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PedroLeiva · 10-28-2015, 12:24 AM

(10-27-2015 10:25 PM)Dieter Wrote:
(10-27-2015 07:32 PM)Dieter Wrote: BTW, at the moment I am looking at an 35s implementation of the HP67 routines.

OK, here is a first experimental version of the Chi² algorithm used in the HP67 Stat Pac. PDF and CDF are calculated simultaneously, and both are returned in Y and X. Here is the code:

Code:

C001 LBL C C002 INPUT J C003 e C004 IP C005 STO T C006 ÷ C007 ! C008 STO G C009 INPUT X C010 XEQ C013 C011 STOP C012 GTO C009 C013 RCL X C014 RCL÷ T C015 +/- C016 e^x C017 LASTx C018 +/- C019 RCL J C020 RCL÷ T C021 y^x C022 x C023 RCL÷ G C024 STO A C025 RCL J C026 STO B C027 SGN C028 ENTER C029 - C030 ENTER C031 ENTER C032 LASTx C033 RCLx X C034 RCL B C035 RCL+ T C036 STO B C037 ÷ C038 + C039 x≠y? C040 GTO C030 C041 RCL A C042 RCL X C043 x=0? C044 e^x C045 ÷ C046 RCLx J C047 RCL÷ T C048 x<>y C049 RCLx A C050 RCL+ A C051 RTN

Usage (results as shown in FIX 9 mode):

Code:

[XEQ] C [ENTER] J=? Enter degrees of freedom 10 [R/S] X=? Enter random variable X 6,3 [R/S] 0,087896860 // =PDF(6,3|10) 0,210539719 // =CDF(6,3|10) Do another calculation [R/S] X=? 20 [R/S] 0,009458319 // =PDF(20|10) 0,970747312 // =CDF(20|10)

As usual, comments, suggestions and error reports are welcome. ;-)

Since the main computation routine (C013 ff.) returns both the PDF and the CDF, adding a function that determines the inverse (quantile) is rather simple. I tried an approach using a second-order (Halley) method and it worked well for me. This required about 30 additional lines of code.

Dieter

Dieter, the program works perfect!. Please include the function for the inverse, no mather how long it takes

PedroLeiva · (This post was last modified: 10-28-2015 12:38 AM by PedroLeiva.)

(10-28-2015 12:11 AM)Thomas Klemm Wrote:
(10-27-2015 11:41 PM)PedroLeiva Wrote: 2° time: x=0.5, y=3.850 E-7; variable X= 0.001

That was helpful. The value of E got copied to X.
Turns out there was a bug in this line:

Code:

I018 GTO I007 ; not yet

Should go to line I007 instead of I006.

Cheers
Thomas

PS: Fixed it in the listing.

Thomas, I erase the STOP instruction, and corrected for "I018 GTO I007" so the program began to work perfect. Thank you

Thomas Klemm · 10-28-2015, 01:05 AM

(10-27-2015 10:25 PM)Dieter Wrote: I tried an approach using a second-order (Halley) method and it worked well for me. This required about 30 additional lines of code.

Or then you can just use the built-in solver.

Cheers
Thomas

Thomas Klemm · 10-28-2015, 01:40 AM

(10-28-2015 12:36 AM)PedroLeiva Wrote: I erase the STOP instruction, and corrected for "I018 GTO I007" so the program began to work perfect.

That was probably the most distant remote debugging session I ever had.

I've added line I005 to improve the initial guess only after writing the listing of the program. This shifted all the following commands down one line. And then I forgot to update the GTO command in the listing. We all know that the Go To Statement is Considered Harmful. Did we really need another proof?

Cheers
Thomas

Dieter · (This post was last modified: 10-28-2015 12:07 PM by Dieter.)

(10-28-2015 01:05 AM)Thomas Klemm Wrote: Or then you can just use the built-in solver.

A dedicated solver is more effective and requires just one initial guess. ;-)

But maybe it is possible to give a lower and higher limit for the quantile and feed this to the solver. For all four distributions, that is. With some slight tweaks (e.g. exit if a certain accuracy level is reached since the CDF is not absolutely exact anyway) this might work as well.

