A Madness for all Seasons

Our fourth post on the  2017 VCE exam madness will be similar to our previous post: a quick whack of a straight-out error. This error was flagged by a teacher friend, David. (No, not that David.)

The 11th multiple choice question on the first Further Mathematics Exam reads as follows:

Which one of the following statistics can never be negative? 

A. the maximum value in a data set

B. the value of a Pearson correlation coefficient

C. the value of a moving mean in a smoothed time series

D. the value of a seasonal index

E. the value of a slope of a least squares line fitted to a scatterplot

Before we get started, a quick word on the question’s repeated use of the redundant “the value of”.

Bleah!

Now, on with answering the question.

It is pretty obvious that the statistics in A, B, C and E can all be negative, so presumably the intended answer is D. However, D is also wrong: a seasonal index can also be negative. Unfortunately the explanation of “seasonal index” in the standard textbook is lost in a jungle of non-explanation, so to illustrate we’ll work through a very simple example.

Suppose a company’s profits and losses over the four quarters of a year are as follows:

    \[ \begin{tabular} {| c | c | c | c |}\hline {\bf\phantom{S}Summer \phantom{I}} &{\bf\phantom{S}Autumn \phantom{I}} &{\bf\phantom{S}Winter \phantom{I}} &{\bf\phantom{S}Spring \phantom{I}} \\  \hline {\bf \$6000} & {\bf -\$1000} & {\bf -\$2000} & {\bf \$5000}\\ \hline \end{tabular}\]

So, the total profit over the year is $8,000, and then the average quarterly profit is $2000. The seasonal index (SI) for each quarter is then that quarter’s profit (or loss) divided by the average quarterly profit:

    \[ \begin{tabular} {| c | c | c | c |}\hline {\bf Summer SI} &{\bf Autumn SI} &{\bf Winter SI} &{\bf Spring SI} \\  \hline {\bf 3} & {\bf -0.5} & {\bf -1.0} & {\bf 2.5}\\ \hline \end{tabular}\]

Clearly this example is general, in the sense that in any scenario where the seasonal data are both positive and negative, some of the seasonal indices will be negative. So, the exam question is not merely technically wrong, with a contrived example raising issues: the question is wrong wrong.

Now, to be fair, this time the VCAA has a defense. It appears to be more common to apply seasonal indices in contexts where all the data are one sign, or to use absolute values to then consider magnitudes of deviations. It also appears that most or all examples Further students would have studied included only positive data.

So, yes, the VCAA (and the Australian Curriculum) don’t bother to clarify the definition or permitted contexts for seasonal indices. And yes, the definition in the standard textbook implicitly permits negative seasonal indices. And yes, by this definition the exam question is plain wrong. But, hopefully most students weren’t paying sufficient attention to realise that the VCAA weren’t paying sufficient attention, and so all is ok.

Well, the defense is something like that. The VCAA can work on the wording.

 

Further Madness

Our third post on the 2017 VCE exam madness will be brief, on a question containing a flagrant error.

The first question in the matrix module of Further Mathematics’ Exam 2 is concerned with a school canteen selling pies, rolls and sandwiches over three separate weeks. The number of items sold is set up as a 3 x 3 matrix, one row for each week and one column for each food choice. The last part, (c)(ii), of the question then reads:

The matrix equation below shows that the total value of all rolls and sandwiches sold in these three weeks is $915.60 

    \[   \boldsymbol{L \times\begin{bmatrix} 491.55 \\ 428.00\\ 487.60 \end{bmatrix} \ = \ [915.60]}\]

Matrix L in this equation is of order 1 x 3.

Write down matrix L.

This 1-mark question is presumably meant to be a gimme, with answer L = [0 1 1]. Unfortunately the question is both weird and wrong. (And lacking in punctuation. Guys, it’s not that hard.) The wrongness comes from the examiners having confused their rows and columns. As is made clear in the the previous part, (c)(i), of the question, the  3 x 1 matrix of numbers indicates the total earnings from each of the three weeks, not from each of the three food choices. So, the equation indicates the total value of all products sold in weeks 2 and 3.

There’s not much to say about such an obvious error. It is very easy to confuse rows and columns, and we’ve all done it on occasion, but if VCAA’s vetting cannot catch this kind of mistake then it cannot be relied upon to catch anything. The only question is how the Examiners’ Report will eventually address the error. The VCAA is well-practised in cowardly silence and weasel-wording, but it would be exceptionally Trumplike to attempt such tactics here.

