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Hi everybody,

I have never done statistics in Maple.  In a simple calculation, I need to calculate the RMS of 55 numbers.  The average of those numbers is 100484.3 and it is given that the RMS is 1.28 counts.  I have the a list of the 55 numbers.  Since I don't need a demonstration, it would help me a lot if you could tell me how to load the data from a *.txt file (one number per line), and use the appropriate commands to obtaine the result that is only given.

For the moment, I only need the steps to proceed with the calculations and how to do it in Maple.

Thank you very much in advance for your help.

 

--------------------------------------
Mario Lemelin
Maple 18 Ubuntu 13.10 - 64 bits
Maple 18 Win 7 - 64 bits messagerie : mario.lemelin@cgocable.ca téléphone :  (819) 376-0987

I am interested in using both in the same graph. And I am aware that i can drag from one to the other manually, but I am interesseted in doing it in a command. 

So i have 3 vectors a, b and an error. 

I can define ExponentialFit function based on a and b, and the plotting both observations, errors and the fitted function.

 

Per Kirkegaard

 

I have a statistics task that is very complicated to me, and I would really appretiate some help!

I believe that I have completed task a, but task b and c seems to be too hard for me to grasp.

Here is a presentation of the task text:

 

xi 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
yi 12.3 12.8 16.8 17.9 20.5 22.7 24.9 27.5 26.9 29.4 30.3 34.2 36.4 36.7 41.1

Assume that the data represents independent realizations of a "linearlnormal" model.

Yi = α + β xi + ei i = 1..15,

where ei˜N(0,σ), where sigma is unknown.

 

(a) create a scatterplot of the data {f(xi,yi)}15 ; i=1
Then calculate the empiric correlation coefficient.

(b) Find the minimum least square estimate for α & β.

(c) Create a 95% confidence interval for the parameter β. What does this confidence interval express?

Is there anyone out there who are able to help me with some maplecode?


thanks in advance,

Soeta

The question was "Let X be the random variable uniformly distributed in the disk centered at the origin O(0,0) with radius 1 and let Y be the random variable uniformly distributed in the square having its vertices A(6,-1), B(9,-2), C(8,-5), and E(5,-4). What is the PDF of the distance between X and Y? Is it possible to find that with Maple?"
Having a long think about the topic, I draw the conclusion that the exact closed form of the PDF/CDF, even the one can be found, would be useless because of its complexity.
Thus, an approximate formula for the CDF/PDF under consideration is a proper way. That formula can be derived in such a way. First,rotating the picture, we may consider the square having its sides horizontal or vertical: K((1/5)*sqrt(1410)-(1/2)*sqrt(10),1/sqrt(10)), L((1/5)*sqrt(1410)+(1/2)*sqrt(10),1/sqrt(10), M((1/5)*sqrt(1410)+(1/2)*sqrt(10),1/sqrt(10)-sqrt(10)), Q((1/5)*sqrt(1410)-(1/2)*sqrt(10),1/sqrt(10)-sqrt(10)). The geometry behind that is omitted.
We randomly choose a point P1 belonging to the square [-1,1] x [-1.1]. If the one belongs to the disk {(x,y):x^2+y^2 <=1}, then we randomly choose a point P2 from the square [K,L] x [Q,L]. Next, we calculate the distance between P1 and P2 (The  LinearAlgebra[Norm] command http://www.maplesoft.com/support/help/Maple/view.aspx?path=LinearAlgebra/Norm is used to this end.) and add it to the set S.This is repeated 2*10^4 times.

Converting S to an Array A, we constuct the empirical distribution X by A and find its mean mu and standard deviation sigma.


7.67568900820260

1.029831470

Let us compare the obtained empirical distribution and the normal distribution with the parameters mu and sigma.

The plot suggests a good fit between these. However, it is only semblance. Applying the Kolmogorov-Smirnov test (for example, see  http://www.mapleprimes.com/posts/119903-The-KolmogorovSmirnov-Test
), we  calculate

and

3.32619143372726

while the critical value equals 1.358098639 at the level 0.05. Thus, the hypothesis about the concordance should be rejected.

