8th Order Polynomial Trendline - Is there an easy way?

[Frank & Pam Hayes]

Excel can handle polynomial trendlines up to 6th order through the Add
Trendline command. I would like to be able to add a trendline up to the 8th
order. Google searches showed me how to compute this, but I am wondering if
there is an easier way. I tried using VBA :

ActiveChart.SeriesCollection(1).Select
ActiveChart.SeriesCollection(1).Trendlines.Add(Type:=xlPolynomial,
Order:=6 _
, Forward:=0, Backward:=0, DisplayEquation:=False, DisplayRSquared:=
_
False).Select

and then changing the Order:=8 but it would not work.

Any great ideas?

Thanks,

Frank

 

[Bernard Liengme]

Use LINEST to generate coefficients - see
www.stfx.ca/people/bliengme/ExcelTips
Use the coefficients to generate trendline data

Do your really have data that can <meaningfully> be fitted to 8th order?

Best wishes

 

[Jon Peltier]

Bernard -

"Do your really have data that can <meaningfully> be fitted to 8th order?"

This was my first thought. I've found most higher than third order are ill-advised.

- Jon
-------
Jon Peltier, Microsoft Excel MVP
Peltier Technical Services
Tutorials and Custom Solutions
http://PeltierTech.com/
_______

 

[Frank & Pam Hayes]

Thank you both for your comments. Let me tell you what I am am doing and
ask your advice. I am constructing a cumulative probability curve of
values, bounded by 0% and 100% probabilities. The curve generally follows a
basic S shape with the very low probabilies and very high probabilities
going asyntotic. It is very important that the trendline does not go below
0% or above 100%.

Here are the data I am using:

X-values (Expected NPV)

(123.981)
(101.476)
(85.174)
(68.565)
(68.002)
(50.994)
(41.568)
(39.868)
(34.903)
(33.363)
(32.627)
(22.587)
(21.014)
(10.907)
(9.361)
(8.694)
(5.820)
(5.659)
(2.764)
0.829
1.342
3.878
5.467
6.090
8.851
10.436
10.795
11.237
13.438
14.695
17.310
19.057
19.765
21.193
21.666
22.925
25.311
34.417
35.560
36.511
38.089
38.796
42.145
44.587
56.408
58.991
59.794
61.564
61.920
66.724
71.064
74.772
77.386
79.294
85.779
89.446
93.157
93.761
93.858
98.944
99.150
99.998
101.652
117.539
129.741
191.151
206.613
219.735
220.505
255.970
256.805
260.091
279.759
289.067
303.237
308.903
312.172
362.389
366.679
384.683
438.860


Y-Values (Cumulative Probability)
0.008
0.023
0.027
0.043
0.051
0.066
0.070
0.078
0.086
0.094
0.102
0.133
0.141
0.156
0.172
0.180
0.195
0.199
0.203
0.234
0.250
0.258
0.262
0.270
0.277
0.293
0.301
0.305
0.309
0.313
0.320
0.336
0.340
0.344
0.352
0.355
0.363
0.371
0.379
0.395
0.398
0.406
0.410
0.418
0.434
0.449
0.457
0.465
0.473
0.480
0.496
0.500
0.508
0.512
0.520
0.527
0.543
0.574
0.582
0.598
0.613
0.645
0.652
0.668
0.684
0.699
0.703
0.711
0.719
0.750
0.781
0.797
0.805
0.813
0.828
0.844
0.859
0.922
0.953
0.969
1.000


Thank you to both of you. I have used much of your work in my own meager
efforts at mastering excel.

Frank

 

[Michael R Middleton]

Frank -

... Let me tell you what I am am doing and ask your advice. I am
constructing a cumulative probability curve of values, bounded by 0% and
100% probabilities. The curve generally follows a basic S shape with the
very low probabilies and very high probabilities going asyntotic. It is
very important that the trendline does not go below 0% or above 100%. <

My advice would depend on what you plan to do with the cumulative curve.

