We say that X has a Fisher distribution, denoted by F(n₁, n₂), if K₁/m₁ K₂/n₂ where K₁ ~ X², and K₂ ~ X₂ are two independent r.v.s and the n, 's are degrees of freedom. we have X Let X₁, X₂~ N(ux, ok) and Y₁,..., Ym~ N(μy, o) be two independent samples where the 4 parameters are unknown. We want to test the hypothesis Ho: o = o versus H₁: oo at the (1-a)% confidence level. Answer the following questions:
We say that X has a Fisher distribution, denoted by F(n₁, n₂), if K₁/m₁ K₂/n₂ where K₁ ~ X², and K₂ ~ X₂ are two independent r.v.s and the n, 's are degrees of freedom. we have X Let X₁, X₂~ N(ux, ok) and Y₁,..., Ym~ N(μy, o) be two independent samples where the 4 parameters are unknown. We want to test the hypothesis Ho: o = o versus H₁: oo at the (1-a)% confidence level. Answer the following questions:
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Transcribed Image Text:We say that X has a Fisher distribution, denoted by F(n₁, n₂), if
K₁/m₁
K₂/1₂
where K₁ ~ X₁ and K₂ ~ X₂ are two independent r.v.s and the n; 's are degrees of freedom.
Let X₁, X~ N(ux, ok) and Y₁,..., Ym~ N(μy, oz) be two independent samples where
the 4 parameters are unknown. We want to test the hypothesis
Ho: o = o versus H₁: oo at the (1-a)% confidence level.
we have
Answer the following questions:
(a) If
n
S = n²x.
n - 1
i=1
X =
(X-X₂)² and S3 =
m 1
-
m
Σ(Υ; - Ym),
j=1
(2)
what is the distribution of
under Ho and why?
(b) If F₁-a (n₁, n₂) denotes the real number for which the area to its left under the F(n1, n₂)
density is equal to 1-a, then find a test statistic to test the hypothesis above and write
explicitly under which condition we would reject Ho.
Note: The support of the F(n₁, n₂) density is (0,00) and the shape of its density is
similar to a x² density. In particular, it is not symmetric.
(c) Suppose we have the following data, assumed to follow the model above:
(xi): 10.62, 10.58, 10.33, 10.72, 10.44, 10.74
(y): 10.50, 10.52, 10.58, 10.62, 10.55, 10.51, 10.53
Test the hypothesis that o = o at the 90% confidence level, using the test from item
(b).
(Note: F0.05,5,6 = 0.20 and Fo.95,5,6 = 4.39.)
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