Re: Friedman test

[Tim Menzies]

you are doing the right thing. as you look down the rows you will see that the values converge to some final figure. using the last row is correct

t

On Tue, Dec 2, 2008 at 7:47 PM, Bryan Lemon <bryan.le...@gmail.com> wrote:

That sheet of majik numbers that you gave us only goes to 99 rows. We need row 270. What do we do? For tonight, I am just going to work with the number furthest down.

Thank you,
Bryan Lemon

--
timm, a/prof, csee, wvu, usa

Arthur C. Clarke

[Tim Menzies]

dear bryan,
regarding getting negative numbers in the calculation of the friedmann test.

the maths of that test impose limitations on the space of possible experimental designs. your k=55 results in a k(k+1)^/4 results in

(55 * ((55 + 1)^2)) / 4 = 43 120

which is highly lifely to generate a negative (and failed) friedmann

t

On Tue, Dec 2, 2008 at 8:50 PM, Bryan Lemon <bryan.le...@gmail.com> wrote:

...We are using 55 k. We can reduce it... Will advise...

Thank you,
Bryan Lemon

On Tue, Dec 2, 2008 at 8:46 PM, Tim Menzies <t...@menzies.us> wrote:

On Tue, Dec 2, 2008 at 8:42 PM, Bryan Lemon <bryan.le...@gmail.com> wrote:

We come up with a result though. Rather depressing. Our magic number was 2.306... and the number that Ff came up with was -25. So, although we passed the second test, we failed the first.

a negative number? that is wacked- can only happen when k(K+1)^2/4 > r^2. how many k you using? can you reduce k?

t

 

Thank you,
Bryan Lemon

George Burns

--
timm, a/prof, csee, wvu, usa

Tom Robbins