For example, if you simply want to display it, then you could add an
arbitrary point for the 0% cumulative probability, plot an XY (Scatter)
chart, and use Format Data Series | Patterns | Line | Smoothed Line.

I don't know where your cumulative probabilities came from, but if your
original data were discrete probabilities, you could display the cumulative
distribution using a stair-step pattern. I think Tushar Mehta has an example
on his Step Chart page at www.tushar-mehta.com.

- Mike

www.mikemiddleton.com

 

[Jerry W. Lewis]

And if the data can meaningfully be fitted to an 8th order polynomial, I
would still worry about numerical problems unless you were using Excel
2003 and no coefficients were estimated to be exactly zero
http://groups.google.com/groups?selm=412980D4.5040305@no_e-mail.com

Jerry

 

[Tushar Mehta]

Complementing Mike's suggestion about a step chart (the specific address
is http://www.tushar-mehta.com/excel/charts/step_chart/index.html), you
may also want to consider stepwise-linear interpolation to estimate
probability values for X0 <= X <- Xn. For one approach see
Interactive Chart
http://www.tushar-
mehta.com/excel/software/interactive_chart_display/index.html

--
Regards,

Tushar Mehta
www.tushar-mehta.com
Multi-disciplinary business expertise
+ Technology skills
= Optimal solution to your business problem
Recipient Microsoft MVP award 2000-2004

 

[Martin Brown]

Thank you both for your comments. Let me tell you what I am am doing and
ask your advice. I am constructing a cumulative probability curve of
values, bounded by 0% and 100% probabilities. The curve generally follows a
basic S shape with the very low probabilies and very high probabilities
going asyntotic. It is very important that the trendline does not go below
0% or above 100%.

You are dead in the water with a polynomial fit then. The higher the
order of polynomial will by its nature make wild excursions in the
nether regions.

Your best bet would be to fit a function that necessarily by
construction matches your prescribed boundary conditions as x tends to
infinity.

A scaled version of (1 + tanh(Bx))/2 might be worth a try.
Where B is to be determined by least squares fitting.

Regards,

 

[David J. Braden]

Another idea:
Generalized inverse Tukey-lambda fit, which requires but 4 parameters, and
is very well behaved at endpoints. The fit is on the inverse cumulative, and
seems to be very stable wrt Excel.

 

[Frank & Pam Hayes]

David,

The Tukey-lambda fit looks like it has promise for my cumulative probability
curve, but a google search on Tukey-lambda and Excel was pretty sparse.
Searching on Tukey-lambda alone brought many more results, most of which
were beyond my statistical competance. The cumulative distribution function
shown at : http://www.itl.nist.gov/div898/handbook/eda/section3/eda366f.htm
looks to be exactly what I am trying to produce.

Can you point me in the right direction on how I would use Tukey-lambda in
Excel to calculate the cumulative probabilty curve?

Frank

 

[Jerry W. Lewis]

See the CRC book "Fitting Statistical Distributions: The Generalized
Lambda Distribution and Generalized Bootstrap Methods" by Zaven A.
Karian and Edward J. Dudewicz

It develops the math and provides Maple code for fitting. You will have
to adapt to Excel.

Jerry

 

[Jerry W. Lewis]

"Long time, no see." Good to have you back.

Jerry

David J. Braden wrote:
....

 

[David J. Braden]

Thanks. Finally, there's a stat question I can add something to before *you*
get too far into it <vbg>.
I didn't know of the CRC source you cited. I know only of a neat article by
Freimer, Mudhoker at al. that really digs into it, and, AFAIK, was the one
to generalize the distribution, though how I don't recall. I'll look into
it, though it will take me a few days yet. Could you plz post the
generalization from the CRC, or at least point us towards it? Then we can
walk the OP through how to fit it to his data using Excel.

What brought it to mind is its simplicity and flexibility. If the OP is
willing to work with cumulatives (my preference as well) then, short of
working directly with the empirical cumulative, I cannot think of a better
proximal distribution in this case for data I haven't seen.

