brain sizes: Einstein's and women's

John Knight johnknight at usa.com
Mon Jul 22 18:20:18 EST 2002


"Jet" <thatjetnospam at yahoo.com> wrote in message
news:3D3A62FB.4A11CBC3 at yahoo.com...
>
>
> Parse Tree wrote:
> >
> > "John Knight" <johnknight at usa.com> wrote in message
> > news:3Jq_8.13603$Fq6.1134149 at news2.west.cox.net...
> > >
> > > "mat" <mats_trash at hotmail.com> wrote in message
> > > news:43525ce3.0207190255.4ebf0ab1 at posting.google.com...
> > > > > Nice try, "mat", but you're way off.  Thanks for giving it a try,
> > > though.
> > > > > P1 = P2 = P3 = P4 = 0.25
> > > > > In other words, the probability of getting one question right by
> > > guessing at
> > > > > a 4 part multiple choice question is 0.25
> > > > > But to figure the probability of getting four of the same
questions
> > > right,
> > > > > not the *same* answer on each question, but the *correct* answer,
you
> > > must
> > > > > add them up, which is P1 + P2 + P3 + P4 = 1.0
> > > > >
> > > >
> > > > It is quite amusing how you spount this anti-"everything other than
> > > > me" dogma, citing test results, when you yourself are apparently
akin
> > > > to two short planks.
> > > >
> > > > Your math suggests that I am CERTAIN to get one answer correct if I
> > > > guess on four multiple choice questions (probability of 1).  Can you
> > > > not see that this is illogical just on the basis of common sense?
> > > > thats like saying if I throw a dice six times I'm sure to throw a
six,
> > > > which of course is totally dumb
> > > >
> > >
> > > No, Mat, it's not, and if you'd have answered a TIMSS question this
way,
> > you
> > > would have been just as wrong as almost 100% of American girls were on
> > some
> > > of these probability and statistics questions.
> > >
> > > The statement is that the probability is 1.0 that you'll get one
answer
> > > correct if you just guess on four different questions with four
multiple
> > > choice answers.  That's much different than you are "CERTAIN to get
one
> > > answer correct".
> >
> > Probability of 1 indicates certainty.  You evidently don't understand
> > statistics.
> >
> > > Obviously you were never taught probabilities and statistics.  This is
the
> > > most basic principle possible.  No, it doesn't happen that way EVERY
time,
> > > but over a long series of questions and answers, it will eventually
end up
> > > this way.
> >
> > Actually, I have already proven you to be wrong here.
> >
> > P(answer a question correctly) = 0.25
> > P(at least one of two correct) = P(both correct) + 2 * P(one correct and
> > other incorrect) = 0.25*0.25 + 2 * 0.25 * 0.75 = 0.4375
> > It's not the 0.5 like you thought.
> > P(at least one of three correct) = 1 - P(all 3 wrong) = 1 -
(0.75*0.75*0.75)
> > = .578125
> > P(at least one of 4 correct) = 1 - P(all 4 wrong) = 1 - (0.75)^4 =
0.6835
> > (etc)
> >
> > Clearly not equal to 1.
> >
> > > > > > Your assumptions are also further invalid in that you calculate
the
> > > > > > number 'guessed correctly' by assuming that all who got it wrong
> > > > > > 'guessed incorrectly'.  Not only is this illogical in that you
are
> > > > > > characterising one group of students on the basis of another
group
> > but
> > > > > > it also leads to strange conclusions such as if 70% answer
> > correctly,
> > > > > > 10 of this 70% of this is accounted for by correct guesses,
whereas
> > if
> > > > > > 100% answer correctly no-one guessed, since there are no
incorrect
> > > > > > answers.
> > > > > >
> > > > >
> > > > > No.  The assumption is valid, particularly since the percent
correct
> > is
> > > > > lower than if they had just all guessed, and when all of the
questions
> > > are
> > > > > answered.  It's true that some of the questions might reflect some
bad
> > > > > instruction in the classroom, but when the responses are spread
across
> > > the
> > > > > spectrum like they were, and still the score was lower than pure
> > > guesses,
> > > > > then the only conclusion can be that they guessed on most of the
> > > questions.
> > > >
> > > > You still have yet to address the conclusion that if 100% answer
> > > > correctly no-one guessed and if 0% answer correctly then something
> > > > very strange happens as the population taking the test suddenly
> > > > increases by a third but their papers are somehow lost.  As you
might
> > > > see from working with the latter case, your analysis does not hold.
> > >
> > > It's not clear what you're saying, but the last part of your statement
> > means
> > > that it's obviously wrong--and it's ceretainly not consistent with the
> > > original statement.  Consider the two extremes:
> > >
> > > If zero percent get a four part multiple choice question wrong, then
> > they're
> > > scoring 25% lower than if they'd just guessed, in which event might
> > conclude
> > > that there's a high probability that they were taught the wrong thing
if
> > > everyone selected just one wrong answer.
> >
> > If zero percent get a question wrong, then 100% got it right.  How is
this
> > 25% lower than if they had guessed?  If they guessed would 125% of
people
> > got it right?
> >
> > >  The other extreme is if 50% get
> > > the answer correct, and the rest of the answers are spread evenly over
the
> > > other three answers, then this *is* an indication that 50% understood
the
> > > problem [50% guessed wrong, x = total percent who guessed, .25x = the
> > total
> > > number who guessed correctly, .75x = the percent who guessed
incorrectly,
> > x
> > > = 66.67 percent = total guesses, .25x = 16.7 percent = percent who
guessed
> > > correctly, and .75x = 50 percent].
> >
> > So even though 50% of people understood the problem, some of those
people
> > still guessed?
> >
> > Your logic is flawed.
>
> John is a dumb ass, I tried to point out to him why is equations are
> nonsense.
>
> Let's look at the 50% case. If 50% knew the answer, and 50% guessed
> randomly, the random guesses would be spread over all 4 answers, not
> just the incorrect ones.
>
> 50/4=12.5
>
> So, 62.5% would get the answer right, the 50 who knew, and the 12.5 who
> made a lucky guess.
>
> 12.5% would guess each of the wrong answers.
>
> J

You're still not setting up the problem properly.  It's impossible that
62.5% got the answer correct, because it's a given that only 50% did.

>From the above, we know that x = total guesses = 66.7%, and that correct
guesses = .25x = 16.7%.

>From the 50% who got the answer correct, we must subtract the 16.7% who got
it correct by guessing to determine that 33.3% got it correct because they
understood the problem.

Viewed another way:

66.7% = total guesses
16.7% = total correct guesses
50% = total incorrect guesses
33.3% = got it correct by solving the problem

So when 50% get the answer correct on a four part multiple choice question,
and if students just guessed at the answers they didn't know or understand
(which is the worst case scenario), then (ignoring the standard error) as
little as one third of them actually demonstrated a knowledge or
understanding of the question.

Let's look at what happens if 90% got it correct.  x = total guesses, .25x =
correct guesses, .75x = incorrect guesses = 10 percent, x = 13.33 percent,
.25x = 3.33 percent = correct guesses, 90% got it correct - 3.33% got it
correct by guessing = 86.67% demonstrated a knowledge or understanding of
the question.








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