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Luis A. Afonso
Posts:
4,277
From:
LIsbon (Portugal)
Registered:
2/16/05
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How Bootstrap is able to attain alpha
Posted:
Oct 3, 2012 11:53 AM
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How Bootstrap is able to attain alpha
In order to have an idea how Bootstraps is able to get worthy confidence intervals for normal data I though to compare:
___[ U, V]
___U = mX - T * sd /sqrt(n)
___V = mX + T * sd /sqrt (n)
___ sd= sqrt [( sumsquares - n * mX^2 ) / (n-1)]
where
____ mX = sample n-size mean value
Procedure:
____We simulate, via Box-Muller algorithm, 10´000 N(0,1) samples each was one the ?source? of 100 Bootstrap pseudo-samples: each mean value is checked it is outside [U, V] and counted. Because this interval was set alpha = 0.05 (2 tails) is expected that a total of 500 can be found outside it. The dependence on sample sizes were studied.
Results : Bootstrap N(0 , 1) samples
________________________________Student Test____
________________outside [U, V]__
_____n=10___________ 1.6%________T=2.262
_____n=25___________ 3.5__________ = 2.064
_____n=50___________ 4.2__________ = 2.010
_____n=100__________ 4.6__________ = 1.984
_____n=200__________ 4.8__________ = 1.972
_____n=300__________ 4.9__________ = 1.968
Here is clearly shown that only for sample sizes greater than 300 the nominal 5% alpha can be attained.
Luis A. Afonso
REM "ONEdata"
CLS : PRINT : PRINT "______ONEDATA____";
PRINT " Bootstrap and Student"
PRINT " ONE SAMPLE for Normal data sources "
REM SOURCE
pi = 4 * ATN(1)
INPUT " Sample size "; n0
INPUT " Student .975 size-1 df "; T2
INPUT " Bootraps/ Source "; ali
INPUT " how many sources "; hm
DIM X(n0), XX(n0), W(8004)
REM
FOR source = 1 TO hm
REM
LOCATE 10, 50
PRINT USING "#########"; hm - source
RANDOMIZE TIMER
REM outbysource = 0
mX = 0: sq = 0
REM THE SOURCE
outbysc = 0
FOR i = 1 TO n0
aa = RND
X(i) = 0 + 1 * aa * COS(2 * pi * RND)
mX = mX + X(i) / n0
sq = sq + X(i) * X(i)
NEXT i
sd = SQR((sq - n0 * mX * mX) / (n0 - 1))
U = mX - T2 * sd / SQR(n0)
V = mX + T2 * sd / SQR(n0)
REM [U,V] is the acceptance interval for means
REM BOOTSTRAP
FOR j = 1 TO ali
mmean = 0
FOR ii = 1 TO n0
g = INT(RND * n0) + 1
mmean = mmean + X(g) / n0
NEXT ii
IF mmean < U THEN below = below + 1 / ali
IF mmean > V THEN above = above + 1 / ali
NEXT j
NEXT source
a = below / hm: ab = above / hm
LOCATE 10, 42: COLOR 14
PRINT USING
"###.#% BOOTSTRAP alpha"; (a + ab) * 100
COLOR 7
END
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