Regards,
dave braden

 

[David J. Braden]

Frank (and Pam?)
I want to get the generalized version first; it is not at hand,
unfortunately, and unless I get help from Jerry or someone else in the
community, it will take me a day or so to retrieve it. Once I get it, I will
be happy to walk you through how to use Excel to fit it. Remember, it works
off of the *inverse* cumulative.

Do you know how to set it up? You also need to determine what you mean by
"closeness of fit". Jerry's CRC suggestion might well do the trick; I
haven't seen it yet, so I don't know how the distribution is generalized,
nor how easy the CRC version is to fit. But we'll get there.

Regards,
dave braden

 

[Jerry W. Lewis]

I presume the Freimer, Mudholkar et al paper you saw was Comm.Stat.A
17:3547-3567, 1988. If you have direct access to the Comm.Stat. series,
you might also look at a couple of Karian & Dudewicz papers from
Comm.Stat.B 25:611-642,1996 and 28:793-819,1999. Another reference
would be Oeztuerk & Dale's Technometrics 27:81-84,1985 paper.

I have access to the Karian & Dudewicz book and Technometrics CDs at the
office. I will bring them home tonight to follow up if the question is
still open.

Jerry

 

[Frank & Pam Hayes]

Jerry,

The question is still open. Thanks for the continuing help.

Frank

 

[Jerry W. Lewis]

The Generalized Lambda Distribution is the 4-parameter distribution with
inverse
GLDinv(p,L1,L2,L3,L4) = L1 + (p^L3+(1-p)^L4)/L2

This represents a valid distribution if and only if
L3*p^(L3-1)+L4*(1-p)^(L4-1)
has the same sign (positive or negative) for all p in [0,1], as long as
L2 takes that sign also (which in particular is true if L2, L3, and L4
all have the same sign). See the Karian & Dudewicz book for extensive
discussion of valid and invalid regions.

When L3>-1/4 and L4>-1/4, then the first four moments are
mean = L1+A/L2
var = (B-A^2)/L2^2
a3 = (C-3*A*B+2*A^3)/(L2*SQRT(var))^3
a4 = (D-4*A*C+6*A^2*B-3*A^4)/(L2*SQRT(var))^4
for
A = 1/(1+L3) -1/(1+L4)
B = 1/(1+2*L3) +1/(1+2*L4) -2*beta(1+L3,1+L4)
C = 1/(1+3*L3) -1/(1+3*L4) -3*beta(1+2*L3,1+L4) +3*beta(1+L3,1+2*L4)
D = 1/(1+4*L3) +1/(1+4*L4) -4*beta(1+3*L3,1+L4)
+6*beta(1+2*L3,1+2*L4) -4*beta(1+L3,1+3*L4)

where
a3 = E(X-mean)^3/sigma^3
a4 = E(X-mean)^4/sigma^4

You can use the method of moments to estimate the parameters
(L1,L2,L3,L4) from data. Alternately, you can estimate the parameters
from 4 sample quantiles. Karian & Dudewicz provide tables and Maple
code for fitting GLD from either approach. They also discuss the
bivariate extension.

(L1,L2,L3,L4) = (0, 0.1975, 0.1349, 0.1349) approximates the standard
normal distribution.
(L1,L2,L3,L4) = (0.5, 1/12, 0, 9/5) is Uniform(0,1).
Karian & Dudewicz discusses approximations to other standard distributions.

Jerry

 

[Frank & Pam Hayes]

Jerry,

First of all, thank you very much for putting so much thought and effort
into your posting. I appreciate your efforts. I am not a Statistics whiz,
so forgive me if some of my questions are a bit basic.

Am I correct in thinking that L1 is the Mean, L2 is the Variance, L3 is the
Skewness and the L4 is the Kurtosis? Given that Excel can calculate all of
these from my data, is the answer as simple as calculating those values and
plugging them into the formula you provided?

