Confidence Interval for ratio of variances Examples

Confidence Interval for ratio of variances

In this tutorial we will discuss some examples on confidence interval for ratio of variances.

Example 1

The same capacity of hard drive is manufactured on two different machines, Machine A and Machine B. Samples are taken from both machines and sample mean manufacturing times and sample variances are recorded as follows:

Machine A: $n_1=13$, $\overline{x}= 127.4$, $s_1^2= 384.16$

Machine B: $n_2=9$, $\overline{y}= 108.3$, $s_2^2 =106.09$

Compute 95% confidence interval for the ratio of variance $\sigma^2_1/\sigma^2_2$.

Solution

Let $X$ denote the manufacturing time for hard drive by machine A and let $Y$ denote the manufacturing time for hard drive by machine B. Given data is as follows:

.	Machine A	Machine B
Sample size	$n_1= 13$	$n_2=10$
Sample variance	$s_1^2 = 384.16$	$s_2^2 = 106.09$

Step 1 Specify the confidence level $(1-\alpha)$

Confidence level $=1-\alpha = 0.95$. Thus $\alpha = 0.05$.

Step 2 Given information

Specify the given information, sample sizes $n_1=13$, $n_2=10$, sample standard deviations $s_1^2=384.16$ and $s_2^2=106.09$.

Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the ratio of variances $\sigma^2_1/\sigma^2_2$ is

$$ \begin{equation*} \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(\alpha/2, n_1-1, n_2-1)}} \leq \frac{\sigma^2_1}{\sigma^2_2} \leq \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(1-\alpha/2, n_1-1, n_2-1)}}. \end{equation*} $$

Step 4 Determine the critical value

The critical values are $F_{(\alpha/2, n_1-1, n_2-1)}$ and $F_{(1-\alpha/2, n_1-1, n_2-1)}$.

The critical values are $F_{(\alpha/2, n_1-1, n2-1)}=F{(0.025, 12,9)}=3.868$ and $F_{(1-\alpha/2, n_1-1, n2-1)}=F{(0.975, 12,9)}=0.291$.

Step 5 Determine the confidence interval

$95$% confidence interval estimate for the ratio of variances is

$$ \begin{eqnarray*} \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(\alpha/2, n_1-1, n_2-1)}} & \leq & \frac{\sigma^2_1}{\sigma^2_2} \leq \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(1-\alpha/2, n_1-1, n_2-1)}}\\ \frac{384.16}{106.09}\cdot\frac{1}{3.868} & \leq & \frac{\sigma^2_1}{\sigma^2_2} \leq \frac{384.16}{106.09}\cdot\frac{1}{0.291}\\ 0.9362& \leq & \frac{\sigma_1^2}{\sigma^2_2} \leq 12.4436. \end{eqnarray*} $$

Interpretation

We can be $95$% confident that the ratio of variance $\sigma^2_1/\sigma^2_2$ is between $0.9362$ and $12.4436$.

Example 2

Two different brands of batteries are tested and the variations of their voltage outputs are noted. The results are given as follows:

Brand A: $n_1=10$, $\overline{x}= 9.31$, $s_1=0.37$

Brand B: $n_2=8$, $\overline{y}= 8.82$, $s=0.31$

Construct 90% confidence interval for $\sigma^2_1/\sigma^2_2$.

Solution

Let $X$ denote the voltage output of batteries manufactred by Brand A and $Y$ denote the voltage output of batteries manufactured by brand B. Given data is as follows:

.	Machine A	Machine B
Sample size	$n_1= 10$	$n_2=8$
Sample sd	$s_1 = 0.37$	$s_2 = 0.31$

Step 1 Specify the confidence level $(1-\alpha)$

Confidence level $=1-\alpha = 0.9$. Thus $\alpha = 0.1$.

Step 2 Given information

Given that samples sizes are $n_1=10$ and $n_2=8$. The sample standard deviations are $s_1=0.37$ and $s_2=0.31$.

Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the ratio of variances $\sigma^2_1/\sigma^2_2$ is

$$ \begin{equation*} \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(\alpha/2, n_1-1, n_2-1)}} \leq \frac{\sigma^2_1}{\sigma^2_2} \leq \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(1-\alpha/2, n_1-1, n_2-1)}}. \end{equation*} $$

Step 4 Determine the critical value

The critical values are $F_{(\alpha/2, n_1-1, n_2-1)}$ and $F_{(1-\alpha/2, n_1-1, n_2-1)}$.

The critical values are $F_{(\alpha/2, n_1-1, n_2-1)}=F_{(0.05, 9,7)}=3.677$ and $F_{(1-\alpha/2, n_1-1, n_2-1)}=F_{(0.95, 9,7)}=0.304$.

Step 5 Determine the confidence interval

$90$% confidence interval estimate for the ratio of variances is

$$ \begin{eqnarray*} \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(\alpha/2, n_1-1, n_2-1)}} & \leq & \frac{\sigma^2_1}{\sigma^2_2} \leq \frac{s_1^2}{s_2^2}\cdot\frac{1}{F_{(1-\alpha/2, n_1-1, n_2-1)}}\\ \frac{0.37^2}{0.31^2}\cdot\frac{1}{3.677} & \leq & \frac{\sigma^2_1}{\sigma^2_2} \leq \frac{0.37^2}{0.31^2}\cdot\frac{1}{0.304}\\ \frac{0.1369}{0.0961}\cdot\frac{1}{3.677} & \leq & \frac{\sigma^2_1}{\sigma^2_2} \leq \frac{0.1369}{0.0961}\cdot\frac{1}{0.304}\\ 0.3874& \leq & \frac{\sigma_1^2}{\sigma^2_2} \leq 4.686. \end{eqnarray*} $$

Interpretation

We can be $90$% confident that the ratio of variance $\sigma^2_1/\sigma^2_2$ is between $0.3874$ and $4.686$.

Do read my article on Confidence interval for ratio of variances to understand step by step procedure of finding confidence interval for ratio of variances.

If you have any doubt or queries on Confidence interval for ratio of variances examples feel free to post them in the comment section.