12th grade math standards
Michael Zarlenga
on the proposals for new standards in education.
Under the proposals for 12th grade math we find :
1. Which is a better fit, a round peg in a square hole, or a
square peg in a round hole?
2. Ann's free-throw %age using her old method is 60%. Using
a new method she makes 9 of her first 10 attempts. Is the
new method better than the old? (Hint: "advanced statisti-
cal formula must be used.")
1 was child's play. I bet most high school sophomores in pri-
vate schools can solve it in under 2 minutes. This is a
10th grade level question, and even then, a simple one.
2 was worded ambiguously (I changed the wording here to make
it less so), but, even so, there's no "advanced statisti-
cal formula" needed, unless multiplication is advanced.
If these are our standards for 12th grade students, we're set-
ting our sights too low.
--
-- Mike Zarlenga
Bob Dole for President in 1996.
Choose the Veteran over the Veteran Liars.
finger zarl...@conan.ids.net for PGP Public key and killfile
George D. Phillies
On 29 Mar 1996, Michael Zarlenga wrote:
> This week's issue of USN&WR ("Dumb and Dumber") had a section
> on the proposals for new standards in education.
>
>
> Under the proposals for 12th grade math we find :
>
> 1. Which is a better fit, a round peg in a square hole, or a
> square peg in a round hole?
What do you mean, "better"? Were there, say, pictures of the two?
Assuredly, one can inscribe a square in a circle, or inscribe a circle in
a square, but what was the question supposed to mean? There seems to be
something missing.
This is a very dangerous wording to the question, since "better" can
include anything from "more firmly wedged in place" to "rattles louder
when shaken".
> 2. Ann's free-throw %age using her old method is 60%. Using
> a new method she makes 9 of her first 10 attempts. Is the
> new method better than the old? (Hint: "advanced statisti-
> cal formula must be used.")
>
<snip>
> 2 was worded ambiguously (I changed the wording here to make
> it less so), but, even so, there's no "advanced statisti-
> cal formula" needed, unless multiplication is advanced.
Wrong! Is 9 out of 10 better than 60%? Saying 9*10/10 => 90% is the
wrong answer to the question. Free throws are random (trust me--mine are
very random). Is 9 out of 10 better than 60%? You can quote odds on
that, but not answer the question with certainty. The method could be
worse, with 9 out of 10 being a fluke. The question is how often a person
who shoots worse than 60% will shoot will shoot exactly 9 out of a
specified 10.
The key word, which you appear to have overlooked, was "first", as in
"first 10 attempts".
> If these are our standards for 12th grade students, we're set-
> ting our sights too low.
>
If (2) represents performance of typical net-posters, perhaps we need to
raise our contributors standards a bit.
> --
> -- Mike Zarlenga
> Bob Dole for President in 1996.
> Choose the Veteran over the Veteran Liars.
Why?
George Phillies
Libertarian Candidate, U.S. Senate
Massachusetts
Gerald McCutcheon
>This week's issue of USN&WR ("Dumb and Dumber") had a section
>on the proposals for new standards in education.
>Under the proposals for 12th grade math we find :
>1. Which is a better fit, a round peg in a square hole, or a
> square peg in a round hole?
>2. Ann's free-throw %age using her old method is 60%. Using
> a new method she makes 9 of her first 10 attempts. Is the
> new method better than the old? (Hint: "advanced statisti-
> cal formula must be used.")
>1 was child's play. I bet most high school sophomores in pri-
> vate schools can solve it in under 2 minutes. This is a
> 10th grade level question, and even then, a simple one.
>2 was worded ambiguously (I changed the wording here to make
> it less so), but, even so, there's no "advanced statisti-
> cal formula" needed, unless multiplication is advanced.
>If these are our standards for 12th grade students, we're set-
>ting our sights too low.
>--
>-- Mike Zarlenga
> Bob Dole for President in 1996.
> Choose the Veteran over the Veteran Liars.
> finger zarl...@conan.ids.net for PGP Public key and killfile
Yes, it's an absolute shame. Education is rapidly regressing to new lows
due to the "feel good" approaches being adopted as a reckless approach to
esteem building which, in the long run, will produce a greater percentage
of "graduates" who will be abject failures in terms of logical reasoning,
reading comprehension, and problem solving skills (especially mathematical).
I am astounded that the American public - not commentators or news media
- fails to object to these measures in a more resolute fashion. This
reminds me of the Great Society Program sponsored by LBJ and The Swimmer,
which wasted trillions of dollars in a reckless approach. Though
different, it's the same shot-in-the-dark fix-it approach that is not
rooted in responsibility, discipline and adequate planning.
