2.
Null hypothesis:
H
0
:
2
= 0.01
3.
Alternative hypothesis:
H
1
:
2
> 0.01

9-4 Hypothesis Tests on the Variance and Standard
Deviation of a Normal Distribution
Example 9-8
4.
Test statistic:
The test statistic is
5.
Reject
H
0
:
Use
=
0.05
,
and reject
H
0
if
.
6.
Computations:
7.
Conclusions:
Since
, we conclude that
there is no strong evidence that the variance of fill volume exceeds 0.01
(fluid ounces)
2
. So there is no strong evidence of a problem with
incorrectly filled bottles.
2
0
2
1
s
n
14
.
30
2
07
.
29
01
.
0
)
0153
.
0
(
19
14
.
30
07
.
29
2

9-5 Tests on a Population Proportion
9-5.1 Large-Sample Tests on a Proportion
Many engineering decision problems include hypothesis testing about
p
.
An appropriate
test statistic
is
(9-10)
0
0
1
0
:
:
H
p
p
H
p
p
)
1
(
0
0
0
0
p
np
np
X
Z

9-5 Tests on a Population Proportion
Another form of the test statistic
Z
0
is
or
n
p
p
p
n
X
Z
/
)
1
(
/
0
0
0
0
n
p
p
p
P
Z
/
)
1
(
ˆ
0
0
0
0

9-5 Tests on a Population Proportion
EXAMPLE 9-10
Automobile Engine Controller
A semiconductor
manufacturer produces controllers used in automobile engine applications.
The customer requires that the process fallout or fraction defective at a critical
manufacturing step not exceed 0.05 and that the manufacturer demonstrate
process capability at this level of quality using
= 0.05. The semiconductor
manufacturer takes a random sample of 200 devices and finds that four of
them are defective. Can the manufacturer demonstrate process capability for
the customer?
We may solve this problem using the seven-step hypothesis-testing procedure
as follows:
1.
Parameter of Interest:
The parameter of interest is the process fraction
defective
p.
2.
Null hypothesis:
H
0
:
p
= 0.05
3.
Alternative hypothesis:
H
1
:
p <
0.05
This formulation of the problem will allow the manufacturer to make a strong
claim about process capability if the null hypothesis
H
0
:
p
= 0.05 is rejected.

9-5 Tests on a Population Proportion
Example 9-10
4.
The test statistic is (from Equation 9-10)
where
x
= 4,
n =
200, and
p
0
=
0.05.
5.
Reject H
0
if:
Reject
H
0
:
p =
0.05 if the p-value is less than 0.05.
6.
Computations:
The test statistic is
7.
Conclusions:
Since z
0
=
1.95, the
P-
value is
(
1.95) = 0.0256, so we reject
H
0
and conclude that the process fraction defective
p
is less than 0.05.
Practical Interpretation: We conclude that the process is capable.
)
1
(
0
0
0
0
p
np
np
x
z
95
.
1
)
95
.
0
)(
05
.
0
(
200
)
05
.
0
(
200
4
0
z

9-5 Tests on a Population Proportion
9-5.2 Type II Error and Choice of Sample Size
For a two-sided alternative
(9-11)
If the alternative is
p
<
p
0
(9-12)
If the alternative is
p
>
p
0
(9-13)
n
p
p
n
p
p
z
p
p
n
p
p
n
p
p
z
p
p
/
)
1
(
/
)
1
(
/
)
1
(
/
)
1
(
0
0
/2
0
0
0
/2
0
n
p
p
n
p
p
z
p
p
/
)
1
(
/
)
1
(
1
0
0
0
n
p
p
n
p
p
z
p
p
/
)
1
(
/
)
1
(
0
0
0

9-5 Tests on a Population Proportion
9-5.3 Type II Error and Choice of Sample Size
For a two-sided alternative
(9-14)
For a one-sided alternative
(9-15)
2
0
0
0
/2
)
1
(
)
1
(
p
p
p
p
z
p
p
z
n
2
0
0
0
)
1
(
)
1
(
p
p
p
p
z
p
p
z
n

9-5 Tests on a Population Proportion
Example 9-11
Automobile Engine Controller
Type II Error
Consider the semiconductor manufacturer from Example 9-10. Suppose that its
process fallout is really
p
= 0.03. What is the
-error for a test of process capability
that uses
n
= 200 and a = 0.05?

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- Normal Distribution, Null hypothesis, Statistical hypothesis testing