On 30/09/2014 2:11 PM, Andre wrote:
Hi Duncan,
No, that's correct. Actually, I have data set below;
Then it seems Excel is worse than I would have expected. I confirmed
R's value in two other pieces of software,
OpenOffice and some software I wrote a long time ago based on an
algorithm published in 1977 in Applied Statistics. (They are probably
all using the same algorithm. I wonder what Excel is doing?)
N= 1223
alpha= 0.05
Then
probability= 0.05/1223=0.0000408831
degree of freedom= 1223-2= 1221
So, TINV(0.0000408831,1221) returns 4.0891672
Could you show me more detail a manual equation. I really appreciate
it if you may give more detail.
I already gave you the expression: abs(qt(0.0000408831/2, df=1221)).
For more detail, I suppose you could look at the help page for the qt
function, using help("qt").
Duncan Murdoch
Cheers!
On Wed, Oct 1, 2014 at 1:01 AM, Duncan Murdoch
<murdoch.dun...@gmail.com <mailto:murdoch.dun...@gmail.com>> wrote:
On 30/09/2014 1:31 PM, Andre wrote:
Dear Sir/Madam,
I am trying to use calculation for two-tailed inverse of the
student`s
t-distribution function presented by Excel functions like
=TINV(probability, deg_freedom).
For instance: The Excel function =TINV(0.0000408831,1221) =
returns
4.0891672.
Would you like to show me a manual calculation for this?
Appreciate your helps in advance.
That number looks pretty far off the true value. Have you got a
typo in your example?
You can compute the answer to your question as
abs(qt(0.0000408831/2, df=1221)), but you'll get 4.117.
Duncan Murdoch
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