Chapter 15:
DistributionFree Procedures
458
3.
We test
39
.
7
:
0
=
m
H
vs.
39
.
7
:
≠
m
a
H
, so a two tailed test is appropriate.
With n =
14 and
025
.
2
/
=
a
, Table A.13 indicates that H
o
should be rejected if either
21
84
≤
≥
+
or
s
.
The
)
39
.
7
(

i
x
’s are .37, .04, .05, .22, .11, .38, .30, .17, .06, .44,
.01, .29, .07, and .25, from which the ranks of the three positive differences are 1, 4, and 13.
Since
21
18
≤
=
+
s
, H
o
is rejected at level .05.
4.
The appropriate test is
30
:
0
=
m
H
vs.
30
:
<
m
a
H
.
With n = 15, and
10
.
=
a
, reject
H
o
if
( 29
37
83
2
16
15
=

≤
+
s
.
i
x
)
30
(

i
x
ranks
i
x
)
30
(

i
x
ranks
30.6
0.6
3*
31.9
1.9
5*
30.1
0.1
1*
53.2
23.2
15*
15.6
14.4
12
12.5
17.5
13
26.7
3.3
7
23.2
6.8
11
27.1
2.9
6
8.8
21.2
14
25.4
4.6
8
24.9
5.1
10
35
5
9*
30.2
0.2
2*
30.8
0.8
4*
S
+
= 39, which is not
37
≤
, so H
o
cannot be rejected.
There is not enough evidence to prove
that diagnostic time is less than 30 minutes at the 10% significance level.
5.
The data is paired, and we wish to test
0
:
0
=
D
H
m
vs.
0
:
≠
D
a
H
m
. With n = 12 and
05
.
=
a
, H
o
should be rejected if either
64
≥
+
s
or if
14
≤
+
s
.
d
i
.3
2.8
3.9
.6
1.2
1.1
2.9
1.8
.5
2.3
.9
2.5
rank
1
10*
12*
3*
6*
5
11*
7*
2*
8*
4*
9*
72
=
+
s
, and
64
72
≥
, so H
o
is rejected at level .05.
In fact for
01
.
=
a
, the critical value
is c = 71, so even at level .01 H
o
would be rejected.