Your formula shows the variable p, but it is not defined in the rest of the
posting. Does this stand for the probability of the occurance from 0 to 1?
How would I produce the trend line that reflects the data I am interested in
.... perhaps calculate GLDinv for 100 points (p) from 0 to 1 and then plot
that line?

Thank you,

Frank

The Generalized Lambda Distribution is the 4-parameter distribution with
inverse
GLDinv(p,L1,L2,L3,L4) = L1 + (p^L3+(1-p)^L4)/L2

This represents a valid distribution if and only if
L3*p^(L3-1)+L4*(1-p)^(L4-1)
has the same sign (positive or negative) for all p in [0,1], as long as L2
takes that sign also (which in particular is true if L2, L3, and L4 all
have the same sign). See the Karian & Dudewicz book for extensive
discussion of valid and invalid regions.

When L3>-1/4 and L4>-1/4, then the first four moments are
mean = L1+A/L2
var = (B-A^2)/L2^2
a3 = (C-3*A*B+2*A^3)/(L2*SQRT(var))^3
a4 = (D-4*A*C+6*A^2*B-3*A^4)/(L2*SQRT(var))^4
for
A = 1/(1+L3) -1/(1+L4)
B = 1/(1+2*L3) +1/(1+2*L4) -2*beta(1+L3,1+L4)
C = 1/(1+3*L3) -1/(1+3*L4) -3*beta(1+2*L3,1+L4) +3*beta(1+L3,1+2*L4)
D = 1/(1+4*L3) +1/(1+4*L4) -4*beta(1+3*L3,1+L4)
+6*beta(1+2*L3,1+2*L4) -4*beta(1+L3,1+3*L4)

where
a3 = E(X-mean)^3/sigma^3
a4 = E(X-mean)^4/sigma^4

You can use the method of moments to estimate the parameters (L1,L2,L3,L4)
from data. Alternately, you can estimate the parameters from 4 sample
quantiles. Karian & Dudewicz provide tables and Maple code for fitting
GLD from either approach. They also discuss the bivariate extension.

(L1,L2,L3,L4) = (0, 0.1975, 0.1349, 0.1349) approximates the standard
normal distribution.
(L1,L2,L3,L4) = (0.5, 1/12, 0, 9/5) is Uniform(0,1).
Karian & Dudewicz discusses approximations to other standard
distributions.

Jerry

I presume the Freimer, Mudholkar et al paper you saw was Comm.Stat.A
17:3547-3567, 1988. If you have direct access to the Comm.Stat. series,
you might also look at a couple of Karian & Dudewicz papers from
Comm.Stat.B 25:611-642,1996 and 28:793-819,1999. Another reference would
be Oeztuerk & Dale's Technometrics 27:81-84,1985 paper.

I have access to the Karian & Dudewicz book and Technometrics CDs at the
office. I will bring them home tonight to follow up if the question is
still open.

Jerry

 

[Jerry W. Lewis]

You are welcome. Moreover, clarifications are not a burden.

L1 is a location parameter, but is equal to the mean only if L3=L4,
since A=0 when L3=L4. More Generally,
mean = L1+A/L2

L2 is a spread parameter, but the value of the variance is given by the
formula in my previous post.

L3 and L4 are shape parameters, but the usual coefficient of skewness is
a3, whose formula is given in my previous post. In particular, note
that when L3=L4, the coefficient of skewness is zero (the distribution
is symmetric), regardless of the actual value of L3

Similarly, the usual coefficient of kurtosis is either a4 or a4-3,
depending on whether you consider the kurtosis of the Normal
distribution to be 3 or 0. The formula for a4 is given in my previous post.

Also, note that for positive values of L3 and L4 that
GLDinv(p,L1,L2,L3,L4) is finite for p=0 and p=1. That means that the
distribution function has a finite domain. Thus, the generalized lambda
distribution may poorly fit the extreme tails of distributions (such as
Normal, Gamma, Chi-Square, etc.) that have have an infinite domain, even
though it may be a good approximation elsewhere.