Mr. Clinton sounds more and more like a republican these days calling for
higher educational standards. This is yet another ephemeral lie, as
earlier in his term he sponsored a debacle in respect to testing - I
reference the SAT's in particular, in terms of making median scores
APPEAR higher, allowing calculators to be used, reducing the number of
questions and providing half an hour longer for sections. This neither
rasies standards nor engenders self-esteem, but is a poor excuse for the
lack of a solid educational policy. This will cause long term
deleterious effects, a terrible, long term price that others will pay
for the sake of the politically correct's experimentation and perceived
political gains.
...yet another reason why this aberration, this president-in-title must
be defeated in the November '96 election.
GM
************************************************************************
My views are not necessarily those of my employer!
************************************************************************
Michael Zarlenga
: > 1. Which is a better fit, a round peg in a square hole, or a
: > square peg in a round hole?
: What do you mean, "better"? Were there, say, pictures of the two?
: Assuredly, one can inscribe a square in a circle, or inscribe a circle in
: a square, but what was the question supposed to mean? There seems to be
: something missing.
There is only one solution which makes sense. And that presumes that
"better" means least amount of wasted space when one is inscribed
within the other."
: > 2. Ann's free-throw %age using her old method is 60%. Using
: > a new method she makes 9 of her first 10 attempts. Is the
: > new method better than the old? (Hint: "advanced statisti-
: > cal formula must be used.")
: <snip>
: > 2 was worded ambiguously (I changed the wording here to make
: > it less so), but, even so, there's no "advanced statisti-
: > cal formula" needed, unless multiplication is advanced.
: Wrong! Is 9 out of 10 better than 60%? Saying 9*10/10 => 90% is the
: wrong answer to the question. Free throws are random (trust me--mine are
: very random). Is 9 out of 10 better than 60%? You can quote odds on
: that, but not answer the question with certainty. The method could be
: worse, with 9 out of 10 being a fluke. The question is how often a person
: who shoots worse than 60% will shoot will shoot exactly 9 out of a
: specified 10.
No, that's not the question. The question is "what are the odds that
a single event with p=0.6 will happen 9 times in a row?"
To solve this problem, the "advanced statistical formula" you need is
p(1,2) = p(1) * p(2). Wow. That's a bit too hard for today's seniors
in high school, eh?
Maybe we'd better stick to "If Suzie has 3 apples and she eats 1, how
many are left?" for our 12th graders?
--
-- Mike Zarlenga
Dole '96. Send King William and Queen Hillary packing in November.
Andrew Rogers
>
> Under the proposals for 12th grade math we find :
> 2. Ann's free-throw %age using her old method is 60%. Using
> a new method she makes 9 of her first 10 attempts. Is the
> new method better than the old? (Hint: "advanced statisti-
> cal formula must be used.")
> 2 was worded ambiguously (I changed the wording here to make
> it less so), but, even so, there's no "advanced statisti-
> cal formula" needed, unless multiplication is advanced.
Calculating the confidence interval based on the sample size
isn't an "advanced statistical formula"?
> If these are our standards for 12th grade students, we're set-
> ting our sights too low.
Thanks for proving it.
Andrew
Andrew Rogers
> Mr. Clinton sounds more and more like a republican these days calling for
> higher educational standards. This is yet another ephemeral lie, as
> earlier in his term he sponsored a debacle in respect to testing - I
> reference the SAT's in particular, in terms of making median scores
> APPEAR higher, allowing calculators to be used, reducing the number of
> questions and providing half an hour longer for sections.
The SATs didn't exactly change overnight; in fact, the latest round of
revisions had been in the planning and testing stages for nearly 10
years before their final implementation. With that minor detail in
mind, please explain how Clinton "sponsored a debacle" that had already
been in the works for several years prior to his taking office. And,
for that matter, please explain how any U.S. President has control over
the actions of a *private* testing firm such as ETS.
> ...yet another reason why this aberration, this president-in-title must
> be defeated in the November '96 election.
Your post is yet another reason why literacy tests weren't such a
bad idea after all.
Andrew
Andrew Hall
Michael> George D. Phillies (phil...@wpi.edu) wrote:
>> > 1. Which is a better fit, a round peg in a square hole, or a
>> > square peg in a round hole?
>> What do you mean, "better"? Were there, say, pictures of the two?
>> Assuredly, one can inscribe a square in a circle, or inscribe a circle in
>> a square, but what was the question supposed to mean? There seems to be
>> something missing.
Michael> There is only one solution which makes sense. And that presumes that
Michael> "better" means least amount of wasted space when one is inscribed
Michael> within the other."
Yes, that was my guess at what the poorly worded question meant.
>> > 2. Ann's free-throw %age using her old method is 60%. Using
>> > a new method she makes 9 of her first 10 attempts. Is the
>> > new method better than the old? (Hint: "advanced statisti-
>> > cal formula must be used.")
