In constructing a 95% interval estimate for the ratio of two population variances, $\sigma _ { 1 } ^ { 2 }$ / $\sigma _ { 2 } ^ { 2 }$ , two independent samples of sizes 41 and 61 are drawn from the populations. If the sample variances are 515 and 920, then the upper confidence limit is:
A) 0.321.
B) 1.009.
C) 0.311.
D) 0.974.

Question

Quizplus · Accepted Answer

The 95% confidence interval for the ratio of two population variances is given by:

$$\left( \frac{s _ {1} ^ {2} }{s _ {2} ^ {2} } \times \frac{F_{\alpha/2, v_1, v_2}}{F_{1-\alpha/2, v_1, v_2}}, \frac{s _ {1} ^ {2} }{s _ {2} ^ {2} } \times \frac{F_{1-\alpha/2, v_2, v_1}}{F_{\alpha/2, v_2, v_1}}\right)$$

where $s _ {1} ^ {2} $ and $s _ {2} ^ {2} $ are the sample variances, and $v_1$ and $v_2$ are the degrees of freedom for the two samples.

Substituting the given values, we have:

$$ \left( \frac{515}{920} \times \frac{F_{0.025, 40, 60}}{F_{0.975, 40, 60}}, \frac{515}{920} \times \frac{F_{0.975, 60, 40}}{F_{0.025, 60, 40}}\right)$$

Using a table of F-distribution or calculator, we can find:

- $F_{0.025, 40, 60} = 0.559$
- $F_{0.975, 40, 60} = 1.751$
- $F_{0.025, 60, 40} = 0.443$
- $F_{0.975, 60, 40} = 2.286$

Substituting these values:

$$ \left( \frac{515}{920} \times \frac{0.559}{1.751}, \frac{515}{920} \times \frac{2.286}{0.443}\right)$$

Simplifying:

$$ \left( 0.187, 0.974\right)$$

Therefore, the upper confidence limit is 0.974, which corresponds to choice (B).

In Constructing a 95% Interval Estimate for the Ratio of Two