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Jim West
 
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Default Blindtest question

In article l2mWa.36515$Ho3.6598@sccrnsc03, Harry Lavo wrote:
"Jim West" wrote in message
news:fnaWa.19572$cF.7720@rwcrnsc53...

You are indeed cherry picking. With 12 individuals the probability that
one would would appear to meet the 95% level is fairly high. Remember
that you can expect 1 in 20 to meet that level entirely by random. It
is not acceptable scientific practice to select specific data sub-sets
out of the complete set. Otherwise you could "prove" anything by simply
running enough trials and ignoring those you don't like. Check any peer
reviewed journal.


Yep it is more probable than one in twenty. But not so high that we have to
accept your assumption that he/she *IS* the one-in-twenty.


Where did I assume that?

In any event, 11 out 15 has a probability of 5.9 % of occuring by chance.
That does not meet the 95 % confidence level. It would be rejected in
a peer reviewed statistical study. (If that was the only data more
trials would called for. But it wasn't the only data.)


My mistake..you are correct. So the number is only 94.1%. But 12 out of 15
yields 98.2%. Which is closer to the 95% standard?


Irrelevant. You can't arbitrarily play with the numbers.


Where are you getting your numbers? The data they posted on the
web page showed that there were 52 correct answers in 96 trials.
At least 52 correct answers will occur entirely by chance 23.8 %
of the time. This is far from statistically significant.


Oops! Late at night and the only book of binomial probabilites I had handy
contained only raw data, not the cumulative error tables that I was used to
from my previous work. So I goofed. I agree that the three-to-one odds are
not statistically significant, and so the full panel results are null for
both cables. My apologies.

The issue with the individuals still stands, however.


No it doesn't. Even with cherry picking (which is not valid anyway)
no individual performed at a statistically signficant level. Period.