E1C04 09/14/2010
14:7:43 Page 138
ASSUMPTIONS Variations between bearings fit a normal distribution.
FIND x
2
a
SOLUTION This problem could be restated as follows: What is the probability that s
2
x based on
20 measurements will not predict s
2
x for the entire batch?
For n ¼ N À 1 ¼ 19, and using Equation 4.26, the x
2 value is
x
2
a ðnÞ ¼ ns
2
x =s
2
¼ 30:16
Inspection of Table 4.5 shows x
2
:05 ð19Þ ¼ 30:1, so we can take a % 0.05. Taking x
2
a as a measure of
discrepancy due to random chance, we interpret this result as a 5% chance that a batch actually
within tolerance will be rejected. So there is a probability, P ¼ 1 À a, of 95% that s
2
x does predict the
s
2 for the batch. Rejecting a batch on this basis is a good decision.
Goodness-of-Fit Test
Just how well does a set of measurements follow an assumed distribution function? In Example 4.4,
we assumed that the data of Table 4.1 followed a normal distribution based only on the rough form of
its histogram (Fig. 4.2). A more rigorous approach would apply the chi-squared test using the
chi-squared distribution. The chi-squared test provides a measure of the discrepancy between the
measured variation of a data set and the variation predicted by the assumed density function.
To begin, construct a histogram of K intervals from a data set of N measurements. This
establishes the number of measured values, n j , that lie within the jth interval. Then calculate the
degrees of freedom in the variance for the data set, n ¼ N À m, where m is the number of restrictions
imposed. From n, estimate the predicted number of occurrences, n
0
j , to be expected from the
distribution function. For this test, the x
2 value is calculated from the entire histogram by
x
2
¼
P
j n j À n
0
j
À
Á 2
n
0 j
J ¼ 1; 2; . . . ; K
ð4:30Þ
The goodness-of-fit test evaluates the null hypotheses that the data are described by the assumed
distribution against the alternative that the data are not sampled from the assumed distribution. For
the given degrees of freedom, the better a data set fits the assumed distribution function, the lower its
x
2 value (left side of Table 4.5), whereas the higher the x
2 value (right side of Table 4.5), the more
dubious is the fit. For example, Pðx
2
a Þ ¼ 1 À a 0:05 leaves only a 5% or less chance that the
discrepancy is due to a systematic effect such as a different distribution, a good (unequivocol) result.
As with all finite samplings of a population, statistics can be used only to suggest what is most
probable. Conclusions are left to the user.
Example 4.7
Test the hypothesis that the variable x as given by the measured data of Table 4.1 is described by a
normal distribution.
KNOWN Table 4.1 and histogram of Figure 4.2
From Example 4.4: x ¼ 1:02; s x ¼ 0.16; N ¼ 20; K ¼ 7
138 Chapter 4 Probability and Statistics
14:7:43 Page 138
ASSUMPTIONS Variations between bearings fit a normal distribution.
FIND x
2
a
SOLUTION This problem could be restated as follows: What is the probability that s
2
x based on
20 measurements will not predict s
2
x for the entire batch?
For n ¼ N À 1 ¼ 19, and using Equation 4.26, the x
2 value is
x
2
a ðnÞ ¼ ns
2
x =s
2
¼ 30:16
Inspection of Table 4.5 shows x
2
:05 ð19Þ ¼ 30:1, so we can take a % 0.05. Taking x
2
a as a measure of
discrepancy due to random chance, we interpret this result as a 5% chance that a batch actually
within tolerance will be rejected. So there is a probability, P ¼ 1 À a, of 95% that s
2
x does predict the
s
2 for the batch. Rejecting a batch on this basis is a good decision.
Goodness-of-Fit Test
Just how well does a set of measurements follow an assumed distribution function? In Example 4.4,
we assumed that the data of Table 4.1 followed a normal distribution based only on the rough form of
its histogram (Fig. 4.2). A more rigorous approach would apply the chi-squared test using the
chi-squared distribution. The chi-squared test provides a measure of the discrepancy between the
measured variation of a data set and the variation predicted by the assumed density function.
To begin, construct a histogram of K intervals from a data set of N measurements. This
establishes the number of measured values, n j , that lie within the jth interval. Then calculate the
degrees of freedom in the variance for the data set, n ¼ N À m, where m is the number of restrictions
imposed. From n, estimate the predicted number of occurrences, n
0
j , to be expected from the
distribution function. For this test, the x
2 value is calculated from the entire histogram by
x
2
¼
P
j n j À n
0
j
À
Á 2
n
0 j
J ¼ 1; 2; . . . ; K
ð4:30Þ
The goodness-of-fit test evaluates the null hypotheses that the data are described by the assumed
distribution against the alternative that the data are not sampled from the assumed distribution. For
the given degrees of freedom, the better a data set fits the assumed distribution function, the lower its
x
2 value (left side of Table 4.5), whereas the higher the x
2 value (right side of Table 4.5), the more
dubious is the fit. For example, Pðx
2
a Þ ¼ 1 À a 0:05 leaves only a 5% or less chance that the
discrepancy is due to a systematic effect such as a different distribution, a good (unequivocol) result.
As with all finite samplings of a population, statistics can be used only to suggest what is most
probable. Conclusions are left to the user.
Example 4.7
Test the hypothesis that the variable x as given by the measured data of Table 4.1 is described by a
normal distribution.
KNOWN Table 4.1 and histogram of Figure 4.2
From Example 4.4: x ¼ 1:02; s x ¼ 0.16; N ¼ 20; K ¼ 7
138 Chapter 4 Probability and Statistics