9-29
Section 9-4
9-63
a)
α
=0.01, n=20, from table V we find the following critical values 6.84 and 38.58
b)
α
=0.05, n=12, from table V we find the following critical values 3.82 and 21.92
c)
α
=0.10, n=15, from table V we find the following critical values 6.57 and 23.68
9-64
a)
α
=0.01, n=20, from table V we find
=
−
2
1
,
n
α
χ
36.19
b)
α
=0.05, n=12, from table V we find
=
−
2
1
,
n
19.68
c)
α
=0.10, n=15, from table V we find
=
−
2
1
,
n
21.06
9-65
a)
α
=0.01, n=20, from table V we find
=
−
−
2
1
,
1
n
7.63
b)
α
=0.05, n=12, from table V we find
=
−
−
2
1
,
1
n
4.57
c)
α
=0.10, n=15, from table V we find
=
−
−
2
1
,
1
n
7.79
9-66
a) 2(0.1)<
P
-value<2(0.5), then 0.2<
P
-value<1
b) 2(0.1)<
P
-value<2(0.5), then 0.2<
P
-value<1
c) 2(0.05)<
P
-value<2(0.1), then 0.1<
P
-value<0.2
9-67
a) 0.1<1-
P
<0.5 then 0.5<
P
-value<0.9
b) 0.1<1-
P
<0.5 then 0.5<
P
-value<0.9
c) 0.99<1-
P
<0.995 then 0.005<
P
-value<0.01
9-68
a) 0.1<
P
-value<0.5
b) 0.1<
P
-value<0.5
c) 0.99<
P
-value<0.995
9-69
a) In order to use the
χ
2
statistic in hypothesis testing and confidence interval construction, we need
to assume that the underlying distribution is normal.
1) The parameter of interest is the true standard deviation of performance time
σ
.
However,
the answer can be found by performing a hypothesis test on
σ
2
.
2) H
0
:
σ
2
= .75
2
3) H
1
:
σ
2
>.75
2
4)
α
= 0.05
5)
χ
0
2
=
()
ns
−
1
2
2
σ
6) Reject H
0
if
2
1
,
2
0
−
>
n
where
30
.
26
2
16
,
05
.
0
=
7) n = 17, s = 0.09
χ
0
2
=
23
.
0
75
.
)
09
.
0
(
16
)
1
(
2
2
2
2
=
=
−
σ
s
n
8) Because 0.23 < 26.30 do not reject H
0
and conclude there is insufficient evidence to indicate
the true variance of performance time content exceeds 0.75
2
at
α
= 0.05.
P
-value:
Because
χ
0
2
=0.23 the
P
-value>0.995
b) The 95% one sided confidence interval given below, includes the value 0.75. Therefore, we are
not be able to conclude that the standard deviation is greater than 0.75.