$\begin{array}{ll}
n_1=16 & n_2=22 \\s_1^2=4.875 & s_2^2=1.449\end{array}$
ces is given by
$\left(\frac{s_1^2}{s_2^2} \cdot \frac{1}{F_{\frac{a}{2}}}\right)<\frac{\sigma_1^2}{\sigma_2^2}<\left(\frac{s_1^2}{s_2^2} \cdot \frac{1}{F_{\left(1-\frac{a}{2}\right)}}\right)$
confidence $c=1-\alpha$ and the $F$-distribution with $n_1$
the ratio of the population variances. Round the endpo
Lower Endpoint = $\square$
Answer (1)
Calculate degrees of freedom: d f 1 = 15 and d f 2 = 21 .
Determine α = 0.05 and find the F-value F 0.025 , 15 , 21 ≈ 3.6716 .
Calculate the lower endpoint: 1.449 4.875 ⋅ 3.6716 1 ≈ 1.3278 .
The lower endpoint of the confidence interval is 1.3278 .
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 lower endpoint of the confidence interval for the ratio of 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 , d f 1 , d f 2 1 ) < σ 2 2 σ 1 2 < ( s 2 2 s 1 2 ⋅ F ( 1 − 2 α ) , d f 1 , d f 2 1 ) where c = 1 − α is the confidence level, and F follows an F-distribution with d f 1 = n 1 − 1 and d f 2 = n 2 − 1 degrees of freedom.
Calculate degrees of freedom First, we calculate the degrees of freedom: d f 1 = n 1 − 1 = 16 − 1 = 15 d f 2 = n 2 − 1 = 22 − 1 = 21
Determine alpha and calculate F-value Since the confidence level is not explicitly given, we assume a standard confidence level of 95%, so c = 0.95 and α = 1 − c = 1 − 0.95 = 0.05 . Therefore, 2 α = 2 0.05 = 0.025 . We need to find the F-value F 0.025 , 15 , 21 .
Find the F-value Using a calculator or statistical software, we find that F 0.025 , 15 , 21 ≈ 3.6716 .
Calculate the lower endpoint Now we calculate the lower endpoint of the confidence interval using the formula: Lower Endpoint = s 2 2 s 1 2 ⋅ F 2 α , d f 1 , d f 2 1 = 1.449 4.875 ⋅ 3.6716 1 ≈ 1.3278
State the final answer Therefore, the lower endpoint of the confidence interval for the ratio of population variances is approximately 1.3278 .
Examples
Understanding the variance ratio is crucial in fields like manufacturing, where consistency in product dimensions is vital. For example, if two machines produce parts, comparing the variance in their output helps determine if one machine is more reliable. A confidence interval for this variance ratio provides a range within which the true ratio likely falls, aiding decisions on machine maintenance or replacement.