EDIT: the built-in solver probably will not work here. It is known that the 35s hat problems when solving programs containing loops. So a dedicated solver seems to be the only solution.

Dieter

Dieter · (This post was last modified: 10-28-2015 12:58 PM by Dieter.)

(10-28-2015 12:24 AM)PedroLeiva Wrote: Dieter, the program works perfect!. Please include the function for the inverse, no mather how long it takes

OK, here is a first version with a rough initial guess. At the X=? prompt simply enter a negative value if you want to compute the quantile for that probabily. If you already have entered the previous C program, simply add lines C010 and C011 and continue with line C054.

Code:

C001  LBL C

C002  INPUT J

C003  e

C004  IP

C005  STO T

C006  ÷

C007  !

C008  STO G

C009  INPUT X

C010  x<0?

C011  GTO C054

C012  XEQ C015

C013  STOP

C014  GTO C009

C015  RCL X

C016  RCL÷ T

C017  +/-

C018  e^x

C019  LASTx

C020  +/-

C021  RCL J

C022  RCL÷ T

C023  y^x

C024  x

C025  RCL÷ G

C026  STO A

C027  RCL J

C028  STO B

C029  SGN

C030  ENTER

C031  -

C032  ENTER

C033  ENTER

C034  LASTx

C035  RCLx X

C036  RCL B

C037  RCL+ T

C038  STO B

C039  ÷

C040  +

C041  x≠y?

C042  GTO C032

C043  RCL A

C044  RCL X

C045  x=0?

C046  e^x

C047  ÷

C048  RCLx J

C049  RCL÷ T

C050  x<>y

C051  RCLx A

C052  RCL+ A

C053  RTN

C054  +/-

C055  STO P

C056  RCLx J

C057  RCL÷ T

C058  RCLx G

C059  RCL T

C060  RCL÷ J

C061  y^x

C062  Pi

C063  x

C064  STO X

C065  XEQ C015

C066  ENTER

C067  RCL÷ P

C068  LN

C069  x

C070  ENTER

C071  ENTER

C072  RCL J

C073  RCL- X

C074  RCL- T

C075  RCL÷ X

C076  RCL÷ T

C077  RCL÷ T

C078  x

C079  R↑

C080  -

C081  ÷

C082  RCL+ X

C083  x<> X

C084  RCL X

C085  STOP

C086  GTO C064

Usage: At the X=? prompt enter the negative probability for which you want the Chi² quantile. The program will then display two estimates in Y and X (the previous and the current), so that you can see the iteration converge. Pressing R/S yields the next approximation.

Code:

[XEQ] C [ENTER]             J=?

Enter degrees of freedom

    10  [R/S]               X=?

Enter negative probability

  -0,95 [R/S]               11,176989523

                            15,310270784

        [R/S]               15,310270784

                            17,970799176

        [R/S]               17,970799176

                            18,305703292

        [R/S]               18,305703292

                            18,307038039

If 18,307 is sufficiently accurate you may stop here.
Otherwise continue:

Code:

        [R/S]               18,307038039

                            18,307038054

        [R/S]               18,307038054

                            18,307038054

When the two approximations are displayed you may also enter your own guess before pressing R/S, and the approximation will continue with this value.

NB: This program is not very sophisticated, and it does not match what's inside the 34s. For instance, the latter evaluates CDFs for large x via P(x) = 1–Q(x), and also the quantile function has a much, much better initial guess. It took me quite a while to get the guess within <10% of the true value. ;-) For instance, the first guess for the case above is 18,2..., compared to the exact value 18,307...

Some time ago I wrote 35s/41C programs for the Normal distribution. They evaluate the PDF, one- and two-sided CDFs and also one- and two-sided quantiles with several tweaks to ensure best possible accuracy and efficiency. But these are about three times the size of the small Chi² progam posted above.

Dieter

PedroLeiva · 10-28-2015, 06:21 PM

I think we have tight-fitting Student t-distribution. I attached the pdf with the program that Thomas wrote, five calculation examples (test cases), an outline of the distribición and an instruction table for input and output of data.

