**Notes:****Binomial Calculations
...**
For an inspection with a probability of a hit, *p*,
there are four possible outcomes of two independent inspections: HH, HM MH, MM. The
probability of HH is *p*�*p*, and the probability of MM is (1-*p*)�(1-*p*).
The probability of HM and MH are each *p*�(*1-p*), so the probability of one
hit and one miss in either sequence is *2�p*�(*1-p*). A similar argument is
made for 3, 4, or *N* independent inspections. Collecting like terms we have _{h+m}C_{h}�p^{h}(1-*p*)^{m}
where *h* and *m* are the number of Hits and Misses, and _{N}C_{n}
, the combination of *N* things taken *n* at a time, is the number of different
sequences in which a particular combination of hits and misses can occur. Now, if *POD*=0.9,
the probability of 29 successes in 29 attempts is 1* *�* *(0.9)^{29 }�*
*(0.1)^{0} = 0.047, so it would be improbable that the string of successes
resulted from a *POD* *less* than 0.9. **[****return to text]**
**Twentnine of twentynine
...**
The 29 of 29 method assumes that (1) all cracks are nominally
identical, (2) the sample is random and representative of the actual application, (3)
inspections are independent, and (4) 90% probability of detection is adequate for
aerospace applications.
In actuality: (1) It is a practical impossibility to
make cracked specimens with a sufficiently small range of sizes that the influence of
cracksize can be ignored. (2) Cracks nominally the same size often exhibit large
differences in detectability. These "hard to find" cracks are representative of
real inspection circumstances. Still, it is not uncommon for them to be ruled somehow
"unfair" and then removed from consideration. Also, since all 29 cracks must be
found to "pass" the test, inspectors soon learn that it pays to guess. (3)
Unfortunately, some NDE "experts" have sanctioned using six specimens, each
inspected five times. Since 6* *�* *5 = 30, the misguided reasoning goes, this must be at least
as good as a scheme based on 29 cracks. And finally, (4) Would you submit to a blood
transfusion if you were __very__ confident that the test to determine the presence of
an Andromeda Strain microbe was at least 90% effective? (I
wouldn't.) **[****return to text]**
**Difficulties
with Lotus 1-2-3 ...**
Polynomial approximations of the inverse normal
function from Abramowitz and Stegun (1965) were also unsuccessful in Lotus 1-2-3 which has
trouble with the implementation of its solver. The Version 4.01 for Windows user's
manual suggests that the solver has difficulties with involved or nested functions
(p.244), and experience corroborates this: larger datasets are more troublesome for Lotus
1-2-3 than smaller ones, and simpler link functions are more likely to reach convergence
than polynomial links. In any event Lotus 1-2-3 allows little user control over the
iteration mechanics unlike EXCEL and Quattro Pro which permit choice of scaling, linear or
nonlinear responses, and central or endpoint numerical differentiation. Lotus 1-2-3
can find mle's for the logit model with smaller size datasets,
however. **[****return to text]**
**Spreadsheet
Tip 1**
This will require that you redefine this sum for each new
dataset. I once estimated parameters using only half the data because I
"conveniently" located the summation function at the top of the column, but
forgot to redefine it for a larger dataset. You can still display the sum in cell
H5, but refer to the cell at the bottom of the column. Then if you forget to redefine,
cell H5 will display only an individual lnlikelihood calling attention to your
oversight. **[return to text]**
**Spreadsheet Tip 2**
I used a loop construction to copy the next row into the iteration
workspace and copy the results back into the appropriate table after each convergence.
Don't forget to use the PASTE VALUES, so the results won't be changed by the subsequent
iterations. One caution, however: iterative numerical methods can be unstable in a steep
gradient and are thus sensitive to initial guesses. Before invoking a macro, experiment
with the guess values so that the starting value of L is not too far its target value. My
macros use the current solution as guesses for next contour point and seem to behave well
when started well. **[return to text]** |