Re: p-value
- From: "Ethan.Johnsons@xxxxxxxxx" <Ethan.Johnsons@xxxxxxxxx>
- Date: 26 Oct 2006 09:31:37 -0700
Reef Fish wrote:
espyrian wrote:
Ethan.Johnsons@xxxxxxxxx wrote:
espyrian wrote:
Ethan.Johnsons@xxxxxxxxx wrote:
The mean was measured in 12 patients 24 hours after they received a
newly proposed antibiotic was 1.2 mg/dL.
If the mean = 1.0 and sd = 0.4 mg/dL in the general population, with CI
0.05, I need to test if the mean level of patients is different from
that of the general population.
So, I came up with:
Ho = the mean level is the same as that of the general population
H1 = the mean level is different from that of the general population
The test statistic is:
(1.2-1.0)/(0.4/sqrt(12)) = 1.732051
The critical value is:
qt(p=0.05, df=11) = -1.795885
Because the t = 1.732051 > -1.795885,
So, we can accept Ho at the 5% level.
My question is how do you calculate the p-value?
I came up with
p = 2* pt(q=1.732051, df=11) = 1.888827but, I know it is off far.
thx
ej
You're close. Using tables, you'll see that your t-stat of 1.732 lies
between p=0.10 and p=0.05 on the 11 d.f. row. That is not enough to
reject your H0 in any event, but your H1 is two tailed (i.e. < or >),
so you must double the p-value, putting it between p=0.20 and p=0.10.
So you are correct that you retain H0 at the alpha=0.05 level.
Using software, such as Excel, will give you a more precise figure:
TDIST(1.732,11,2)=0.111
HTH
Thank you for reply.
I am still confused.
If the test statistic is negative, (i.e. -1.52) and n =7
then, it would be TDIST(-1.52,6,2)?
Excel gives no value with TDIST(-1.52,6,2).....
It seems that p-value is:
pt(q=-1.52, df=6)[1] 0.089661
thx
ej
P(T<-1.52) is equivalent to P(T>1.52). So ignore the minus sign.
This comment is fine with respect to a TWO-tailed Alternative.
The p-value of a one-sided test=TDIST(1.52,6,1)=0.0897.
This point has a LENGTH history of discussion in sci.stat.math.
What you said is incorrect, relative to the DEFINITION of a p-value:
p-value = Pr( Test Statistic T is "more extreme" than the observed T*)
where "more extreme" is defined by the Alternative Hypothesis!
Thus, if the Alternative Hypothesis is Ha: mu < given value,
and the observed T = -1.52, then the p-value is P(T < -1.52) =.089661
and if the Alternative is Ha: mu > given value, then
the p-value is P( T > -1.52) = .91034
Similarly for an observed T of 1.52.
The fact that one seldom sees a p-value greater than 0.5 is one of
the evils of statistical computing software. Most of them give
p-values
correponding to a TWO-tailed test, in which case the removal of the
sign of the observed T makes no difference.
But for ONE-tailed tests, the INEQUALITY in Ho is of paramount
importance.
These points can be found in Google by doing a search on
Definition of "p-value"
returned 14 threads, totally over 300 posts. :-) Not all 300 posts
are strictly on p-value definition, but merely contained the
keywords. In particular, one of the 14 threads was
"Now a RoadMap (was Re: m00es on NOISE; and m00es's
NOISE; and a DATA QUIZ!!!) "
which contained 75 posts, nearly all NOISE by m00es. :-)
Below is a pure post of statistical substance of relevance:
http://groups.google.com/group/sci.stat.math/msg/0677873d9b959a86?hl=en&
which gave many numerical examples as well as the summary:
RF> the key element of a P value is that you CANNOT determine
RF> a P value unless you know the Alternative Hypothesis,
RF> whether it is one-tailed or two-tailed, and if one-tailed, WHICH
tail.
-- Reef Fish Bob.
I am a little confused as usual :-)
To dobule-check,
If we assume that the sample sd is 0.6 and the sd of the general
population is unknown:
then, the test statistic is:
(1.2-1.0)/(0.6/sqrt(12))[1] 1.154701
We need to use t-test because "If the population sd is not known but is
estimated by the sample standard deviation then a t-test should be
used."
then, the p-value is pt(q=1.155, df=11) = 0.863718 ????
thx
ej
.
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