Comments are welcome. Pedro

PedroLeiva · 10-28-2015, 06:59 PM

(10-28-2015 12:34 PM)Dieter Wrote:
(10-28-2015 12:24 AM)PedroLeiva Wrote: Dieter, the program works perfect!. Please include the function for the inverse, no mather how long it takes

OK, here is a first version with a rough initial guess. At the X=? prompt simply enter a negative value if you want to compute the quantile for that probabily. If you already have entered the previous C program, simply add lines C010 and C011 and continue with line C054.

Code:

C001  LBL C C002  INPUT J C003  e C004  IP C005  STO T C006  ÷ C007  ! C008  STO G C009  INPUT X C010  x<0? C011  GTO C054 C012  XEQ C015 C013  STOP C014  GTO C009 C015  RCL X C016  RCL÷ T C017  +/- C018  e^x C019  LASTx C020  +/- C021  RCL J C022  RCL÷ T C023  y^x C024  x C025  RCL÷ G C026  STO A C027  RCL J C028  STO B C029  SGN C030  ENTER C031  - C032  ENTER C033  ENTER C034  LASTx C035  RCLx X C036  RCL B C037  RCL+ T C038  STO B C039  ÷ C040  + C041  x≠y? C042  GTO C032 C043  RCL A C044  RCL X C045  x=0? C046  e^x C047  ÷ C048  RCLx J C049  RCL÷ T C050  x<>y C051  RCLx A C052  RCL+ A C053  RTN C054  +/- C055  STO P C056  RCLx J C057  RCL÷ T C058  RCLx G C059  RCL T C060  RCL÷ J C061  y^x C062  Pi C063  x C064  STO X C065  XEQ C015 C066  ENTER C067  RCL÷ P C068  LN C069  x C070  ENTER C071  ENTER C072  RCL J C073  RCL- X C074  RCL- T C075  RCL÷ X C076  RCL÷ T C077  RCL÷ T C078  x C079  R↑ C080  - C081  ÷ C082  RCL+ X C083  x<> X C084  RCL X C085  STOP C086  GTO C064

Usage: At the X=? prompt enter the negative probability for which you want the Chi² quantile. The program will then display two estimates in Y and X (the previous and the current), so that you can see the iteration converge. Pressing R/S yields the next approximation.

Code:

[XEQ] C [ENTER]             J=? Enter degrees of freedom     10  [R/S]               X=? Enter negative probability   -0,95 [R/S]               11,176989523                             15,310270784         [R/S]               15,310270784                             17,970799176         [R/S]               17,970799176                             18,305703292         [R/S]               18,305703292                             18,307038039

If 18,307 is sufficiently accurate you may stop here.
Otherwise continue:

Code:

        [R/S]               18,307038039                             18,307038054         [R/S]               18,307038054                             18,307038054

When the two approximations are displayed you may also enter your own guess before pressing R/S, and the approximation will continue with this value.

NB: This program is not very sophisticated, and it does not match what's inside the 34s. For instance, the latter evaluates CDFs for large x via P(x) = 1–Q(x), and also the quantile function has a much, much better initial guess. It took me quite a while to get the guess within <10% of the true value. ;-) For instance, the first guess for the case above is 18,2..., compared to the exact value 18,307...

Some time ago I wrote 35s/41C programs for the Normal distribution. They evaluate the PDF, one- and two-sided CDFs and also one- and two-sided quantiles with several tweaks to ensure best possible accuracy and efficiency. But these are about three times the size of the small Chi² progam posted above.

Dieter

Dieter, all perfect. I got the same figures as you!. I will prepare a PDF with programme keytroke sequence, test cases and instructions for in and output data

Dieter · (This post was last modified: 10-28-2015 08:00 PM by Dieter.)

(10-28-2015 06:21 PM)PedroLeiva Wrote: I think we have tight-fitting Student t-distribution. I attached the pdf with the program that Thomas wrote, five calculation examples (test cases), an outline of the distribición and an instruction table for input and output of data.

For the record, here is a version without using the Integrate function. It is based on algorithm AS 3, published as a Fortran program in Applied Statistics (1968). This version includes an improved input routine for the degrees of freedom in that it checks whether the input is a positive integer. Again, both the PDF and CDF are returned on the stack (in Y and X). Code for the inverse is not yet included.

