Let S2X and S2Y be the respective variances of two independent random samples of

Question

Let S2X and S2Y be the respective variances of two independent random samples of sizes n and m from N(mu X, theta2X) and N(mu Y, theta 2Y). Use the fact that F=[S2Y/theta 2Y]/[S2X/theta 2X] has an F distribution, with parametersr1 = m-1 and r2 = n-1, to rewrite P(c lessthanorequalto F lessthanorequalto d) = 1-a, where c = F1 -a/2(r1, r2) and d = Fa/2(r1, r2), so that P(c S^2_X/S^2_Y lessthanorequalto sigma^2_X/sigma^2_Y lessthanorequalto d S^2_X/S^2_Y) = 1 - alpha. If the observed values are n = 13, m = 9, 12s2x= 128.41, and 8s2y= 36.72, show that a 98% confidence interval for the ratio of the two variances, theta 2X/theta 2Y, is [0.41, 10.49], so that [0.64, 3.24] is a 98% confidence interval for theta X/theta Y.

Explanation / Answer

Result:

Variance ratio = (128.41/12)/(36.72/8) =2.3313

F(12,8) critical value at 98% level = (0.22, 5.67)

Lower limit for variance = 2.3313/5.67 =0.41

upper limit for variance = 2.3313/0.22 =10.59

Lower limit for sd = sqrt(0.41) =0.64

upper limit for sd   =sqrt(10.59) =3.25

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.