Also we draw the approximation to the PDF:


CDF.mw

Dear Maple users

The Statistics package contain a Histogram command which can produce a linegraph (or what you call it). Ex:

 

with(Statistics):

A:=[3,5,7,2,11,2,1,10,6,15]:

Histogram(A,discrete)

 

It will calculate the frequency of each of the outcomes as they appear in the list and will plot them accordingly: 

would like to draw the pdf, however after tried two methods, still can not draw it,

how to draw?

 

l := [binomial(349,20), binomial(349,8),binomial(349,22),binomial(349,38),binomial(349,49),binomial(349,36),

binomial(349,59),binomial(349,41),binomial(349,32),binomial(349,21),binomial(349,23)];

sl := gfun[listtoseries](l, x, egf);

g := convert(sl, ratpoly);

g := g - subs(x=0, g); 

if you have four teams a b c and d playing in a tournament and teams a and d have not played each other but all other teams have played .. typing this from my phone but its not too friendly ...  soccer scores how would you determine who would win and by what score

I need to integrate the Student-T function which is a function of two variables (nu and t). Maple allows me to use this function through this command:

with(Statistics);

PDF(RandomVariable(StudentT(nu)),t);

The integration of this function gives me the probability (p) of certain event (nu is already known):

p=Int(Student,t=-infinity..X)

If I provide the value of X I can compute the integral easily, now the problem appears when I want to...

as i do not know what is nu and omega in beta distribution, can i solve it by equating with mean of real data's mean and stardard derivation, to find nu and omega, but solution of nu is negative, 

i guess this method may be wrong, as result diagram is not fitted with real data

and weibull using this method, can not be solved and said solution lost

what is the correct way to draw it?

How to update this code to the Statistic Package for the graph below?for n from 1 to 15 do num:=convert(evalf(n), string): tracker[n]:=textplot([18,0.3,`Lambda is `.num],color=blue): H[n]:=ProbHist(PoissonPDF(n,x),0..25,26): N[n]:=plot(NormalPDF(n,n,x),x = 0..25):  P[n]:=display({H[n],N[n],tracker[n]}): od: display([seq(P[n], n=1..15)], insequence=true,title="Normal Approx. to the Poisson. Lambda is increasing from 1 to 15");

One of the most basic decisions a baseball manager has to make is how to arrange the batting order.  There are many heuristics and models for estimating the productivity of a given order.  My personal favourite is the use of simulation, but by far the most elegant solution from a mathematical perspective uses probability matrices and Markov chains.  An excellent treatment of this topic can be found in Dr. Joel S. Sokol's article,

restart;

with(ExcelTools):

with(ListTools):

with(DynamicSystems):

filename := "MSFT";

close3 := Import(cat(cat("C://US//",filename),".xls"), filename, "E2:E100");

this i usually use ln(close3[n]/close[n-1]) as the original series

as i do not know whether Sample in statistics package can use this directly,

or 

need to use counting to classify them into some group and got a distribution graph

restart;with(Statistics):

MartinPoisson := (p/((p+2*q*T)/(1+p*T+q*T^2)-(p*T+q*T^2)*(p+2*q*T)/(1+p*T+q*T^2)^2))^r*r^2*T^n/factorial(n);

Dist := subs(T=t,MartinPoisson) assuming t > 0;

MartinDist := Distribution(CDF = unapply(piecewise(t>1,Dist,0),t));

MartinDist := Distribution(CDF = unapply(Dist, t)) assuming t > 0; #Change 2

X:=RandomVariable(MartinDist);

MartinDensity := PDF(X,t);

MartinDensity := subs(t=x, MartinDensity);

Using the Quandl API, we have created a Quandl library for Maple.  The Quandl library for Maple provides easy access to Quandl’s repository of over 5 million time-series datasets from directly inside Maple, allowing you to utilize Maple’s robust tools for mathematical statistics and data analysis on Quandl’s extensive collection of data.  This library features a similar set of functionalities to the Quandl

Hi,

I'm trying to compute the mean and variance of a custom discrete random variable, as indicated in the file below (well, I do know its mean and variance, but I need the variable in a more complex context). Defining it is no problem at all, but when I parse Mean(Y) and Variance(Y), a weird expression shows up, despite my random variable being really simple. Does somebody know what goes wrong?

 

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