Code:

T001  LBL T

T002  ALL

T003  INPUT J

T004  ENTER

T005  ABS

T006  IP

T007  x=y?

T008  x=0?

T009  GTO T001

T010  e

T011  IP

T012  STO T

T013  SGN

T014  -

T015  RCL÷ T

T016  !

T017  LASTx

T018  RCL T

T019  1/x

T020  -

T021  !

T022  ÷

T023  Pi

T024  RCLx J

T025  √x

T026  ÷

T027  STO G

T028  FIX 9

T029  INPUT X

T030  RCL J

T031  √x

T032  ÷

T033  STO A

T034  RCL J

T035  RCL X

T036  x²

T037  RCL+ J

T038  ÷

T039  STO B

T040  RCL J

T041  RCL T

T042  RMDR

T043  RCL+ T

T044  STO K

T045  SGN

T046  STO C

T047  ENTER

T048  ENTER

T049  RCL K

T050  RCL- J

T051  x≥0?

T052  GTO T066

T053  SGN

T054  RCL+ K

T055  RCLx B

T056  RCL÷ K

T057  RCLx C

T058  STO C

T059  +

T060  x=y?

T061  GTO T066

T062  RCL T

T063  STO+ K

T064  R↓

T065  GTO T047

T066  R↓

T067  STO C

T068  RCL J

T069  RCL T

T070  RMDR

T071  x=0?

T072  GTO T090

T073  RCL- J

T074  x=0?

T075  STO C

T076  RCL A

T077  RCLx B

T078  RCLx C

T079  Pi

T080  ÷

T081  RCL A

T082  ATAN

T083  RCL T

T084  +/-

T085  SGN

T086  ACOS

T087  ÷

T088  +

T089  GTO T095

T090  RCL B

T091  √x

T092  RCLx A

T093  RCLx C

T094  RCL÷ T

T095  RCL T

T096  1/x

T097  +

T098  RCL X

T099  x²

T100  RCL÷ J

T101  RCL T

T102  SGN

T103  +

T104  LASTx

T105  RCL+ J

T106  RCL÷ T

T107  +/-

T108  y^x

T109  RCLx G

T110  x<>y

T111  STOP

T112  GTO T028

Here is Thomas' sample case:

Code:

[XEQ] T [ENTER]             J=?

Enter degrees of freedom

     7  [R/S]               X=?

Enter random variable X

    1,5 [R/S]               0,126263061   // =PDF(1,5|7)

                            0,911350757   // =CDF(1,5|7)

Do another calculation

        [R/S]               X=?

    -2  [R/S]               0,063135337   // =PDF(-2|7)

                            0,042809664   // =CDF(-2|7)

Final remark: like the other programs discussed in this thread, this is a rather simple and straightforward implementation that works fine for most practical cases. However, far out in the distribution tails the results may lose accuracy due to rounding, and eventually values may round to 0 or 1. For instance, for 7 d.o.f. and x=–100 you will get a CDF of 2,000...E–12 while the exact result is 1,318 E–12, and for x=0 the returned result is zero while it actually is 1,69 E–17. If you want more or less exact values even for such extreme arguments, a different approach is required.

Dieter

PedroLeiva · 10-28-2015, 10:34 PM

(10-28-2015 07:54 PM)Dieter Wrote:
(10-28-2015 06:21 PM)PedroLeiva Wrote: I think we have tight-fitting Student t-distribution. I attached the pdf with the program that Thomas wrote, five calculation examples (test cases), an outline of the distribición and an instruction table for input and output of data.

For the record, here is a version without using the Integrate function. It is based on algorithm AS 3, published as a Fortran program in Applied Statistics (1968). This version includes an improved input routine for the degrees of freedom in that it checks whether the input is a positive integer. Again, both the PDF and CDF are returned on the stack (in Y and X). Code for the inverse is not yet included.

