12
Chapter 1
A. Anderson-Darling Test for Normality
The MILHDBK 5F [1.8] pages 9-185 to 9-188 discusses this test which
requires the calculation of the mean, ~, and standard deviation, s, after
the raw data is processed by plotting or computer calculation. A variable
is developed
Z I = (X i -- 2)/S i = 1 ..... n
(1.44)
The Anderson-Darling Test, AD, statistic is
1~-~ 1 - 2i [ln(Fo[Zi]) ln(1 - Fo [Z(n ÷ 1 - i)] )~- n
(1.45)
AD
Li=I
where
Fo is the area Fo(x) under the Gaussian curve to the left of x
Then if
AD > 0.75211 + 0.75/n + 2.25/n2]
-~
(1.46)
The data is not normally distributed from the calculation for a 95% confidence level.
B. Anderson-Darling Test for Weilbullness
This [1.18] is a test for a three parameter Weibull fit of raw data and a similar
variable is
Zi = [(xi - ~50)/~50] 1~5° i = 1 ..... n
(1.47)
However/350, c~50, rso require data processing. The Anderson-Darling test
statistic
if
D = [~ 1 - 2i [ln(1Li== --n
- exp[Zi])+exp[Z(n+l_i)])]-n
(1.48)
AD > 0.757{1 + 0.2/v%]
-L
(1.49)
It is concluded the raw data is not part of a three parameter Weibull distribution for a 95% confidence level.
C. Qualification of Tests
When using the goodness of fit tests there is a five percent error on the test.
Further the tests may reject data even when a reasonable approximation
Chapter 1
A. Anderson-Darling Test for Normality
The MILHDBK 5F [1.8] pages 9-185 to 9-188 discusses this test which
requires the calculation of the mean, ~, and standard deviation, s, after
the raw data is processed by plotting or computer calculation. A variable
is developed
Z I = (X i -- 2)/S i = 1 ..... n
(1.44)
The Anderson-Darling Test, AD, statistic is
1~-~ 1 - 2i [ln(Fo[Zi]) ln(1 - Fo [Z(n ÷ 1 - i)] )~- n
(1.45)
AD
Li=I
where
Fo is the area Fo(x) under the Gaussian curve to the left of x
Then if
AD > 0.75211 + 0.75/n + 2.25/n2]
-~
(1.46)
The data is not normally distributed from the calculation for a 95% confidence level.
B. Anderson-Darling Test for Weilbullness
This [1.18] is a test for a three parameter Weibull fit of raw data and a similar
variable is
Zi = [(xi - ~50)/~50] 1~5° i = 1 ..... n
(1.47)
However/350, c~50, rso require data processing. The Anderson-Darling test
statistic
if
D = [~ 1 - 2i [ln(1Li== --n
- exp[Zi])+exp[Z(n+l_i)])]-n
(1.48)
AD > 0.757{1 + 0.2/v%]
-L
(1.49)
It is concluded the raw data is not part of a three parameter Weibull distribution for a 95% confidence level.
C. Qualification of Tests
When using the goodness of fit tests there is a five percent error on the test.
Further the tests may reject data even when a reasonable approximation
