Question

(Round all intermediate calculations to at least 4 decimal places.) Consider the following measures based on independently drawn samples from normally distributed populations: Use Table 4.

Sample 1: s21s12 = 250, and n1 = 16

Sample 2: s22s22 = 231, and n2 = 11

  

a.	Construct the 90% interval estimate for the ratio of the population variances. (Round "F" value and final answers to 2 decimal places.)

Confidence interval

to

  

b.	Using the confidence interval from Part a, test if the ratio of the population variances differs from one at the 10% significance level. Explain.

The 90% confidence interval (Click to select)containsdoes not contain the value 1. Thus, we (Click to select)cannotcanconclude that the population variances differ at the 10% significance level.

Answer 1

Solution

Back-up Theory

Given X ~ N(μ₁, σ₁²), Y ~ N(μ₂, σ₂²), n₁ ≠ n₂

100(1 - α) % Confidence Interval for σ₁²/ σ₂² is:

{(s₁²/ s₂²)/ (F_{n1 – 1, n2 – 1,} α/2), (s₁²/ s₂²) /(F_{n1 – 1, n2 – 1, 1 -} α/2)} where s₁, s₂ are respective sample standard deviations;   F_{n1 – 1, n2 – 1,α/2} and F_{n1 – 1, n2 – 1, 1 -α/2} are respectively upper and lower (α/2) percent point of F-distribution with degrees of freedom (n₁ – 1) and (n₂ – 1); and n₁ , n₂ are sample sizes.

Now to work out the solution,

Part (a)

Here α = 0.1, s₁² = 250, s₂² = 231, n₁ = 16, n₂ = 11.

Substituting these values,

90% Confidence Interval for ratio of population variances: 0.38 to 2.75 Answer

Details of calculations

R = (s12/ s22); F1 = (Fn1 – 1, n2 – 1, α/2); F2 = (Fn1 – 1, n2 – 1, 1 - α/2)

n1	16
n2	11
s1^2	250
S2^2	231
α	0.1
R	1.082251
F1	2.845017
F2	0.393125
LB	0.380402
UB	2.752942
n1 - 1	15
n2 - 1	10
α/2	0.05
1 - (α/2)	0.95

Part (b)

The 90% Confidence Interval, 0.38 to 2.75 does contain the value 1. Thus, we cannot conclude that the population variances differ. Answer

DONE