Re: CONFIDENCE LIMITS AROUND A PERCENTILE
- From: Jack Tomsky <jtomsky@xxxxxxxxxxxxx>
- Date: Wed, 05 Dec 2007 11:14:12 EST
Date: Dec 4, 2007 8:48 PM
Author: Jack Tomsky
Subject: Re: CONFIDENCE LIMITS AROUND A PERCENTILE
Hi everyone...I am a rank amateur when it comes
tostatistics so> asking for your expertise. > I have
a dataset of 30 values. I am being asked to find the
95th> percentile of this data and then construct a
95%> confidence interval> around that 95th percentile
value. To add insult to> injury I am asked> to do
this both parametrically and nonparametrically. >
Could anyone> supply some insight as to how this is
s done. I have> tried to do some> research on this
and have found that in statistical> quality control
terms this methodology is called creating tolerance>
limits. Or at> least it looks that way. I can
figure out how to> apply that formula> but the
nonparametric corollary is nowhere to be> found. If
anyone> could help me, direct me in the right
direction or> provide any> guidance at all it would
be most appreciated.
[Response
Fot the normal distribution, it's based on the
noncentral t distribution. Nonparametrically, it's
based on order statistics.Jack]
MY RESPONSE
If (the OP) consult a credible text–book) as
Conover´s Practical Nonparametric Statistics h would
find that his problem has a very simple solution
CONFIDENCE INTERVALS FOR QUANTILES
The interval [rth, sth] that contains 1-alpha of the
data (significance = alpha) is such that the orders
are
______r = n*p + w(alpha/2) * sqrt (n*p*(1-p))
______s = n*p + w(1-alpha/2) * sqrt(n*p*(1-p))
(respectively the lower and upper bonds)
With n the size, p=0.95,
w(alpha/2)= -1.645, w(1-alpha/2)= 1.645
This algorithm is based on the Normal Distribution
Approximation well fitted if n*p >5 and n*(1-p)>5
Since p=0.95 we have n> 5/0.95 and n> 5 / 0.05 = 100.
Putting n=100 the 95% interval is bounded by the 95
th value +/- 3.6 or finally by the 91th and 99th
ordered sample values
For n=10000 we have 9500 th +/- 22 or [9478 th,
9522th].
Jack Tomsky insist in to post nonsense, the answers
have misdirect proposes or doesn’t lead nowhere.
Incredibly!
__________
Luis Amaral Afonso
The exact confidence interval for the quantile of a normal is based on the noncentral t distribution and is given by
L = Xbar + s*t'(Zp*sqrt(N), N-1; 1-alpha/2)/sqrt(N)
U = Xbar + s*t'(Zp*sqrt(N), N-1; alpha/2)/sqrt(N),
where N is the sample size, Xbar is the sample mean, s is the sample standard deviation, Zp is the standard normal quantile, 1-alpha is the confidence level, and t' is the noncentral t distribution.
Because of his lack of education, Afonso considers exact analytic solutions to be nonsense and prefers the wrong answers.
Jack
.
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