Code:

T001 LBL T T002 ALL T003 INPUT J T004 ENTER T005 ABS T006 IP T007 x=y? T008 x=0? T009 GTO T001 T010 e T011 IP T012 STO T T013 SGN T014 - T015 RCL÷ T T016 ! T017 LASTx T018 RCL T T019 1/x T020 - T021 ! T022 ÷ T023 Pi T024 RCLx J T025 √x T026 ÷ T027 STO G T028 FIX 9 T029 INPUT X T030 RCL J T031 √x T032 ÷ T033 STO A T034 RCL J T035 RCL X T036 x² T037 RCL+ J T038 ÷ T039 STO B T040 RCL J T041 RCL T T042 RMDR T043 RCL+ T T044 STO K T045 SGN T046 STO C T047 ENTER T048 ENTER T049 RCL K T050 RCL- J T051 x≥0? T052 GTO T066 T053 SGN T054 RCL+ K T055 RCLx B T056 RCL÷ K T057 RCLx C T058 STO C T059 + T060 x=y? T061 GTO T066 T062 RCL T T063 STO+ K T064 R↓ T065 GTO T047 T066 R↓ T067 STO C T068 RCL J T069 RCL T T070 RMDR T071 x=0? T072 GTO T090 T073 RCL- J T074 x=0? T075 STO C T076 RCL A T077 RCLx B T078 RCLx C T079 Pi T080 ÷ T081 RCL A T082 ATAN T083 RCL T T084 +/- T085 SGN T086 ACOS T087 ÷ T088 + T089 GTO T095 T090 RCL B T091 √x T092 RCLx A T093 RCLx C T094 RCL÷ T T095 RCL T T096 1/x T097 + T098 RCL X T099 x² T100 RCL÷ J T101 RCL T T102 SGN T103 + T104 LASTx T105 RCL+ J T106 RCL÷ T T107 +/- T108 y^x T109 RCLx G T110 x<>y T111 STOP T112 GTO T028

Here is Thomas' sample case:

Code:

[XEQ] T [ENTER] J=? Enter degrees of freedom 7 [R/S] X=? Enter random variable X 1,5 [R/S] 0,126263061 // =PDF(1,5|7) 0,911350757 // =CDF(1,5|7) Do another calculation [R/S] X=? -2 [R/S] 0,063135337 // =PDF(-2|7) 0,042809664 // =CDF(-2|7)

Final remark: like the other programs discussed in this thread, this is a rather simple and straightforward implementation that works fine for most practical cases. However, far out in the distribution tails the results may lose accuracy due to rounding, and eventually values may round to 0 or 1. For instance, for 7 d.o.f. and x=–100 you will get a CDF of 2,000...E–12 while the exact result is 1,318 E–12, and for x=0 the returned result is zero while it actually is 1,69 E–17. If you want more or less exact values even for such extreme arguments, a different approach is required.

Dieter

Dieter, perfect!. I will try this option also. Thanks
I am preparing a compiled program pdf for Chi-Square to publish in the Forum

PedroLeiva · 10-28-2015, 10:45 PM

I leave the program compilation that Dieter wrote for Chi-Square Distribution, with examples and a list of programming steps.

Much I wish you could make the attempt with F Distribution, just to complete de Statistical Distribution Functions. Remember that Normal and Inverse Distribution is already included in the HP 35s Manual!

PedroLeiva · 10-29-2015, 12:09 AM

(10-28-2015 10:45 PM)PedroLeiva Wrote: I leave the program compilation that Dieter wrote for Chi-Square Distribution, with examples and a list of programming steps.

Much I wish you could make the attempt with F Distribution, just to complete de Statistical Distribution Functions. Remember that Normal and Inverse Distribution is already included in the HP 35s Manual!

Here is the program for Normal&Inverse-Normal Distributions. Just for those who will be interested in this subject

Dieter · 10-29-2015, 07:01 AM

(10-29-2015 12:09 AM)PedroLeiva Wrote: Here is the program for Normal&Inverse-Normal Distributions. Just for those who will be interested in this subject

This is a sample program from the 35s manual. It works, but there is room for improvement. For instance the HP15C Advanced Functions Handbook has a nice 15C program that uses the Integrate function with variable transformation which works much better in the distribution tails.

Personally, I'd prefer a different approach with the classic combination of series expansion and continued fraction to evaluate the Gauss integral. For the inverse, rational approximations with various accuracy levels can be given (yes, there is much more that Hastings' three-digit method as implemented in several HP software pacs). Maybe I will eventually post an example here.

