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6
Suggested solution
Question 1
Answer: For 99% certainty, z(.995)=2.575.
(a)
Since P can be any number between 0 and 1, P(1P) maxizes at 0.5x0.5=0.25.
n=(2.575/0.02)2 P(1P)=(2.575/0.02)2 0.5(10.5)=16576.56x0.25=4144.14=4145
(b)
Since P is between 0 and 0.1, P(1P) maximizes at 0.1(10.1)=0.09
n=16576.56x0.09=1491.9=1492
Question 2
Question 3:
Note that
H
0
:
p
= 0.5 vs. H
1
:
p
< 0.5
where
p
is the probability of getting heads.
(a) Let
X
be the number of heads in 10 tossings. Then, the probability of Type I error
is given by Pr{X <
1} under H
0
. That is, X ~ Bin(10, 0.5). Therefore,
Pr{X <
1} = Pr{X = 0} + Pr{X = 1}
= 0.000977 + 0.009766
= 0.010743
(b) If CLT holds for
n
= 10, then
ˆ
p
= X/10 ~ N(0.5, 0.5(0.5)/10) under H
0
. In that
case, the probability of Type I error is given by:
Pr{X <
1} = Pr{
ˆ
p
<
0.1} = Pr{Z <
(0.1
−
0.5) /
0.5(0.5)/10 }
= Pr{Z <
−
2.53} = 0.0057
(c) The discrepancy is due to the fact that
n
= 10 is too small for CLT. Therefore,
0.010743 is more close to the true probability of Type I error.
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This note was uploaded on 08/26/2010 for the course ISOM ISOM111 taught by Professor Anthonychan during the Spring '09 term at HKUST.
 Spring '09
 AnthonyChan

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