Error aside, the question is artificial, and it is not clear that the matrix equation “shows” much of anything. Yes, 0-1 and on-or-off matrices are important and useful, but the use of such a matrix in this context is contrived and confusing. Not a hanging offence, and benign by VCAA’s standards, but the question is pretty silly. And, not forgetting, wrong.

Some Special Madness

Our second post on the 2017 VCE exam madness concerns a question on the first Specialist Mathematics exam. Typically Specialist exams, particularly the first exams, don’t go too far off the rails; it’s usually more “meh” than madness. (Not that “meh” is an overwhelming endorsement of what is nominally a special mathematics subject.) This year, however, the Specialist exams have some notably Methodsy bits. The following nonsense was pointed out to us by John, a friend and colleague.

The final question, Question 10, on the first Specialist exam concerns the function \boldsymbol{f(x) = \sqrt{\arccos(x/2)}}, on its maximal domain [-2,2]. In part (c), students are asked to determine the volume of the solid of revolution formed when the region under the graph of f is rotated around the x-axis. This leads to the integral

    \[V \ = \ \pi \int\limits_{-2}^{2}  \arccos(x/2)\, {\rm d}x\,.\]

Students don’t have their stupifying CAS machines in this first exam, so how to do the integral? It is natural to consider integration by parts, but unfortunately this standard and powerful technique is no longer part of the VCE curriculum. (Why not? You’ll have to ask the clowns at ACARA and the VCAA.)

No matter. The VCAA examiners love to have the students to go through a faux-parts computation. So, in part (a) of the question, students are asked to check the derivative of \boldsymbol{x\arccos(x/a)}. Setting a = 2 in the resulting equation, this gives

    \[ \frac{{\rm d}\phantom{x}}{{\rm d}{x}}\left(x\arccos(x/2)\right)= \arccos(x/2) - \dfrac{x}{\sqrt{4-x^2}}\,.\]

We can now integrate and rearrange, giving

    \[ \aligned V \ &= \ \pi\left\Big[\!x\arccos(x/2)\!\!\right\Big]\limits_{-2}^{2} \quad +\quad \pi \int\limits_{-2}^{2} \dfrac{x}{\sqrt{4-x^2}}\, {\rm d}x\\[2\jot] \ &= \ 2\pi^2\quad +\quad \pi \int\limits_{-2}^{2} \dfrac{x}{\sqrt{4-x^2}}\, {\rm d}x\,.\endaligned\]

So, all that remains is to do that last integral, and … uh oh.

It is easy to integrate \boldsymbol{x/\sqrt{4-x^2}} indefinitely by substitution, but the problem is that our definite(ish) integral is improper at both endpoints. And, unfortunately, improper integrals are not part of the VCE curriculum. (Why not? You’ll have to ask the clowns at ACARA and the VCAA.) Moreover, even if improper integrals were available, the double improperness is fiddly: we are not permitted to simply integrate from some –b to b and then let b tend to 2.

So, what is a Specialist student to do? One can hope to argue that the integral is zero by odd symmetry, but the improperness is again an issue. As an example indicating the difficulty, the integral \boldsymbol{\int\limits_{-2}^2 x/(4-x^2)\,{\rm d}x} is not equal to 0. (The TI Inspire falsely computes the integral to be 0, which is less than inspiring.) Any argument which arrives at the answer 0 for integrating \boldsymbol{x/(4-x^2)} is invalid, and is thus prima facie invalid for integrating \boldsymbol{x/\sqrt{4-x^2}} as well.

Now, in fact \boldsymbol{\int\limits_{-2}^2 x/\sqrt{4-x^2}\,{\rm d}x} is equal to zero, and so \boldsymbol{V = 2\pi^2}. In particular, it is possible to argue that the fatal problem with \boldsymbol{x/(4-x^2)} does not occur for our integral, and so both the substitution and symmetry approaches can be made to work. The argument, however, is subtle, well beyond what is expected in a Specialist course.

Note also that this improperness could have been avoided, with no harm to the question, simply by taking the original domain to be, for example, [-1,1]. Which was exactly the approach taken on Question 5 of the 2017 Northern Hemisphere Specialist Exam 1. God knows why it wasn’t done here, but it wasn’t and the consequently the examiners have trouble ahead.