Dieter

Dieter · (This post was last modified: 10-29-2015 07:29 AM by Dieter.)

(10-28-2015 10:45 PM)PedroLeiva Wrote: I leave the program compilation that Dieter wrote for Chi-Square Distribution, with examples and a list of programming steps.

Looks very nice. FTR: There are special fonts available that resemble the keys and the display of various HP calculators (simple 7 segment LCD, dot matrix, 14 segment HP41, etc.). I remember a font package by Luiz Viera and there also is a zip-file at educalc.net that inclues a 33s key font. Simply search for HP calculator fonts

I am not quite happy with the initial guess of the Chi² quantile. This is a very stripped down version of what's implemented in the 34s. But a more sophisticated version that usually converges within 3...5 iterations would require an estimate for the Normal quantile, which is also true for the Student inverse. So this can be done in a unified package that includes all three (or four) distributions. For the time being you may include the improved J input routine from the Student program.

Dieter

Dieter · 10-29-2015, 11:56 PM

(10-29-2015 07:18 AM)Dieter Wrote: I am not quite happy with the initial guess of the Chi² quantile. This is a very stripped down version of what's implemented in the 34s. But a more sophisticated version that usually converges within 3...5 iterations would require an estimate for the Normal quantile, which is also true for the Student inverse. So this can be done in a unified package that includes all three (or four) distributions.

I now tried this with the Chi² and Student programs. Both calculate a quite good first guess for the quantile which is based on a simple approximation of the Normal quantile. The previous Chi² example (10 dof and p=0,95) converges after just three iterations. The initial estimate was 18,24 with 18,307 being the correct result. In most cases, both for Chi² and Student, merely 3...5 iterations will do. This approach requires one or two additional subroutine(s) for the Normal estimate, and the complete Student program (including the inverse) finally has about 200 steps.

Dieter

PedroLeiva · 10-30-2015, 02:36 AM

(10-28-2015 07:54 PM)Dieter Wrote:
(10-28-2015 06:21 PM)PedroLeiva Wrote: I think we have tight-fitting Student t-distribution. I attached the pdf with the program that Thomas wrote, five calculation examples (test cases), an outline of the distribición and an instruction table for input and output of data.

For the record, here is a version without using the Integrate function. It is based on algorithm AS 3, published as a Fortran program in Applied Statistics (1968). This version includes an improved input routine for the degrees of freedom in that it checks whether the input is a positive integer. Again, both the PDF and CDF are returned on the stack (in Y and X). Code for the inverse is not yet included.

Code:

T001 LBL T T002 ALL T003 INPUT J T004 ENTER T005 ABS T006 IP T007 x=y? T008 x=0? T009 GTO T001 T010 e T011 IP T012 STO T T013 SGN T014 - T015 RCL÷ T T016 ! T017 LASTx T018 RCL T T019 1/x T020 - T021 ! T022 ÷ T023 Pi T024 RCLx J T025 √x T026 ÷ T027 STO G T028 FIX 9 T029 INPUT X T030 RCL J T031 √x T032 ÷ T033 STO A T034 RCL J T035 RCL X T036 x² T037 RCL+ J T038 ÷ T039 STO B T040 RCL J T041 RCL T T042 RMDR T043 RCL+ T T044 STO K T045 SGN T046 STO C T047 ENTER T048 ENTER T049 RCL K T050 RCL- J T051 x≥0? T052 GTO T066 T053 SGN T054 RCL+ K T055 RCLx B T056 RCL÷ K T057 RCLx C T058 STO C T059 + T060 x=y? T061 GTO T066 T062 RCL T T063 STO+ K T064 R↓ T065 GTO T047 T066 R↓ T067 STO C T068 RCL J T069 RCL T T070 RMDR T071 x=0? T072 GTO T090 T073 RCL- J T074 x=0? T075 STO C T076 RCL A T077 RCLx B T078 RCLx C T079 Pi T080 ÷ T081 RCL A T082 ATAN T083 RCL T T084 +/- T085 SGN T086 ACOS T087 ÷ T088 + T089 GTO T095 T090 RCL B T091 √x T092 RCLx A T093 RCLx C T094 RCL÷ T T095 RCL T T096 1/x T097 + T098 RCL X T099 x² T100 RCL÷ J T101 RCL T T102 SGN T103 + T104 LASTx T105 RCL+ J T106 RCL÷ T T107 +/- T108 y^x T109 RCLx G T110 x<>y T111 STOP T112 GTO T028

