My major cuestions that I repeat bellow are these:

1.- I had to obtain the ACOV matrix and I use the inverse of the Information matrix for the Maximum Likelihood method (-inverse of the Fisher observed information matrix)number of data regarding the articles and forum revised. s this a correct way?

2.- When multiplying my vector or 1x1 matrix, as it is the partial derivate on only one variable, by the ACOV 4x4 matrix the software gives me another matrix, as if one multiplies a number times a matrix. Instead, we can obtain a simple number by adding three Zeros to the gradient vector, but I think that this is not correct. Then, what can we do if the number of columns in our vector is different that the number of rows in our ACOV matrix?.

I attached some concepts from the theory in the file "1.jpg".

This is the standard method that I want to apply in Mathematica, as we can find out in TableCurve 2 D or other software (see 2.jpg attached).

I put here my DELTA METHOD TO OBTAIN THE VARIANCE OF A FUNCTION STEP BY STEP AND THEN THEIR CONFIDENCE AND PREDICTION INTERVALS OR BANDS: It seems that the key to carry out this method is in find the variance of the function that give us the prediction values*) I find an example where I could test a part of the procces and I develop following:*) Example http://www.montana.edu/rotella/502/DeltaExample.pdf in the case of female sex only. (See equation to the variance of a arbitrary function in the file 3.jpg attached).

Code:

```
VectorGrad = {1, -1, Li};
ACOVMatrix = {{0.04080, 0.0044646, -0.0054777}, {0.0044646082,
0.0438542, -0.0127939}, {-0.0054777, -0.0127939, 0.0486278733}};
(*Applying the dot product to obtain the variance of the function*)
Var = VectorGrad.ACOVMatrix.VectorGrad;(*Mathematica detect the same \
vector to the right side and transposes it automatically*)
(*The function to female rate,i.e., with sex=-1.*)
fem = Exp[-0.416831 + 0.540171 + 0.564244 Li]/(1 +
Exp[-0.416831 + 0.540171 + 0.564244 Li]);
uc = -0.416831 + 0.540171 + 0.564244 Li + 1.96 Sqrt[Var];
uctran = Exp[uc]/(1 + Exp[uc]);
ic = -0.416831 + 0.540171 + 0.564244 Li - 1.96 Sqrt[Var];
ictran = Exp[ic]/(1 + Exp[ic]);
Show[Plot[uctran, {Li, -3, 3}], Plot[ictran, {Li, -3, 3}],
Plot[fem, {Li, -3, 3}], PlotRange -> {{-3, 3}, {0, 1}},
AxesOrigin -> {-3, 0}, Frame -> True,
FrameStyle -> Directive[Orange, 12]]
```

Code:

`acov[data_, dist_, paramlist_, mleRule_] := Block[{len, infmat, cov}, len = Length[data]; infmat = -D[LogLikelihood[dist, data], {paramlist, 2}]/len /. mleRule; cov = Inverse[infmat]];`

Code:

`data = {31, 46, 70, 87, 87, 93, 114, 128, 133, 134, 143, 155, 161, 161, 163, 177, 181, 207, 207, 226, 302, 315, 319, 347, 347, 362, 375, 377, 413, 440, 447, 461, 464, 511, 524, 556, 800, 860, 880, 954, 5200, 12000}`

Code:

```
dist = ProbabilityDistribution[{"CDF",
Exp[-a Exp[-x b] - c Exp[-x d]]}, {x, 0, Infinity},
Assumptions -> {{0 < a < 10}, {0 < b < 1}, {0 < c < 10}, {0 < d <
1}}];
res = FindDistributionParameters[data,
dist, {{a, 3.8}, {b, 0.006}, {c, 0.08}, {d, 0.0002}}]
```

To obtain the ACOV we can use the acov function.

Code:

`asincov = acov[data, dist, {a, b, c, d}, res];`

Code:

`Varian = grad.asincov.grad;`

Code:

`Varian /. {x -> 1};`

In Mathematica this is tantamount to the quantile for a t with 38 degree of fit and 0.95 confidence level:

Code:

`Quantile[StudentTDistribution[41], 0.95];`