>> <snip>
>> > 2 was worded ambiguously (I changed the wording here to make
>> > it less so), but, even so, there's no "advanced statisti-
>> > cal formula" needed, unless multiplication is advanced.
>> Wrong! Is 9 out of 10 better than 60%? Saying 9*10/10 => 90% is the
>> wrong answer to the question. Free throws are random (trust me--mine are
>> very random). Is 9 out of 10 better than 60%? You can quote odds on
>> that, but not answer the question with certainty. The method could be
>> worse, with 9 out of 10 being a fluke. The question is how often a person
>> who shoots worse than 60% will shoot will shoot exactly 9 out of a
>> specified 10.
Michael> No, that's not the question. The question is "what are the odds that
Michael> a single event with p=0.6 will happen 9 times in a row?"
This is a quite different problem than the badly posed one above.
As posed above, the question is not answerable. One can, from the
information given, and knowledge of advanced (for high school)
statistics determining the confidence interval for the new technique being
better, but of course, not reject the hypothesis that the new technique
is the same at 100% confidence. Was their no mention of the confidence
level in the original question?
Michael> To solve this problem, the "advanced statistical formula" you need is
Michael> p(1,2) = p(1) * p(2). Wow. That's a bit too hard for today's seniors
Michael> in high school, eh?
Nope. If her average is still .6 the probability of getting 9 out of 10
is 4% (.6^9 x .4^1 x 10).
Michael> Maybe we'd better stick to "If Suzie has 3 apples and she eats 1, how
Michael> many are left?" for our 12th graders?
Perhaps you should review some statistics.
ah
Michael Zarlenga
: > 2. Ann's free-throw %age using her old method is 60%. Using
: > a new method she makes 9 of her first 10 attempts. Is the
: > new method better than the old? (Hint: "advanced statisti-
: > cal formula must be used.")
: > 2 was worded ambiguously (I changed the wording here to make
: > it less so), but, even so, there's no "advanced statisti-
: > cal formula" needed, unless multiplication is advanced.
: Calculating the confidence interval based on the sample size
: isn't an "advanced statistical formula"?
Yawn.
Calculating the odds of a string of 9 successive events when the odds
of each individual event is known is child's play, or at least it SHOULD
be. I bet Asian 10th graders could solve this one with ease.
p(1) = 0.6. Therefore, p(1,2,3,4,5,6,7,8,9) = (0.6)**9 = 0.01.
The odds of Ann making 9 in a row, if her new method is just as good as
her old method are 1 in 100. Her new method is better.
And pay attention : the students wouldn't even have to FIGURE THE
ODDS ... the odds are SO skewed (1:100), and the choices so limited
(the new way is better, worse, the same) that almost ANYONE could
GUESS the right answer.
Do you need help with the circle/square problem, too?
Michael Zarlenga
: Nope. If her average is still .6 the probability of getting 9 out of 10
You presume she missed the 10th shot, but reread the problem. The only
thing we know about her shots with the new method is that she made her
first 9. The 10th is irrelevant - we have no data. Did she miss or
hit? We don't know.
The odds of the sequence described in the problem happening, if the new
method is as good as the old, are 0.010, or 1 in 100.
But look ... the student doesn't even need to figure this out for that
problem. The odds are so unlikely that almost anyone can guess right.
Her new method is better.
This is a puff problem.
Andrew Rogers
> Michael> No, that's not the question. The question is "what are the odds that
> Michael> a single event with p=0.6 will happen 9 times in a row?"
>
> This is a quite different problem than the badly posed one above.
Here is the original question, straight from the USN&WR web site:
http://www.usnews.com/usnews/issue/grades.htm
| 2. Ann tells you that under her old method of shooting free throws
| in basketball, her average was 60 percent. Using a new method of
| shooting, she hit on 9 out of her first 10 throws. Should she
| conclude that the new method really is better than the old method?
| (Hint: Advanced statistical formula must be used.)
If anything, the original wording is a lot less ambiguous than the
mangled paraphrasing posted here.
> As posed above, the question is not answerable. One can, from the
> information given, and knowledge of advanced (for high school)
> statistics determining the confidence interval for the new technique being
> better, but of course, not reject the hypothesis that the new technique
> is the same at 100% confidence. Was their no mention of the confidence
> level in the original question?
"Should she" isn't what I'd call a specification of confidence interval.
> Michael> To solve this problem, the "advanced statistical formula" you need is
> Michael> p(1,2) = p(1) * p(2). Wow. That's a bit too hard for today's seniors
> Michael> in high school, eh?
No, but understanding the question as actually posed is obviously too
hard for you.
> Michael> Maybe we'd better stick to "If Suzie has 3 apples and she eats 1, how
> Michael> many are left?" for our 12th graders?
