brain sizes: Einstein's and women's

Parse Tree parsetree at hotmail.com
Mon Jul 22 19:03:50 EST 2002


"John Knight" <johnknight at usa.com> wrote in message
news:Sq0%8.15470$Fq6.1572196 at news2.west.cox.net...
> "Jet" <thatjetnospam at yahoo.com> wrote in message
> news:3D3A62FB.4A11CBC3 at yahoo.com...
> >
> >
> > Parse Tree wrote:
> > >
> > > "John Knight" <johnknight at usa.com> wrote in message
> > > news:3Jq_8.13603$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.

How are you calculating the number of people that guessed?  I would like to
see how this is possible, since I know it is not.





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