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An UMP Test for the Mean of a Normal DistributionStat 305 Spring Semester 2006

A random sample of size n is taken from a normal random variable X with unknownmean but with known variance 2. Formulate H0 and H1 as follows.

H0: = 0H1: > 0

We would like to devise a UMP test for the above set of hypotheses. To do this, let us rstdevise a test for the following set of simple hypotheses where

1>

0.

H0: = 0H1: = 1

We will apply the Neyman-Pearson Lemma to this set of simple hypotheses.Recall that the likelihood function evaluated at is

fX(x1;:::;xn j ) = 1

p2n

exp

1

2

nXi=1

xi

2!

.

Therefore, we will reject H0 if

fX(x1;:::;xn j 0)fX(x1;:::;xn j 1)

=

1p2

nexp

1

2

nPi=1

xi0

2

1p2

nexp

1

2

nPi=1

xi1

2 < 1k .

Therefore, simplifying, we reject H0 if

exp 1

22

nXi=1

(xi 0)2 nXi=1

(xi 1)2!!

22

ln1

k

.

Expanding the sums, we have thatnXi=1

x2i

! 2nxn0 + n20

nXi=1

x2i

!+ 2nxn1 n21 > 22 ln

1

k

1

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or that

2nxn(1 0) > 22 ln

1

k

n2

0+ n2

1.

Therefore,

xn >22 ln(1=k) n2

0+ n2

1

2n(1

0)

= k.

Thus, the critical region has the form xn > k. If we choose that () = 0,0 = (

; 0)

= P(Xn > k j =

0)

= P

Xn 0

=p

n>

k 0=p

n

= P

Z >

k 0

=p

n

.

Therefore,

k 0=pn

= z0

or, equivalently,

k = 0

+ z0

p

n

.

Therefore, we have derived the form of the decision rule.

:Decision Rule

Reject H0 if Xn > 0 + z0 pn (or equivalently, z > z0)

Note that the decision rule does not depend on 1. Therefore, any 1 > 0 would result inthe exactly the same critical region. Therefore, above decision rule constitutes an UMP testfor the mean.

The power of the test can be easily calculated. At = 1

> 0, the power is calculated

by

(; 1) = P

Xn > 0 + z0

p

n

j =

1

= P

Xn 1

=p

n>

0 1=p

n+ z0

= P

Z >

0

1

=p

n+ z0

= 1 0 1=pn + z0 .Therefore, () =

0

1

=pn

+ z0

. Notice that when

1=

0, the power is

1

0

1

=p

n+ z0

= 1 (z0) = 1 (1 0) = 0 = ()

as expected of course.

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