Jerry

Jerry,

First of all, thank you very much for putting so much thought and effort
into your posting. I appreciate your efforts. I am not a Statistics whiz,
so forgive me if some of my questions are a bit basic.

Am I correct in thinking that L1 is the Mean, L2 is the Variance, L3 is the
Skewness and the L4 is the Kurtosis? Given that Excel can calculate all of
these from my data, is the answer as simple as calculating those values and
plugging them into the formula you provided?

Your formula shows the variable p, but it is not defined in the rest of the
posting. Does this stand for the probability of the occurance from 0 to 1?
How would I produce the trend line that reflects the data I am interested in
... perhaps calculate GLDinv for 100 points (p) from 0 to 1 and then plot
that line?

Thank you,

Frank

The Generalized Lambda Distribution is the 4-parameter distribution with
inverse
GLDinv(p,L1,L2,L3,L4) = L1 + (p^L3+(1-p)^L4)/L2

This represents a valid distribution if and only if
L3*p^(L3-1)+L4*(1-p)^(L4-1)
has the same sign (positive or negative) for all p in [0,1], as long as L2
takes that sign also (which in particular is true if L2, L3, and L4 all
have the same sign). See the Karian & Dudewicz book for extensive
discussion of valid and invalid regions.

When L3>-1/4 and L4>-1/4, then the first four moments are
mean = L1+A/L2
var = (B-A^2)/L2^2
a3 = (C-3*A*B+2*A^3)/(L2*SQRT(var))^3
a4 = (D-4*A*C+6*A^2*B-3*A^4)/(L2*SQRT(var))^4
for
A = 1/(1+L3) -1/(1+L4)
B = 1/(1+2*L3) +1/(1+2*L4) -2*beta(1+L3,1+L4)
C = 1/(1+3*L3) -1/(1+3*L4) -3*beta(1+2*L3,1+L4) +3*beta(1+L3,1+2*L4)
D = 1/(1+4*L3) +1/(1+4*L4) -4*beta(1+3*L3,1+L4)
+6*beta(1+2*L3,1+2*L4) -4*beta(1+L3,1+3*L4)

where
a3 = E(X-mean)^3/sigma^3
a4 = E(X-mean)^4/sigma^4

You can use the method of moments to estimate the parameters (L1,L2,L3,L4)
from data. Alternately, you can estimate the parameters from 4 sample
quantiles. Karian & Dudewicz provide tables and Maple code for fitting
GLD from either approach. They also discuss the bivariate extension.

(L1,L2,L3,L4) = (0, 0.1975, 0.1349, 0.1349) approximates the standard
normal distribution.
(L1,L2,L3,L4) = (0.5, 1/12, 0, 9/5) is Uniform(0,1).
Karian & Dudewicz discusses approximations to other standard
distributions.

Jerry

Jerry W. Lewis wrote:

I presume the Freimer, Mudholkar et al paper you saw was Comm.Stat.A
17:3547-3567, 1988. If you have direct access to the Comm.Stat. series,
you might also look at a couple of Karian & Dudewicz papers from
Comm.Stat.B 25:611-642,1996 and 28:793-819,1999. Another reference would
be Oeztuerk & Dale's Technometrics 27:81-84,1985 paper.

I have access to the Karian & Dudewicz book and Technometrics CDs at the
office. I will bring them home tonight to follow up if the question is
still open.

Jerry

David J. Braden wrote:


Frank (and Pam?)
I want to get the generalized version first; it is not at hand,
unfortunately, and unless I get help from Jerry or someone else in the
community, it will take me a day or so to retrieve it. Once I get it, I
will be happy to walk you through how to use Excel to fit it. Remember,
it works off of the *inverse* cumulative.

