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Topic: Intra-permutations to test two different mean values
Replies: 6   Last Post: Feb 15, 2013 12:44 PM

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Luis A. Afonso

Posts: 4,526
From: LIsbon (Portugal)
Registered: 2/16/05
Re: Intra-permutations to test two different mean values
Posted: Feb 14, 2013 4:50 PM
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Deception and joy. . .

The only motivation leading me to try to find out a algorithm able to *multiply* data was the repugnance to concede, in spite of fully used nowadays by Statisticians, that Bootstrapping is a illogic method to do it. In fact to repeat items at a sample, furthermore with no control how much are they, seems to me implausible or odd. So a modified mean, mm, was defined as the sum of the products of the items by l(j) = j/(1+2+?+n) : mm= sum (l(i)*x( )) where x( ) is a randomly item chosen without replacement among the n sample items exhaustively. In terms of expectation is E(mm)= sample mean. I was persuaded, till today, an improvement was found. Surprisingly Bootstrap and this Intra-Permutation method when applied to evaluate Confidence Intervals for the difference of two sample means leads to the same results . . .
___95% and 99% C.I. (fractiles .025, .975 and .005, .995) samples regarding the difference of mean values of 30 females and 30 males of a spider´s species____(1´000´000 replications)
___Intra-permutations______
___[1.76, 2.67]__[1.63, 2.81]
___[1.76, 2.67]__[1.63, 2.81]
___[1.76, 2.67]__[1.63, 2.81]
___[1.76, 2.67]__[1.63, 2.81]

___Bootstrap______________
___[1.747, 2.668]__[1.598, 2.810]
___[1.747, 2.668]__[1.598, 2.842]
___[1.747, 2.670]__[1.598, 2.812]
___[1.746, 2.668]__[1.598, 2.810]

This is deceptive conclusion in practical terms, but conceptually interesting. In fact, the method demands a very weak feature: the items obtained in each observed sample must be completely independent, i.i.d..

Luis A. Afonso



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