>
> Perhaps you should review some statistics.
Not to mention some reading comprehension.
Andrew
Andrew Rogers
[snip]
> Michael> Maybe we'd better stick to "If Suzie has 3 apples and she eats 1, how
> Michael> many are left?" for our 12th graders?
> Perhaps you should review some statistics.
Let's see our resident math genius solve this question, intended for
*8th* graders:
| 1. If, in a school of 1,000 lockers, one student opens
| every locker, a second student closes every other locker
| (second, fourth, sixth, etc.), a third student changes every
| third locker (opens closed lockers and closes open lockers)
| and so on until the 1,000th student changes the 1,000th
| locker, which lockers are open?
Jeez, I'm glad this one didn't appear on my GREs! E-mail me
if you're stumped...
Andrew
George D. Phillies
On 29 Mar 1996, Andrew Hall wrote:
> >>>>> Michael Zarlenga writes:
>
> Michael> George D. Phillies (phil...@wpi.edu) wrote:
> >> > 1. Which is a better fit, a round peg in a square hole, or a
> >> > square peg in a round hole?
>
> >> What do you mean, "better"? Were there, say, pictures of the two?
> >> Assuredly, one can inscribe a square in a circle, or inscribe a circle in
> >> a square, but what was the question supposed to mean? There seems to be
> >> something missing.
>
> Michael> There is only one solution which makes sense. And that presumes that
> Michael> "better" means least amount of wasted space when one is inscribed
> Michael> within the other."
>
> Yes, that was my guess at what the poorly worded question meant.
Well, we can certainly tell where we have a group of students who have
never worked with tools, assembled plastic models, or done anything else
that would incline them toward science or engineering. "Fit" is a
measurement of how well two parts intermesh, for example, a large square
peg not fitting into a small round hole. By the reasonable definition of
"better fit", apparent to anyone who has had to assemble a child's toy out
of parts, a circle with inscribed square, and a square with inscribed
circle, are line drawings of equally good engineering fits.
> > >> > 2. Ann's free-throw %age using her old method is 60%. Using > >>
> a new method she makes 9 of her first 10 attempts. Is the > >> > new
method better than the old? (Hint: "advanced statisti- > >> > cal formula
must be used.") > >> <snip> > >> > 2 was worded ambiguously (I changed the
wording here to make > >> > it less so), but, even so, there's no
"advanced statisti- > >> > cal formula" needed, unless multiplication is
advanced. > > >> Wrong! Is 9 out of 10 better than 60%? Saying 9*10/10
=> 90% is the > >> wrong answer to the question. Free throws are random
(trust me--mine are > >> very random). Is 9 out of 10 better than 60%?
You can quote odds on > >> that, but not answer the question with
certainty. The method could be > >> worse, with 9 out of 10 being a
fluke. The question is how often a person > >> who shoots worse than 60%
will shoot will shoot exactly 9 out of a > >> specified 10. > > Michael>
No, that's not the question. The question is "what are the odds that >
Michael> a single event with p=0.6 will happen 9 times in a row?"
Well, no. Making "9 of her first 10 attempts" and "making her first 9
attempts and missing the tenth" are seriously different.
Also, the question that has been asked is whether the new method is
better than the old method, not whether the new method delivered better
than 60%. Even if the new method were worse than the old method (say, P
= 0.5) there would be some likelihood that the new method would deliver 9
hits out of 10 shots. That is, you are told that Ann made *some* 9
of the first 10 attempts, not being told which. The new method has some P
of making a shot, which you do not know. You would like to know the
probability that the new P is larger than 0.6.
>
> This is a quite different problem than the badly posed one above.
> As posed above, the question is not answerable.
One needs to know the distribution of P. Assuming flatness is probably
enough, but perhaps this is not a good answer. Your answer is a probability
that the new method is better than the old method.
One can, from the
> information given, and knowledge of advanced (for high school)
> statistics determining the confidence interval for the new technique being
> better, but of course, not reject the hypothesis that the new technique
> is the same at 100% confidence. Was their no mention of the confidence
> level in the original question?
Excellent point. However, this is apparently a question that every
Japanese (male) schoolchild is taught to answer in high school, including
*in particular* children going on to "unskilled" assembly-line work,
because this ("based on finite statistics, what is the likelihood that the
new method is superior to the old method") is the precise question that a
shop-worker-level quality circle using Deming,... methods must be able to
answer to decide whether a new production method is superior to an old
production method.
A Thurow-style economist, which I am not, would argue that we are
touching on a reason why the pay ratio between shop employees and senior
managers is much smaller in Japan than it is here. Details are left to
the individual reader.