The blunt fact is, Specialist students cannot validly compute \boldsymbol{\int\limits_{-2}^2 x/\sqrt{4-x^2}\,{\rm d}x} with any technique they would have seen in a standard Specialist class. They must either argue incompletely by symmetry or ride roughshod over the improperness. The Examiners’ Report will be a while coming out, though presumably the examiners will accept either argument. But here is a safe prediction: the Report will either contain mealy-mouthed nonsense or blatant mathematical falsehoods. The only alternative is for the examiners to make a clear admission that they stuffed up. Which won’t happen.

Finally, the irony. Look again at the original integral for V. Though this integral arose in the calculation of a volume, it can still be interpreted as the area under the graph of the function y = arccos(x/2):

But now we can consider the corresponding area under the inverse function y = 2cos(x):

It follows that

    \[V \ = \  \pi \int\limits_{-2}^{2}  \arccos(x/2)\, {\rm d}x \ = \ \pi \int\limits_{0}^{\pi} \left[  2\cos(x) - (-2)\right]  \, {\rm d}x \ = \ 2\pi^2\,.\]

Done.

This inverse function trick is standard for Specialist (and Methods) students, and so the students can readily calculate the volume V in this manner. True, reinterpreting the integral for V as an area is a sharp conceptual shift, but with appropriate wording it could have made for a very good Specialist question.

In summary, the Specialist Examiners guided the students to calculate V with a jerry-built technique, leading to an integral that the students cannot validly compute, all the while avoiding a simpler approach well within the students’ grasp. Well played, Examiners, well played.

 

There’s Madness in the Methods

Yes, we’ve used that title before, but it’s a damn good title. And there is so much madness in Mathematical Methods to cover. And not only Methods. Victoria’s VCE exams are coming to an end, the maths exams are done, and there is all manner of new and astonishing nonsense to consider. This year, the Victorian Curriculum and Assessment Authority have outdone themselves.

Over the next week we’ll put up a series of posts on significant errors in the 2017 Methods, Specialist Maths and Further Maths exams, including in the mid-year Northern Hemisphere examsBy “significant error” we mean more than just a pointless exercise in button-pushing, or tone-deaf wording, or idiotic pseudomodelling, or aimless pedantry, all of which is endemic in VCE maths exams. A “significant error” in an exam question refers to a fundamental mathematical flaw with the phrasing, or with the intended answer, or with the (presumed or stated) method that students were supposed to use. Not all the errors that we shall discuss are large, but they are all definite errors, they are errors that would have (or at least should have) misled some students, and none of these errors should have occurred. (It is courtesy of diligent (and very annoyed) maths teachers that I learned of most of these questions.) Once we’ve documented the errors, we’ll post on the reasons that the errors are so prevalent, on the pedagogical and administrative climate that permits and encourages them.

Our first post concerns Exam 1 of Mathematical Methods. In the final question, Question 9, students consider the function \boldsymbol{ f(x) =\sqrt{x}(1-x)} on the closed interval [0,1], pictured below. In part (b), students are required to show that, on the open interval (0,1), “the gradient of the tangent to the graph of f” is (1-3x)/(2\sqrt{x}). A clumsy combination of calculation and interpretation, but ok. The problem comes when students then have to consider tangents to the graph.

In part (c), students take the angle θ in the picture to be 45 degrees. The pictured tangents then have slopes 1 and -1, and the students are required to find the equations of these two tangents. And therein lies the problem: it turns out that the “derivative”  of f is equal to -1 at the endpoint x = 1. However, though the natural domain of the function \sqrt{x}(1-x)} is [0,∞), the students are explicitly told that the domain of f is [0,1].

This is obvious and unmitigated madness.

Before we hammer the madness, however, let’s clarify the underlying mathematics.

Does the derivative/tangent of a suitably nice function exist at an endpoint? It depends upon who you ask. If the “derivative” is to exist then the standard “first principles” definition must be modified to be a one-sided limit. So, for our function f above, we would define

    \[f'(1) = \lim_{h\to0^-}\frac{f(1+h) - f(1)}{h}\,.\]

This is clearly not too difficult to do, and with this definition we find that f'(1) = -1, as implied by the Exam question. (Note that since f naturally extends to the right of =1, the actual limit computation can be circumvented.) However, and this is the fundamental point, not everyone does this.

At the university level it is common, though far from universal, to permit differentiability at the endpoints. (The corresponding definition of continuity on a closed interval is essentially universal, at least after first year.) At the school level, however, the waters are much muddier. The VCE curriculum and the most popular and most respected Methods textbook appear to be completely silent on the issue. (This textbook also totally garbles the related issue of derivatives of piecewise defined (“hybrid”) functions.) We suspect that the vast majority of Methods teachers are similarly silent, and that the minority of teachers who do raise the issue would not in general permit differentiability at an endpoint.

