Re: Calculation of critical p-, z-, t- and F-values



On 13 Nov., 09:34, "Nasser Abbasi" <n...@xxxxxxxxx> wrote:
<schuetz-dan...@xxxxxxxxxxx> wrote in message

news:1194941721.124021.86460@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Hallo,

could you give me the formulas for calculating critical values for z,
p, t and F for a significance level (e. g. 99 %)? Or could you say me
where I can find those?

Please excuse my bad English.

Thanks
Daniel


Ofcourse one can calculate these from the distribution itself. but it is
easier to use tables. (our text book has them)

You can look at these tables on-line, some are on wikipedia and google
book
has it. do google book search on
Elementary Statistics Tables By Henry R. Neave

Here is also a web site that has them
on-linehttp://www.statsoft.com/textbook/stathome.html

Scroll down the right hand side to the 'distribution tables' button and
click on it.

Nasser


<schuetz-daniel@xxxxxxxxxxx> wrote in message
news:1194946596.904993.24770@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Yes, surely. But if I have very much populations that should be
compared then it would be better to make values automatically
computed. This would be time-saving.

(please do not top post, makes it harder to follow).

Ok, lets look at this problem a little more carefully.

To make like easier, let look at the Z distribution (or standard), i..e
normal with zero mean, and sigma 1.

For confidence interval say 95% what this means is to find the x value such
that are the area under the pdf from -x to +x is .95 (since the area is the
probability). So the problem is to find x such that the area is some value.

For example, we know that the critical value for Z for 95% is 1.96, this is
because if we integrate from -1.96 to +1.96 the pdf we get .95:

Integrate[PDF[NormalDistribution[0, 1], x], {x, -1.96, 1.96}]
0.9500042092453821

Ok, but what we want is the reverse problem, we want x such as the area is
such and such. Then how to find x in this case?

One could find the indefinite integral of the pdf which will be a function
of x, and then solve for x by setting the integral to .95 or whatever C.I.
one wants.

But the indefinite integral can be a nonlinear form as for example here with
standard pdf:

area = Integrate[PDF[NormalDistribution[0, 1], x], x]

(1/2)*Erf[x/Sqrt[2]]

You see, the integral is expressed in terms of erf() function. not a nice
function to work with.

Using CDF is no help help. The same problem.

So the easy way is to make a TABLE. i.e. pick x, and then find the area for
that x. and this is what these tables do.

For example, here is one line of Mathematica code to generate few critical
values for the normal distribution.
Table[{to, 2*Integrate[PDF[NormalDistribution[0, 1], x], {x, 0, to}]}, {to,
1.9, 2, 0.01}]

{{1.9, 0.9425668801186563}, {1.91, 0.9438667864345321},
{1.92, 0.9451421003498051}, {1.93, 0.9463931620067536},
{1.94, 0.9476203098833256}, {1.95, 0.9488238807243032},
{1.96, 0.9500042094744706}, {1.97, 0.9511616292137779},
{1.98, 0.9522964710944976}, {1.99, 0.9534090642803622},
{2., 0.9544997358876777}}

So, you give me the PDF, I can give you back the tables. I can make the
tables as detailed as you want.
It would be easy to write a function then to do a 'lookup' on these tables
to find the critical value for the C.I. you want.

Nasser


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