George Phillies
Don Porges
Michael Zarlenga <zarl...@conan.ids.net> wrote:
>Andrew Rogers (rog...@hi.com) wrote:
>: > 2. Ann's free-throw %age using her old method is 60%. Using
>: > a new method she makes 9 of her first 10 attempts. Is the
>: > new method better than the old? (Hint: "advanced statisti-
>: > cal formula must be used.")
>
>: > it less so), but, even so, there's no "advanced statisti-
>: > cal formula" needed, unless multiplication is advanced.
>
>: isn't an "advanced statistical formula"?
>
>Yawn.
>
>Calculating the odds of a string of 9 successive events when the odds
>of each individual event is known is child's play, or at least it SHOULD
>be. I bet Asian 10th graders could solve this one with ease.
>
>p(1) = 0.6. Therefore, p(1,2,3,4,5,6,7,8,9) = (0.6)**9 = 0.01.
>
>The odds of Ann making 9 in a row, if her new method is just as good as
>
And if the problem had said "makes the first 9 tries in row", you'd even
be right about the math. But she made some collection of 9 out of the
first 10, which isn't quite the same.
Michael Zarlenga
: Let's see our resident math genius solve this question, intended for
: *8th* graders:
: | 1. If, in a school of 1,000 lockers, one student opens
: | every locker, a second student closes every other locker
: | (second, fourth, sixth, etc.), a third student changes every
: | third locker (opens closed lockers and closes open lockers)
: | and so on until the 1,000th student changes the 1,000th
: | locker, which lockers are open?
Another puff problem.
Every locker which has an even number of unique factors will be closed.
Every locker which has an odd number of unique factors will be open.
[ 1, 4, 9, 16, 25, 36 ... ]
: Jeez, I'm glad this one didn't appear on my GREs! E-mail me
: if you're stumped...
I'm stumped. How did you graduate high school?
Andrew Rogers
> Every locker which has an odd number of unique factors will be open.
> [ 1, 4, 9, 16, 25, 36 ... ]
>
> : Jeez, I'm glad this one didn't appear on my GREs! E-mail me
> : if you're stumped...
>
> I'm stumped. How did you graduate high school?
Three and a half years early, with an 800 on the Math SAT
(690 Verbal) nine days after my thirteenth birthday. And you?
Andrew
glindahl
: Michael Zarlenga wrote:
Fascinating discussion, guys. Simply riveting. Now, can we move on?
Michael Zarlenga
: If we are truely going to contribute at an international level we must
: address this problem. A national standardized test would be a great
: step in this direction. As long as individual states are left to
: address this problem the standards will not be the same. We have
: standardized plumbing. Yet, we can't agree on how much knowledge in
: math and science a high school graduate should have to qualify as a
: graduate.
We already have national standardized tests. PSAT, SAT, GRE, GMAT,
etc etc etc.
As far as national standards vs local standards, that's how we got
into this mess in the first place - allowing a national union, the
NEA to set the standards and cater to the lowest common denominator.
Let states set their own standards. Those with higher standards will
attract the better students, faculty and jobs. And some states with
large isolated industries, like the midwest and Hawaii, can set their
own, more applicable standards. I wouldn't expect a student in NYC to
know as much about irrigation or soil erosion as one in Kansas.
Andrew Rogers
> We already have national standardized tests. PSAT, SAT, GRE, GMAT,
> etc etc etc.
They are national and they are standardized, but with the exception
of tests like the MCATs and SAT/GRE subject tests they are not
intended to measure factual knowledge.
In fact, the SATs were originally designed to AVOID this as much as
possible; the idea was that by measuring "scholastic aptitude"
(whatever that is) instead of booklearning per se, these tests would
measure how smart - as opposed to how well-educated - each candidate
was. In that sense, they were intended to be more like IQ tests (is
anyone else here old enough to remember the "they're APTITUDE tests
so you CAN'T study for them" line?) than final exams in a specific
subject, and at least theoretically two equally bright students - say,
one from a prestigious Eastern prep school and the other from a rural
or inner-city public school - would achieve the same scores on the SATs.
(You can read more about the history of the SATs in a two-part article
in _Atlantic Monthly_; I don't have the cover dates, but it was within
the past few months.)
N.B.: the doomsayers who point to "declining SATs" as a symptom of all
that's wrong with the American educational system obviously blew the
reasoning questions on their own SATs - you can't draw any valid
conclusions about the educational system as a whole from a test that
a) is self-selected, and b) was originally designed specifically to
factor out quality of education. Necessary but not sufficient, guys.
> As far as national standards vs local standards, that's how we got
> into this mess in the first place - allowing a national union, the
> NEA to set the standards and cater to the lowest common denominator.
>
> Let states set their own standards.
If they don't, then why is there such a variation in high school
graduation requirements from state to state (and even school to
school)?
