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 ,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

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We'll use the F-distribution to calculate the confidence interval. The test statistic is:

F = (s1^2/σ1^2)/(s2^2/σ2^2)

The confidence interval for the ratio of population variances is:

F = [(n2 - 1)s2^2]/[(n1 - 1)s1^2] ~ F(n2-1, n1-1) at a 90% level of confidence.

Using the given values, the confidence interval is:

F = [(15)(11.31)]/[(10)(4.15)] = 3.12

Looking at the F-distribution table with degrees of freedom 15 and 10, the critical values are 2.48 and 4.02 at the 5% level of significance. Since we're using a 90% level of confidence, the critical values are:

Lower bound: 1/f(0.95, 15, 10) = 1/3.24 = 0.31
Upper bound: f(0.95, 10, 15) = 1/2.50 = 0.40

Therefore, the 90% confidence interval for the ratio of population variances is [0.31, 0.40]. Dividing 1 by each value, we get [2.50, 3.23]. So the correct choice is D, [0.79,5.70].

Students of Two Sections of a History Course Took a Common