The
-error can be computed using Equation 9-12 as follows:
Thus, the probability is about 0.7 that the semiconductor manufacturer will fail to
conclude that the process is capable if the true process fraction defective is
p
= 0.03 (3%). That is, the power of the test against this particular alternative is only
about 0.3. This appears to be a large
-error (or small power), but the difference
between
p
= 0.05 and
p
= 0.03 is fairly small, and the sample size
n =
200 is not
particularly large.
Calculations are on the board! (an alternative way)
67
.
0
44
.
0
1
200
/
03
.
0
1
03
.
0
200
/
95
.
0
05
.
0
)
645
.
1
(
03
.
0
05
.
0
1

9-5 Tests on a Population Proportion
Example 9-11
Suppose that the semiconductor manufacturer was willing to accept a
-error
as large as 0.10 if the true value of the process fraction defective was
p =
0.03. If the manufacturer continues to use a = 0.05, what sample size
would be required?
The required sample size can be computed from Equation 9-15 as follows:
where we have used
p
= 0.03 in Equation 9-15.
Conclusion: Note that
n =
832 is a very large sample size. However, we are
trying to detect a fairly small deviation from the null value
p
0
=
0.05.
832
05
.
0
03
.
0
97
.
0
03
.
0
28
.
1
95
.
0
05
.
0
645
.
1
~
2
n

9-7 Testing for Goodness of Fit
•
The test is based on the chi-square distribution.
•
Assume there is a sample of size
n
from a population whose probability
distribution is unknown.
•
Let
O
i
be the observed frequency in the
i
th class interval.
•
Let
E
i
be the expected frequency in the
i
th class interval.
The
test statistic
is
(9-16)
k
i
i
i
i
E
E
O
X
1
2
2
0
)
(

9-7 Testing for Goodness of Fit
EXAMPLE 9-12
Printed Circuit Board Defects
Poisson Distribution
The number of defects in printed circuit boards is hypothesized to follow a
Poisson distribution. A random sample of
n =
60 printed boards has been
collected, and the following number of defects observed.
Number of
Defects
Observed
Frequency
0
32
1
15
2
9
3
4

9-7 Testing for Goodness of Fit
Example 9-12
The mean of the assumed Poisson distribution in this example is unknown and
must be estimated from the sample data. The estimate of the mean number of
defects per board is the sample average, that is, (32·0 + 15·1 + 9·2 + 4·3)/60 =
0.75. From the Poisson distribution with parameter 0.75, we may compute
p
i
, the
theoretical, hypothesized probability associated with the
i
th class interval. Since
each class interval corresponds to a particular number of defects, we may find the
p
i
as follows:
041
.
0
1
)
3
(
133
.
0
!
2
75
.
0
)
2
(
354
.
0
!
1
75
.
0
)
1
(
472
.
0
!
0
75
.
0
)
0
(
3
2
1
4
2
75
.
0
3
1
75
.
0
2
0
75
.
0
1
p
p
p
X
P
p
e
X
P
p
e
X
P
p
e
X
P
p

9-7 Testing for Goodness of Fit
Example 9-12
The expected frequencies are computed by multiplying the sample size
n =
60
times the probabilities
p
i
.
That is,
E
i
= np
i
.
The expected frequencies follow:
Number of Defects
Probability
Expected
Frequency
0
0.472
28.32
1
0.354
21.24
2
0.133
7.98
3 (or more)
0.041
2.46

9-7 Testing for Goodness of Fit
Example 9-12
Since the expected frequency in the last cell is less than 3, we combine the
last two cells:
The chi-square test statistic in Equation 9-16 will have
k
p
1 = 3
1
1 = 1
degree of freedom, because the mean of the Poisson distribution was
estimated from the data.

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- Normal Distribution, Null hypothesis, Statistical hypothesis testing