whose PDF is η «η IULIII d Oill sa ле поп an f(x|x): It can be shown that 1 X₁ ~ Gamma(n, λ) with PDF =1 = Ae-, for x > 0, A > 0. n-1 Xn f(x|n, x) = ·xn e I(n) By the property of Gamma distributions, it follows that n 2X ➤ Xi ~ Gamma i=1 2n 1 " 2 2 - Xx = x²(2n). with late ₂ (a) Based on the fact that 2X1 X₁ ~ x²(2n), derive a (1 − a) × 100% confidence interval for X. (b) Suppose that we have observed the following data of n = 8 7.8 2.4 3.0 20.0 4.2 4.8 7.4 6.7 from exponential (A). Construct a 95% confidence interval for A based on the formula that you have derived in Part (a).

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5. Suppose that X₁,..., Xn form a random sample from an exponential distribution with rate X,
whose PDF is
-xx
ƒ(x|X) = Ae¯¤, for x > 0, λ > 0.
Gamma(n, A) with PDF
It can be shown that Σ₁ X₁ ~
Xi
In
I(n)
By the property of Gamma distributions, it follows that
f(x|n, x)
=
n
20 ΣΧ ~ Gamma
i=1
xn-¹e-xx
2n 1
22
= x²(2n).
(a) Based on the fact that 2X₁ X; ~ x²(2n), derive a (1 − a) × 100% confidence interval
2λ
for X.
(b) Suppose that we have observed the following data of n = 8
7.8 2.4 3.0 20.0 4.2 4.8 7.4 6.7
from exponential (X). Construct a 95% confidence interval for A based on the formula that
you have derived in Part (a).
Transcribed Image Text:5. Suppose that X₁,..., Xn form a random sample from an exponential distribution with rate X, whose PDF is -xx ƒ(x|X) = Ae¯¤, for x > 0, λ > 0. Gamma(n, A) with PDF It can be shown that Σ₁ X₁ ~ Xi In I(n) By the property of Gamma distributions, it follows that f(x|n, x) = n 20 ΣΧ ~ Gamma i=1 xn-¹e-xx 2n 1 22 = x²(2n). (a) Based on the fact that 2X₁ X; ~ x²(2n), derive a (1 − a) × 100% confidence interval 2λ for X. (b) Suppose that we have observed the following data of n = 8 7.8 2.4 3.0 20.0 4.2 4.8 7.4 6.7 from exponential (X). Construct a 95% confidence interval for A based on the formula that you have derived in Part (a).
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