brain sizes: Einstein's and women's
Jet
thatjetnospam at yahoo.com
Sun Jul 21 02:17:39 EST 2002
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
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