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Six Sigma for Electronics Design and Manufacturing
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dence; k is the number of boundaries, and DOF is the degrees of free-
dom. Selected values of the table are given in Table 5.3.
2
2.4.3 Example of goodness of fit to normal
distribution test
Thirty parts were selected from a production line that was assumed to
be normally distributed, and lengths were measured in m. The data
was sorted in ascending order and five boundaries were created for
the sorted data set. 2
Table 2.5 shows the original as well as the sorted data set of 30
measurements. The average is calculated at 8843.43 and the standard
deviation is 743. The boundary limits are a minimum of less then
8000 to a maximum of more than 9500 in 500 increments. The first
two boundary calculations are shown for illustration:
Boundary 1
P 1 = P( < 8000) = P{( – )/ }
z = (8000 – 8843.43)/743 = –1.135
P 1 = f(z) = 0.128 from normal distribution tables
Expected frequency = NP i = 30 · 0.128 = 3.84
2
2
2
contribution = (m 1 – nP 1 ) /nP 1 = (4 – 3.84) /3.84 = 0.0067
Boundary 2
P 2 = P(8000 < < 8500)
z 2 = (8500 – )/ = (8500 – 8843.43)/743 = –0.462
z 1 = –1.135 (from previous boundary)
P 2 = f(z 2 ) – f(z 1 ) = 0.3228 – 0.128 = 0.1948
Expected frequency = NP i = 30 · 0.1948 = 6.27
2
contribution = (m i – nP i ) /nP i = (8 – 5.844) /5.844 = 0.795
2
2
Other results for the remaining boundaries are shown in Table 2.5.
It can be seen that the total number of observed and expected fre-
quencies in all of the boundaries should be equal to the data set total
of 30. The total probability P i should also equal to 1, and the total ex-
pected frequency should equal 30.
The total chi-square value for the data set is 2.36, which falls be-
tween the limits of = 0.10 (90% confidence) of 1.064 and = 0.5 (50%
2
confidence) of 3.357 for the distribution with degrees of freedom
DOF = 4 (5 boundaries – 1), from Table 5.3. That implies that the data
set is normal since it corresponds with the normal distribution expec-
tations. A plot is shown of the data set values versus their normal