Confidence Interval for Ratio of variances

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 for the 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.

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 $(1-\alpha)$

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 $$\begin{aligned} \bigg(\frac{s_1^2}{s_2^2}\frac{1}{F_{(\alpha/2,n_1-1,n_2-1)}}, \frac{s_1^2}{s_2^2}\frac{1}{F_{(1-\alpha/2,n_1-1,n_2-1)}}\bigg) \end{aligned}$$

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