If we do not know if the mimic is larger or smaller that the model, then we use a twosided test. Below is the
R
commands for the power function with
↵
= 0
.
05
and
n
= 16
observations.
> zalpha = qnorm(.975)
> mu0<10
> sigma0<3
> mu<(600:1400)/100
> n<16
> pi<1pnorm(zalpha(mumu0)/(sigma0/sqrt(n)))
+pnorm(zalpha(mumu0)/(sigma0/sqrt(n)))
> plot(mu,pi,type="l")
We shall see in the the next topic how these tests follow from extensions of the likelihood ratio test for simple
hypotheses.
The next example is unlikely to occur in any genuine scientific situation. It is included because it allows us to
compute the power function explicitly from the distribution of the test statistic. We begin with an exercise.
Exercise 18.6.
For
X
1
, X
2
, . . . , X
n
independent
U
(0
,
✓
)
random variables,
✓
2
⇥
= (0
,
1
)
. The density
f
X
(
x

✓
) =
⇢
1
✓
if
0
< x
✓
,
0
otherwise
.
Let
X
(
n
)
denote the maximum of
X
1
, X
2
, . . . , X
n
, then
X
(
n
)
has distribution function
F
X
(
n
)
(
x
) =
P
✓
{
X
(
n
)
x
}
=
⇣
x
✓
⌘
n
.
Example 18.7.
For
X
1
, X
2
, . . . , X
n
independent
U
(0
,
✓
)
random variables, take the null hypothesis that
✓
lands in
some normal range of values
[
✓
L
,
✓
R
]
. The alternative is that
✓
lies outside the normal range.
H
0
:
✓
L
✓
✓
R
versus
H
1
:
✓
<
✓
L
or
✓
>
✓
R
.
283
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