Do you know how to set it up? You also need to determine what you mean
by "closeness of fit". Jerry's CRC suggestion might well do the trick; I
haven't seen it yet, so I don't know how the distribution is
generalized, nor how easy the CRC version is to fit. But we'll get
there.

Regards,
dave braden



David,

The Tukey-lambda fit looks like it has promise for my cumulative
probability curve, but a google search on Tukey-lambda and Excel was
pretty sparse. Searching on Tukey-lambda alone brought many more
results, most of which were beyond my statistical competance. The
cumulative distribution function shown at :
http://www.itl.nist.gov/div898/handbook/eda/section3/eda366f.htm
looks to be exactly what I am trying to produce.

Can you point me in the right direction on how I would use Tukey-lambda
in Excel to calculate the cumulative probabilty curve?

Frank



Another idea:
Generalized inverse Tukey-lambda fit, which requires but 4 parameters,
and is very well behaved at endpoints. The fit is on the inverse
cumulative, and seems to be very stable wrt Excel.



And if the data can meaningfully be fitted to an 8th order
polynomial, I would still worry about numerical problems unless you
were using Excel 2003 and no coefficients were estimated to be
exactly zero
http://groups.google.com/groups?selm=412980D4.5040305@no_e-mail.com

Jerry

Bernard Liengme wrote:



Use LINEST to generate coefficients - see
www.stfx.ca/people/bliengme/ExcelTips
Use the coefficients to generate trendline data

Do your really have data that can <meaningfully> be fitted to 8th
order?

 

[Frank & Pam Hayes]

Jerry,

Thank you for the clarifications. I am going to dig into it a bit and see
how I can apply it to my application.

Frank

You are welcome. Moreover, clarifications are not a burden.

L1 is a location parameter, but is equal to the mean only if L3=L4, since
A=0 when L3=L4. More Generally,
mean = L1+A/L2

L2 is a spread parameter, but the value of the variance is given by the
formula in my previous post.

L3 and L4 are shape parameters, but the usual coefficient of skewness is
a3, whose formula is given in my previous post. In particular, note that
when L3=L4, the coefficient of skewness is zero (the distribution is
symmetric), regardless of the actual value of L3

Similarly, the usual coefficient of kurtosis is either a4 or a4-3,
depending on whether you consider the kurtosis of the Normal distribution
to be 3 or 0. The formula for a4 is given in my previous post.

Also, note that for positive values of L3 and L4 that
GLDinv(p,L1,L2,L3,L4) is finite for p=0 and p=1. That means that the
distribution function has a finite domain. Thus, the generalized lambda
distribution may poorly fit the extreme tails of distributions (such as
Normal, Gamma, Chi-Square, etc.) that have have an infinite domain, even
though it may be a good approximation elsewhere.

Jerry

Jerry,

First of all, thank you very much for putting so much thought and effort
into your posting. I appreciate your efforts. I am not a Statistics
whiz, so forgive me if some of my questions are a bit basic.

Am I correct in thinking that L1 is the Mean, L2 is the Variance, L3 is
the Skewness and the L4 is the Kurtosis? Given that Excel can calculate
all of these from my data, is the answer as simple as calculating those
values and plugging them into the formula you provided?

Your formula shows the variable p, but it is not defined in the rest of
the posting. Does this stand for the probability of the occurance from 0
to 1? How would I produce the trend line that reflects the data I am
interested in ... perhaps calculate GLDinv for 100 points (p) from 0 to 1
and then plot that line?

Thank you,

Frank

The Generalized Lambda Distribution is the 4-parameter distribution with
inverse
GLDinv(p,L1,L2,L3,L4) = L1 + (p^L3+(1-p)^L4)/L2

This represents a valid distribution if and only if
L3*p^(L3-1)+L4*(1-p)^(L4-1)
has the same sign (positive or negative) for all p in [0,1], as long as
L2 takes that sign also (which in particular is true if L2, L3, and L4
all have the same sign). See the Karian & Dudewicz book for extensive
discussion of valid and invalid regions.

