"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.
