2. Consider X1, X2, X, iid N(4,0²) where o is known. In this problem we show that among the (1 – a) confidence intervals of the form , P1 + P2 = a, the one with the choice pi = P2 = is the shortest. This explains why the symmetric confidence interval is preferred among all possible choices (unless we want to explicitly construct a one sided confidence interval). a. Without loss of generality, suppose pi < P2. Draw a picture of the standard Normal pdf with z,ı: p2; Za/2: Use it to show that Zp2 < Za/2 < Zpı•

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 32E
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2. Consider X1, X2,. , X, iid N(u, o2) where o is known. In this problem we show
that among the (1 - a) confidence intervals of the form
P1+P2 a,
the one with the choice pi = p2 = is the shortest. This explains why the symmetric
confidence interval is preferred among all possible choices (unless we want to explicitly
construct a one sided confidence interval).
a. Without loss of generality, suppose pi < P2. Draw a picture of the standard
Normal pdf with zp1, žp2, Za/2. Use it to show that
Zp2 < Za/2 < žp.
b. Show that
Za/2
T" Sz(dz) =
fz(dz),
Za/2
Zp2
where fz is the standard normal pdf.
c. The key here is fz(2) is decreasing in z for z > 0. Use this and part b to show
с.
that Zpi
Za/2
Za/2 - 2p2.
d. Conclude that the length of the 2 sides CI with choice pi = P2 = a/2 is the
shortest.
Transcribed Image Text:2. Consider X1, X2,. , X, iid N(u, o2) where o is known. In this problem we show that among the (1 - a) confidence intervals of the form P1+P2 a, the one with the choice pi = p2 = is the shortest. This explains why the symmetric confidence interval is preferred among all possible choices (unless we want to explicitly construct a one sided confidence interval). a. Without loss of generality, suppose pi < P2. Draw a picture of the standard Normal pdf with zp1, žp2, Za/2. Use it to show that Zp2 < Za/2 < žp. b. Show that Za/2 T" Sz(dz) = fz(dz), Za/2 Zp2 where fz is the standard normal pdf. c. The key here is fz(2) is decreasing in z for z > 0. Use this and part b to show с. that Zpi Za/2 Za/2 - 2p2. d. Conclude that the length of the 2 sides CI with choice pi = P2 = a/2 is the shortest.
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