Andrew
Andrew Hall
Don> In article <4jhqf0$p...@paperboy.ids.net>,
Don> Michael Zarlenga <zarl...@conan.ids.net> wrote:
>> Andrew Rogers (rog...@hi.com) wrote:
>>> > 2. Ann's free-throw %age using her old method is 60%. Using
>>> > a new method she makes 9 of her first 10 attempts. Is the
>>> > new method better than the old? (Hint: "advanced statisti-
>>> > cal formula must be used.")
>>
>>> > 2 was worded ambiguously (I changed the wording here to make
>>> > it less so), but, even so, there's no "advanced statisti-
>>> > cal formula" needed, unless multiplication is advanced.
>>
>>> Calculating the confidence interval based on the sample size
>>> isn't an "advanced statistical formula"?
>>
>> Yawn.
>>
>> Calculating the odds of a string of 9 successive events when the odds
>> of each individual event is known is child's play, or at least it SHOULD
>> be. I bet Asian 10th graders could solve this one with ease.
>>
>> p(1) = 0.6. Therefore, p(1,2,3,4,5,6,7,8,9) = (0.6)**9 = 0.01.
>>
>> The odds of Ann making 9 in a row, if her new method is just as good as
>> her old method are 1 in 100. Her new method is better.
>>
Don> And if the problem had said "makes the first 9 tries in row", you'd even
Don> be right about the math. But she made some collection of 9 out of the
Don> first 10, which isn't quite the same.
And the question was not even what are the odds, but is
the new method better than the old. Mr. Zarlenga has
a lot of egg on his face with this one.
ah
Andrew Hall
Michael> Andrew Hall (ah...@cs.uml.edu) wrote:
>> Nope. If her average is still .6 the probability of getting 9 out of 10
>> is 4% (.6^9 x .4^1 x 10).
I note you deleted the original problem. Here it is again for
context:
: > 2. Ann's free-throw %age using her old method is 60%. Using
: > a new method she makes 9 of her first 10 attempts. Is the
: > new method better than the old? (Hint: "advanced statisti-
: > cal formula must be used.")
Michael> You presume she missed the 10th shot, but reread the problem. The only
You reread it. It says noting about which of the ten she missed,
only that she made 9 of ten. My formula is correct.
Michael> thing we know about her shots with the new method is that she made her
Michael> first 9. The 10th is irrelevant - we have no data. Did she miss or
Michael> hit? We don't know.
She could have missed any of the first ten. Reread the problem you typed
in.
Michael> The odds of the sequence described in the problem happening, if the new
Michael> method is as good as the old, are 0.010, or 1 in 100.
Michael> But look ... the student doesn't even need to figure this out for that
Michael> problem. The odds are so unlikely that almost anyone can guess right.
Michael> Her new method is better.
Michael> This is a puff problem.
Reread the problem. UIt is not anserabe as stated. Calculating
a confidence interval is not trivial for high school students.
ah
LocalWords: UIt anserabe
Andrew Hall
Michael> Andrew Hall (ah...@cs.uml.edu) wrote:
>> Nope. If her average is still .6 the probability of getting 9 out of 10
>> is 4% (.6^9 x .4^1 x 10).
I note you deleted the original problem. Here it is again for
context:
: > 2. Ann's free-throw %age using her old method is 60%. Using
: > a new method she makes 9 of her first 10 attempts. Is the
: > new method better than the old? (Hint: "advanced statisti-
: > cal formula must be used.")
Michael> You presume she missed the 10th shot, but reread the problem. The only
You reread it. It says noting about which of the ten she missed,
only that she made 9 of ten. My formula is correct.
Michael> thing we know about her shots with the new method is that she made her
Michael> first 9. The 10th is irrelevant - we have no data. Did she miss or
Michael> hit? We don't know.
She could have missed any of the first ten. Reread the problem you typed
in.
Michael> The odds of the sequence described in the problem happening, if the new
Michael> method is as good as the old, are 0.010, or 1 in 100.
Michael> But look ... the student doesn't even need to figure this out for that
Michael> problem. The odds are so unlikely that almost anyone can guess right.
Michael> Her new method is better.
At what confidence level? If the new method is exactly as good as
the old, one out of 25 times she will get 9 out of 10.
Michael> This is a puff problem.
Then why are you having so much trouble with it?
ah
Andrew Hall
Andrew> Andrew Hall wrote:
Michael> No, that's not the question. The question is "what are the odds that
Michael> a single event with p=0.6 will happen 9 times in a row?"
>>
>> This is a quite different problem than the badly posed one above.
Andrew> Here is the original question, straight from the USN&WR web site:
http> www.usnews.com/usnews/issue/grades.htm
>> 2. Ann tells you that under her old method of shooting free throws
>> in basketball, her average was 60 percent. Using a new method of
>> shooting, she hit on 9 out of her first 10 throws. Should she
>> conclude that the new method really is better than the old method?