When L3>-1/4 and L4>-1/4, then the first four moments are
mean = L1+A/L2
var = (B-A^2)/L2^2
a3 = (C-3*A*B+2*A^3)/(L2*SQRT(var))^3
a4 = (D-4*A*C+6*A^2*B-3*A^4)/(L2*SQRT(var))^4
for
A = 1/(1+L3) -1/(1+L4)
B = 1/(1+2*L3) +1/(1+2*L4) -2*beta(1+L3,1+L4)
C = 1/(1+3*L3) -1/(1+3*L4) -3*beta(1+2*L3,1+L4) +3*beta(1+L3,1+2*L4)
D = 1/(1+4*L3) +1/(1+4*L4) -4*beta(1+3*L3,1+L4)
+6*beta(1+2*L3,1+2*L4) -4*beta(1+L3,1+3*L4)

where
a3 = E(X-mean)^3/sigma^3
a4 = E(X-mean)^4/sigma^4

You can use the method of moments to estimate the parameters
(L1,L2,L3,L4) from data. Alternately, you can estimate the parameters
from 4 sample quantiles. Karian & Dudewicz provide tables and Maple code
for fitting GLD from either approach. They also discuss the bivariate
extension.

(L1,L2,L3,L4) = (0, 0.1975, 0.1349, 0.1349) approximates the standard
normal distribution.
(L1,L2,L3,L4) = (0.5, 1/12, 0, 9/5) is Uniform(0,1).
Karian & Dudewicz discusses approximations to other standard
distributions.

Jerry

Jerry W. Lewis wrote:


I presume the Freimer, Mudholkar et al paper you saw was Comm.Stat.A
17:3547-3567, 1988. If you have direct access to the Comm.Stat. series,
you might also look at a couple of Karian & Dudewicz papers from
Comm.Stat.B 25:611-642,1996 and 28:793-819,1999. Another reference
would be Oeztuerk & Dale's Technometrics 27:81-84,1985 paper.

I have access to the Karian & Dudewicz book and Technometrics CDs at the
office. I will bring them home tonight to follow up if the question is
still open.

Jerry

David J. Braden wrote:


Frank (and Pam?)
I want to get the generalized version first; it is not at hand,
unfortunately, and unless I get help from Jerry or someone else in the
community, it will take me a day or so to retrieve it. Once I get it, I
will be happy to walk you through how to use Excel to fit it. Remember,
it works off of the *inverse* cumulative.

Do you know how to set it up? You also need to determine what you mean
by "closeness of fit". Jerry's CRC suggestion might well do the trick;
I haven't seen it yet, so I don't know how the distribution is
generalized, nor how easy the CRC version is to fit. But we'll get
there.

Regards,
dave braden



David,

The Tukey-lambda fit looks like it has promise for my cumulative
probability curve, but a google search on Tukey-lambda and Excel was
pretty sparse. Searching on Tukey-lambda alone brought many more
results, most of which were beyond my statistical competance. The
cumulative distribution function shown at :
http://www.itl.nist.gov/div898/handbook/eda/section3/eda366f.htm
looks to be exactly what I am trying to produce.

Can you point me in the right direction on how I would use
Tukey-lambda in Excel to calculate the cumulative probabilty curve?

Frank



Another idea:
Generalized inverse Tukey-lambda fit, which requires but 4
parameters, and is very well behaved at endpoints. The fit is on the
inverse cumulative, and seems to be very stable wrt Excel.



And if the data can meaningfully be fitted to an 8th order
polynomial, I would still worry about numerical problems unless you
were using Excel 2003 and no coefficients were estimated to be
exactly zero
http://groups.google.com/groups?selm=412980D4.5040305@no_e-mail.com

Jerry

Bernard Liengme wrote:



Use LINEST to generate coefficients - see
www.stfx.ca/people/bliengme/ExcelTips
Use the coefficients to generate trendline data

Do your really have data that can <meaningfully> be fitted to 8th
order?