Here is Thomas' sample case:

Code:

[XEQ] T [ENTER] J=? Enter degrees of freedom 7 [R/S] X=? Enter random variable X 1,5 [R/S] 0,126263061 // =PDF(1,5|7) 0,911350757 // =CDF(1,5|7) Do another calculation [R/S] X=? -2 [R/S] 0,063135337 // =PDF(-2|7) 0,042809664 // =CDF(-2|7)

Final remark: like the other programs discussed in this thread, this is a rather simple and straightforward implementation that works fine for most practical cases. However, far out in the distribution tails the results may lose accuracy due to rounding, and eventually values may round to 0 or 1. For instance, for 7 d.o.f. and x=–100 you will get a CDF of 2,000...E–12 while the exact result is 1,318 E–12, and for x=0 the returned result is zero while it actually is 1,69 E–17. If you want more or less exact values even for such extreme arguments, a different approach is required.

Dieter

Dear Dieter, this is your t Student program. When I tryed with Thomas examples it works perfect, and the same when I try again with HP Stat Pack.

Dieter · 10-30-2015, 01:56 PM

(10-30-2015 02:36 AM)PedroLeiva Wrote: Dear Dieter, this is your t Student program. When I tryed with Thomas examples it works perfect, and the same when I try again with HP Stat Pack.

There is room for improvement. I am working on an integrated solution that contains Normal, Chi² and Student's t distribution and calculates PDFs, CDFs and their inverses, the latter with a much improved initial guess. The F case is a bit more complicated, but I'll see.

BTW: quoting the whole previous post really is not required. If everyone would do so, the messages got longer and longer and longer as the thread continues and you would have to scroll down for a while until you find the new content. So when replying, you should simply delete everything that you do not directly refer to.

BTW2: do you have the 35s emulator? In this case I could send an .ep file that you can load into the emulator directly. This way you would not have to enter a few hundred lines of code.

BTW3: does someone know if there is a way to produce a listing or something similar from programs inside the 35s emulator?

Dieter

PedroLeiva · 10-30-2015, 05:20 PM

[/quote]
Dieter, you are absolutely right; from now on when I quote, I will erase the above text. Sorry, it would not hapend again.

I use of the Hewlett Packard HP 35s emulator, which enormously facilitates joining my work programs, and in turn I make far fewer errors. A remark about it, I wanted to send the files of the summarized previous programs in PDF, but the MoHPC system will not let me, so we should send them by email (leiva.pedro@inta.gob.ar).

It would be very useful to send a text file with instructions from the emulator program, but I do not know if that is really possible.

Dieter · 10-30-2015, 07:16 PM

(10-30-2015 05:20 PM)PedroLeiva Wrote: Dieter, you are absolutely right; from now on when I quote, I will erase the above text.

Of course not all text. Otherwise it's not a quote. ;-)

(10-30-2015 05:20 PM)PedroLeiva Wrote: I use of the Hewlett Packard HP 35s emulator, which enormously facilitates joining my work programs, and in turn I make far fewer errors.

So you could use a .ep file and load it via the emulator menu (Calculator > Load).
I now have a collection with programs for the Normal, Student and Chi² distribution and their inverses. But there still is some work to do.

(10-30-2015 05:20 PM)PedroLeiva Wrote: A remark about it, I wanted to send the files of the summarized previous programs in PDF, but the MoHPC system will not let me,

?!? – you have successfully attached several PDF files to your posts, e.g. the Chi² and Student programs. Where exactly is the problem?

Dieter

PedroLeiva · (This post was last modified: 10-30-2015 07:45 PM by PedroLeiva.)

?!? – you have successfully attached several PDF files to your posts, e.g. the Chi² and Student programs. Where exactly is the problem?

Dieter
[/quote]
My apologies, I expressed badly. The system don´t let me send the * .EP files of the emulator programs that I type while testing what you were designing. Try your self; what the system warned me is that do not accept that file format for sending.