>> (Hint: Advanced statistical formula must be used.)
Andrew> If anything, the original wording is a lot less ambiguous than the
Andrew> mangled paraphrasing posted here.
Yes, they are probably implying that a 95% confidence level
should be used.
>> As posed above, the question is not answerable. One can, from the
>> information given, and knowledge of advanced (for high school)
>> statistics determining the confidence interval for the new technique being
>> better, but of course, not reject the hypothesis that the new technique
>> is the same at 100% confidence. Was their no mention of the confidence
>> level in the original question?
Andrew> "Should she" isn't what I'd call a specification of confidence interval.
With the hint, I have no idea what else she should use.
Michael> To solve this problem, the "advanced statistical formula" you need is
Michael> p(1,2) = p(1) * p(2). Wow. That's a bit too hard for today's seniors
Michael> in high school, eh?
Andrew> No, but understanding the question as actually posed is obviously too
Andrew> hard for you.
Michael> Maybe we'd better stick to "If Suzie has 3 apples and she eats 1, how
Michael> many are left?" for our 12th graders?
>> Perhaps you should review some statistics.
Andrew> Not to mention some reading comprehension.
ah
Michael Zarlenga
: And the question was not even what are the odds, but is
: the new method better than the old.
Didn't I say that? It's another reason why this problem is too simnple.
Most people can guess at the right answer.
: Mr. Zarlenga has a lot of egg on his face with this one.
In my haste I misread the problem. Big deal. The point is that it's a simple
problem for most 12th graders educated in a good private school. These news
standards aren't hard, they're average to easy for any student with a real
education.
--
-- Mike Zarlenga
Senator Kennedy has killed more people with his car than I have
with my gun.
Andrew Rogers
> : And the question was not even what are the odds, but is
> : the new method better than the old.
>
> Didn't I say that?
Sort of, but only after you were corrected by at least four
people - and judging from your responses you still don't have a
clue as to what any of us were talking about when we brought up
that mysterious "confidence interval".
> It's another reason why this problem is too simnple.
The bottom line is that this is a stock question from any intro
Probability and Statistics textbook - so why don't you impress us
all by borrowing one and answering it correctly for all to see.
> Most people can guess at the right answer.
I'm sure they can, but in this case they are expected to show
how they arrived at the answer.
> : Mr. Zarlenga has a lot of egg on his face with this one.
He can always pull one of Rush's favorite stunts when backed into
a corner on a factual matter: bray all the louder that he's right
and the "liberals" are just picking on him.
> In my haste I misread the problem. Big deal. The point is that it's a simple
> problem for most 12th graders educated in a good private school.
It's a simple problem for anyone who's taken (or at least remembers
a quarter-century later :-) ) an intro Probability and Statistics
course, which is still not widely offered at the high-school level
(public *or* private) in the US. (In my day, it was virtually
unheard of - only one of my P&S classmates had taken it in high
school, and he had attended one of the NYC exam schools; even the
preppies hadn't seen the material before.)
It's probably a simple problem for most 10th graders educated in
Japan - but then again when Japan tried to introduce their version
of our "New Math" in the 60s, the parents took evening courses to
keep up with this newfangled stuff their kids were learning instead
of pressuring the school boards to "get back to the basics". Parents
who decry "declining standards" might want to avoid mirrors...
> These news
> standards aren't hard, they're average to easy for any student with a real
> education.
Which still doesn't include you.
Incidentally, if I had been the editor of USN&WR, I would have
added the following comment to the article on proposed standards:
"Letters to the editor concerning this topic will not be
printed unless accompanied by correct solutions to problems
blah, blah, and blah."
Andrew
Michael Zarlenga
: > These new
: > standards aren't hard, they're average to easy for any student with a real
: > education.
: Which still doesn't include you.
Yawn.
Do you agree or not? Are these proposed standards set too low, too
high, or just about right for what we should expect from our 12th
graders?
J.D. Baldwin
[Followups redirected appropriately.]
In article <4jhijg$j...@paperboy.ids.net>, Michael Zarlenga
<zarl...@conan.ids.net> wrote:
>: > 2. Ann's free-throw %age using her old method is 60%. Using
>: > a new method she makes 9 of her first 10 attempts. Is the
>: > new method better than the old? (Hint: "advanced statisti-
>: > cal formula must be used.")
>: [...]
>: Wrong! Is 9 out of 10 better than 60%? Saying 9*10/10 => 90% is the
>: wrong answer to the question. Free throws are random (trust me--mine are
>: very random). Is 9 out of 10 better than 60%? You can quote odds on
>: that, but not answer the question with certainty. The method could be
>: worse, with 9 out of 10 being a fluke. The question is how often a person
>: who shoots worse than 60% will shoot will shoot exactly 9 out of a
>: specified 10.
