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 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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