91
CHAPTER 9
Section 91
91
a)
25
:
,
25
:
1
0
≠
=
μ
μ
H
H
Yes, because the hypothesis is stated in terms of the parameter of
interest, inequality is in the alternative hypothesis, and the value in the null and alternative
hypotheses matches.
b)
10
:
,
10
:
1
0
=
>
σ
σ
H
H
No, because the inequality is in the null hypothesis.
c)
50
:
,
50
:
1
0
≠
=
x
H
x
H
No, because the hypothesis is stated in terms of the statistic rather
than the parameter.
d)
3
.
0
:
,
1
.
0
:
1
0
=
=
p
H
p
H
No, the values in the null and alternative hypotheses do not match and
both of the hypotheses are equality statements.
e)
30
:
,
30
:
1
0
>
=
s
H
s
H
No, because the hypothesis is stated in terms of the statistic rather than the
parameter.
92
The conclusion does not provide strong evidence that the critical dimension mean equals 100nm.
The conclusion is that we don’t have enough evidence to reject the null hypothesis.
93
a)
nm
H
nm
H
20
:
,
20
:
1
0
<
=
σ
σ
b)
This result does not provide strong evidence that the standard deviation has not been reduced.
This result says that we do not have enough evidence to reject the null hypothesis. It is not
support for the null hypothesis.
94
a)
Newtons
H
Newtons
H
25
:
,
25
:
1
0
<
=
μ
μ
b)
No, this results only means that we do not have enough evidence to support
H
1
95
a)
α
= P(reject H
0
when H
0
is true)
= P(
X
≤
11.5 when
μ
= 12) =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
≤
−
4
/
5
.
0
12
5
.
11
/
n
X
P
σ
μ
= P(Z
≤
−
2)
= 0.02275.
The probability of rejecting the null hypothesis when it is true is 0.02275.
b)
β
= P(accept H
0
when
μ
= 11.25) =
(
)
25
.
11
5
.
11
=
>
μ
when
X
P
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
>
−
4
/
5
.
0
25
.
11
5
.
11
/
n
X
P
σ
μ
=
P(Z > 1.0)
= 1
−
P(Z
≤
1.0) = 1
−
0.84134 = 0.15866
The probability of accepting the null hypothesis when it is false is 0.15866.
c)
β
= P(accept H
0
when
μ
= 11.25) =
=
(
)
5
.
11
5
.
11
=
>
μ
when
X
P
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
>
−
4
/
5
.
0
5
.
11
5
.
11
/
n
X
P
σ
μ
= P(Z > 0) = 1
−
P(Z
≤
0) = 1
−
0.5 = 0.5
The probability of accepting the null hypothesis when it is false is 0.5
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92
96
a)
α
= P(
X
≤
11.5 
μ
= 12) =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
≤
−
16
/
5
.
0
12
5
.
11
/
n
X
P
σ
μ
= P(Z
≤
−
4) = 0.
The probability of rejecting the null, when the null is true, is approximately 0 with a sample size
of 16.
b)
β
= P(
X
> 11.5 
μ
=11.25) =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
>
−
16
/
5
.
0
25
.
11
5
.
11
/
n
X
P
σ
μ
= P(Z > 2) = 1
−
P(Z
≤
2)= 1
−
0.97725 = 0.02275.
The probability of accepting the null hypothesis when it is false is 0.02275.
c)
β
= P(
X
> 11.5 
μ
=11.5) =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
>
−
16
/
5
.
0
5
.
11
5
.
11
/
n
X
P
σ
μ
= P(Z > 0) = 1
−
P(Z
≤
0)
= 1
−
0.5 = 0.5
The probability of accepting the null hypothesis when it is false is 0.5.
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 Spring '09
 KAILASHKAPUR
 Null hypothesis, Statistical hypothesis testing

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