CONF=.95, N=1432, n=1
cdfbet("XY", n+1, N+1-n, CONF, 1-CONF)
is doing it just fine. Correct and with high precision...
h
> On 18.05.2020, at 19:08, Rafael Guerra wrote:
>
> Hi Heinz,
>
> Fyi, the following site provides Matlab code that may be translated to Scilab:
>
CONF=.95, N=1432, n=1
cdfbet("XY", n+1, N+1-n, CONF, 1-CONF)
is doing it just fine. Correct and with high precision...
h
On 18.05.2020, at 18:09, Tim Wescott wrote:
>
> So you have \beta(x, n+1, N+1-n) = 0.95, and you want to solve for x?
>
> fsolve will do this for a single value of the
> On 18.05.2020, at 14:18, Dang Ngoc Chan, Christophe
> wrote:
>
> Hello,
>
>> De Heinz Nabielek
>> Envoyé : dimanche 17 mai 2020 23:50
>>
>> CONF=0.95; N=1432; n=1:
>> One-sided 95% upper confidence limit fraction = BETA.INV(CONF, n+1, N+1-n)
>> = 0.003306121
>>
>> How would I do that in
Hi Heinz,
Fyi, the following site provides Matlab code that may be translated to Scilab:
https://people.sc.fsu.edu/~jburkardt/m_src/asa109/asa109.html
betain.m : incomplete Beta function ratio
xinbta.m : inverse of the incomplete Beta function
The Scilab Atoms 'Distfun' package contains:
So you have \beta(x, n+1, N+1-n) = 0.95, and you want to solve for x?
fsolve will do this for a single value of the confidence. Is that
sufficient?
On Sun, 2020-05-17 at 23:49 +0200, Heinz Nabielek wrote:
> Dear SciLabers:
>
> can Scilab compute the inverse of the regularized Incomplete Beta
>
Hi,
as far I remember, the underlying code is based on a very good
ACM TOMS (DiDinato, A. R. and Morris, A. H. Algorithm 708:
Significant
Digit Computation of the Incomplete Beta Function Ratios. ACM
Trans. Math. Softw. 18 (1993), 360-373).
Bruno
--
Non à l'augmentation
Hello,
> De Heinz Nabielek
> Envoyé : dimanche 17 mai 2020 23:50
>
> CONF=0.95; N=1432; n=1:
> One-sided 95% upper confidence limit fraction = BETA.INV(CONF, n+1, N+1-n)
> = 0.003306121
>
> How would I do that in Scilab?
Would it be :
[X,Y]=cdfbet("XY", n+1, N+1-n, CONF, 1-CONF)
X =
