_2-1)}}, \frac{s_1^2}{s_2^2}\frac{1}{F_{(1-\alpha/2,n_1-1,n_2-1)}}\bigg) \end{aligned} $$
Assumptions
a. The two populations are independent.
b. The two samples are simple random samples.
c. The two populations are normally distributed.
Step by step procedure
Step by step procedure to estimate the confidence interval for the ratio of two population variances is as follows:
Step 1 Specify the confidence level
Specify the confidence level $(1-\alpha)$.
Step 2 Given information
Specify the given information, sample sizes $n_1$, $n_2$, sample standard deviations $s_1$ and $s_2$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval for the ratio of variances $\sigma^2_1/\sigma^2_2$ is
$$ \begin{aligned} \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(\alpha/2, n_1-1, n_2-1)}} \leq \frac{\sigma^2_1}{\sigma^2_2} \leq \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(1-\alpha/2, n_1-1, n_2-1)}}. \end{aligned} $$
Step 4 Determine the critical value
Find the critical values $F_{(\alpha/2, n_1-1, n_2-1)}$ and $F_{(1-\alpha/2, n_1-1, n_2-1)}$ for desired confidence level and degrees of freedoms.
Step 5 Determine the confidence interval
$100(1-\alpha)$% confidence interval estimate for the mean of the difference is
$$ \begin{aligned} \bigg(\frac{s_1^2}{s_2^2}\frac{1}{F_{(\alpha/2,n_1-1,n_2-1)}}, \frac{s_1^2}{s_2^2}\frac{1}{F_{(1-\alpha/2,n_1-1,n_2-1)}}\bigg) \end{aligned} $$
Hope you enjoyed the step by step procedure of finding confidence interval for ratio of variances.
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