>
>No, that's not the question. The question is "what are the odds that
>a single event with p=0.6 will happen 9 times in a row?"
That's not the question either. To start with, who said anything
about "in a row"? The question is more like, "is it a reasonable
conclusion that P(x) > 0.6?" (Where "x" is "successful free throw
attempt.") The intuitive answer is Yes, the new P(x) can be inferred
to be 0.9, which is > 0.6. It's inconceivable that the question had
any other intent for high school seniors, as a literal interpretation
along the lines you suggest makes for quite an advanced problem,
definitely beyond college freshman math.
I used to teach combinatorics and probability theory at a university
with one of the top student bodies in the U.S., and I would never have
asked them to solve the problem the way you suggest without a
calculator and a set of the appropriate tables.
>To solve this problem, the "advanced statistical formula" you need is
>p(1,2) = p(1) * p(2). Wow. That's a bit too hard for today's seniors
>in high school, eh?
Totally different problem, the answer to which is indeed easy (0.0101),
as answers to the wrong problems so often are. But your formula won't
get you there, either. (And by the way, the probability of the
actual event described is more like 0.004, given a P(x) of 0.6.)
>Maybe we'd better stick to "If Suzie has 3 apples and she eats 1, how
>many are left?" for our 12th graders?
As long as you're willing to give credit for "three," I guess so.
(Since we're being hyper-literal, we must concede that she still "has"
the "consumed" apple, albeit in a different form.)
--
From the catapult of J.D. Baldwin |+| "If anyone disagrees with anything I
_,_ Finger bal...@netcom.com |+| say, I am quite prepared not only to
_|70|___:::)=}- for PGP public |+| retract it, but also to deny under
\ / key information. |+| oath that I ever said it." --T. Lehrer
***~~~~-----------------------------------------------------------------------
Andrew Hall
Michael> Andrew Hall (ah...@cs.uml.edu) wrote:
>> And the question was not even what are the odds, but is
>> the new method better than the old.
Michael> Didn't I say that? It's another reason why this problem is too simnple.
Michael> Most people can guess at the right answer.
Perhaps, but as it barely squeaks into significance at
the 95% level, they could not be too sure. Remember there
are only two degrees of freedom in the Chi Squared test and
there are only ten samples. If she had gotten eight of the
first 10 it would not be better at the 95% level.
>> Mr. Zarlenga has a lot of egg on his face with this one.
Michael> In my haste I misread the problem. Big deal. The point is that it's a simple
Michael> problem for most 12th graders educated in a good private school. These news
Michael> standards aren't hard, they're average to easy for any student with a real
Michael> education.
But you also said you did not need to know the Chi Squared test, which
is a bigger mistake than not reading the question that you typed in
properly.
ah
Michael Zarlenga
: But you also said you did not need to know the Chi Squared test, which
: is a bigger mistake than not reading the question that you typed in
: properly.
The original problem did not require a numeric solution, only a gross
evaluation of the new shooting method.
Back to the my question, still unanswered:
Do you believe these proposed standards are too tough, too easy or just
about right for what we should expect from out 12th graders?
Andrew Hall
Michael> Andrew Hall (ah...@cs.uml.edu) wrote:
>> But you also said you did not need to know the Chi Squared test,
>> which is a bigger mistake than not reading the question that you
>> typed in properly.
Michael> The original problem did not require a numeric solution,
Michael> only a gross evaluation of the new shooting method.
Then the problem was totally unfair, as the problem
cannot be solved without numbers. Remember, it
was barely significant at the 95% level, so a gross
evaluation was not going to be useful at all.
Michael> Back to the my question, still unanswered:
Michael> Do you believe these proposed standards are too tough, too
Michael> easy or just about right for what we should expect from
Michael> out 12th graders?
It seems like a very poorly written test and should be rewritten.
The stats question, if a correct answer is required, seems above what
one would expect from the average 12th grader. Perhaps it would
be appropriate for an AP math placement test.
ah
Gino Cerro
college. There doesn't have to be a standard set for students, since not
everyone is capable of high level math and science. And for some students
they are stronger in the humanties and arts, and the langauge arts. This
is not a perfect world, not everyone coming out of the 12th grade will be
proficent in Calculus, Trigonmerty, or Alegbra. Every school has a
section devoted to high level math and science classes, not everyone has
the discipline or intelligence for it.
-------------------------------
Gino Cerro
http://www.wpi.edu/~gino
DBCx
>This is not a perfect world, not everyone coming out of the 12th grade
>will be proficent in Calculus, Trigonmerty, or Alegbra.