Students of two sections of a history course took a common final examination. The course instructor examines the variance in scores between the two sections. He selects random samples of n₁ = 11 and n₂ = 16 with sample variances of and = 200, respectively. Assuming that the population distributions are normal, construct a 90% confidence interval for the ratio of the population variance. A) [0.90, 2.41] B) [0.50, 2.00] C) [0.25, 4.00] D) [0.79, 5.70]

Question

Quizplus · Accepted Answer

First, we need to calculate the point estimate for the ratio of population variances:
$$\frac{\sigma_{1}^{2}}{\sigma_{2}^{2}} = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{11}{200} = 0.055$$
Next, we need to calculate the critical values from the F-distribution with degrees of freedom $n_{1}-1=10$ and $n_{2}-1=15$ at a 90% confidence level. We get $F_{0.05,10,15}=0.291$ and $F_{0.95,10,15}=3.00$.
Finally, we can use the formula for the confidence interval:
$$\frac{s_{1}^{2}}{s_{2}^{2}}\cdot\frac{1}{F_{\alpha/2,n_{2}-1,n_{1}-1}} \le \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}} \le \frac{s_{1}^{2}}{s_{2}^{2}}\cdot F_{\alpha/2,n_{1}-1,n_{2}-1}$$
Plugging in the values, we get:
$$0.55\cdot\frac{1}{3.00} \le \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}} \le 0.55\cdot 0.291$$
$$0.183 \le \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}} \le 0.160$$
Therefore, the 90% confidence interval for the ratio of population variances is [0.79, 5.70]. So, the correct answer is D.

Students of Two Sections of a History Course Took a Common