$\begin{array}{ll}
n_1=16 & n_2=22 \\s_1^2=4.875 & s_2^2=1.449\end{array}$
The confidence interval for the ratio of two population variances is given by
$\left(\frac{s_1^2}{s_2^2} \cdot \frac{1}{F_{\frac{a}{2}}}\right)<\frac{\sigma_1^2}{\sigma_2^2}<\left(\frac{s_1^2}{s_2^2} \cdot \frac{1}{F_{\left(1-\frac{a}{2}\right)}}\right)$
where $F_{\frac{-}{2}}$ and $F_{\left(1-\frac{4}{2}\right)}$ are the critical values for the level of confidence $c=1-\alpha$ and the $F$-distribution with $n_1-1$ numerator degrees of freedom and $n_2-1$ denominator degrees of freedom.
Use the $F$-distribution table to find the confidence interval for the ratio of the population variances. Round the endpoints of the interval to four decimal places, if necessary.
Lower Endpoint = $\square$
Upper Endpoint = $\square$
Answer (1)
Calculate degrees of freedom: d f 1 = 15 and d f 2 = 21 .
Find F-critical values: F 0.025 ( 15 , 21 ) ≈ 2.41 and F 0.975 ( 15 , 21 ) ≈ 0.39 .
Compute the ratio of sample variances: s 2 2 s 1 2 ≈ 3.3644 .
Determine the confidence interval: Lower Endpoint ≈ 1.3960 , Upper Endpoint ≈ 8.6267 , resulting in ( 1.3960 , 8.6267 ) .
Explanation
Understand the problem and provided data We are given the sample sizes n 1 = 16 and n 2 = 22 , and the sample variances s 1 2 = 4.875 and s 2 2 = 1.449 . We want to find the confidence interval for the ratio of the population variances σ 2 2 σ 1 2 . The formula for the confidence interval is given by
( s 2 2 s 1 2 ⋅ F 2 α 1 ) < σ 2 2 σ 1 2 < ( s 2 2 s 1 2 ⋅ F ( 1 − 2 α ) 1 )
where F 2 α and F ( 1 − 2 α ) are the critical values for the level of confidence c = 1 − α and the F -distribution with n 1 − 1 numerator degrees of freedom and n 2 − 1 denominator degrees of freedom.
Calculate degrees of freedom and alpha First, we calculate the degrees of freedom for the numerator and denominator:
Numerator degrees of freedom: d f 1 = n 1 − 1 = 16 − 1 = 15 Denominator degrees of freedom: d f 2 = n 2 − 1 = 22 − 1 = 21
We assume a confidence level of c = 0.95 , so α = 1 − c = 1 − 0.95 = 0.05 . Thus, 2 α = 0.025 and 1 − 2 α = 0.975 .
Find the F-critical values Next, we need to find the F-critical values F 2 α = F 0.025 ( 15 , 21 ) and F ( 1 − 2 α ) = F 0.975 ( 15 , 21 ) from the F-distribution table. Also, we know that F 1 − α /2 ( d f 1 , d f 2 ) = F α /2 ( d f 2 , d f 1 ) 1 . Therefore, F 0.975 ( 15 , 21 ) = F 0.025 ( 21 , 15 ) 1 .
From the F-distribution table (which I cannot directly access, so I will use estimated values), we have:
F 0.025 ( 15 , 21 ) ≈ 2.41 F 0.025 ( 21 , 15 ) ≈ 2.57
Therefore, F 0.975 ( 15 , 21 ) = 2.57 1 ≈ 0.39
Calculate the endpoints of the confidence interval Now, we calculate the ratio of the sample variances: s 2 2 s 1 2 = 1.449 4.875 ≈ 3.3644
Then, we calculate the lower and upper endpoints of the confidence interval:
Lower Endpoint: s 2 2 s 1 2 ⋅ F 2 α 1 = 3.3644 ⋅ 2.41 1 ≈ 1.3960
Upper Endpoint: s 2 2 s 1 2 ⋅ F ( 1 − 2 α ) 1 = 3.3644 ⋅ 0.39 1 ≈ 8.6267
State the final answer Therefore, the 95% confidence interval for the ratio of the population variances is approximately ( 1.3960 , 8.6267 ) .
Lower Endpoint = 1.3960 Upper Endpoint = 8.6267
Examples
Understanding the variance ratio is crucial in fields like finance. For instance, comparing the volatility (variance) of two stocks can help investors assess relative risk. If the ratio is significantly different from 1, it suggests one stock is considerably more volatile than the other, influencing portfolio diversification strategies. Confidence intervals provide a range within which the true variance ratio likely falls, aiding in making informed decisions despite inherent market uncertainty. This concept extends to other areas, such as comparing the consistency of manufacturing processes or the effectiveness of different treatments in medical trials.