In summary, it is perfectly acceptable to permit derivatives/tangents to graphs at their endpoints, and it is perfectly acceptable to proscribe them. It is also perfectly acceptable, at least at the school level, to avoid the issue entirely, as is done in the VCE curriculum, by most teachers and, in particular, in part (b) of the Exam question above.

What is blatantly unacceptable is for the VCAA examiners to spring a completely gratuitous endpoint derivative on students when the issue has never been raised. And what is pure and unadulterated madness is to spring an endpoint derivative after carefully and explicitly avoiding it on the immediately previous part of the question.

The Victorian Curriculum and Assessment Authority has a long tradition of scoring own goals. The question above, however, is spectacular. Here, the VCAA is like a goalkeeper grasping the ball firmly in both hands, taking careful aim, and flinging the ball into his own net.

Malcolm the Mathematician

Australia’s Prime Minister tends to be pretty pleased with himself, and plenty of other people seem to think of Malcolm Turnbull as the smartest guy in the room. Perhaps he sometimes he is.* Malcolm didn’t appear so smart, however, when presenting Australia’s proposal to require the tech giants to decrypt their customers’ encrypted messages. When ZDnet reporter Asha McLean suggested that “the laws of mathematics [might] trump the laws of Australia”, Malcolm was unfazed:

The laws of Australia prevail in Australia, I can assure you of that. The laws of mathematics are very commendable but the only law that applies in Australia is the law of Australia.

And yes, the Government’s plan (for want of a better word) is as clueless as Malcolm makes it sound.

We already knew that Malcolm was a scientific clown, an economic illiterate, a coward, a Luddite, an Orwellian thug and a moral midget. So, maybe it shouldn’t be a great surprise when Malcolm also turns out to be an anti-mathematical git.

* If the other people in the room are Peter Dutton and Barnaby Joyce.

NAPLAN’s Mathematical Nonsense, and What it Means for Rural Peru

The following question appeared on Australia’s Year 9 NAPLAN Numeracy Test in 2009:

y = 2x – 1

y = 3x + 2

Which value of x satisfies both of these equations?

It is a multiple choice question, but unfortunately “The question is completely stuffed” is not one of the available answers.

Of course the fundamental issue with simultaneous equations is the simultaneity. Both equations and both variables must be considered as a whole, it simply making no sense to talk about solutions for x without reference to y. Unless y = -7 in the above equations, and there is no reason to assume that, then no value of x satisfies both equations. The NAPLAN question is way beyond bad.

It is always worthwhile pointing out NAPLAN nonsense, as we’ve done before and will continue to do in the future. But what does this have to do with rural Peru?

In a recent post we pointed out an appalling question from a nationwide mathematics exam in New Zealand. We flippantly remarked that one might expect such nonsense in rural Peru but not in a wealthy Western country such as New Zealand. We were then gently slapped in the comments for the Peruvian references: Josh queried whether we knew anything of Peru’s educational system; and, Dennis questioned the purpose of bringing up Peru, since Australia’s NAPLAN demonstrates a “level of stupidity” for all the World to see. These are valid points.

It would have been prudent to have found out a little about Peru before posting, but we seem to be safe. Peru’s economy has been growing rapidly but is not nearly as strong as New Zealand’s or Australia’s. Peruvian school education is weak, and Peru seems to have no universities comparable to the very good universities in New Zealand and Australia. Life and learning in rural Peru appears to be pretty tough.

None of this is surprising, and none of it particularly matters. Our blog post referred to “rural Peru or wherever”. The point was that we can expect poorer education systems to throw up nonsense now and then, or even typically; in particular, lacking ready access to good and unharried mathematicians, it is unsurprising if exams and such are mathematically poor and error-prone.

But what could possibly be New Zealand’s excuse for that idiotic question? Even if the maths ed crowd didn’t know what they were doing, there is simply no way that a competent mathematician would have permitted that question to remain as is, and there are plenty of excellent mathematicians in New Zealand. How did a national exam in New Zealand fail to be properly vetted? Where were the mathematicians?

Which brings us to Australia and to NAPLAN. How could the ridiculous problem at the top of this post, or the question discussed here, make it into a nationwide test? Once again: where were the mathematicians?

One more point. When giving NAPLAN a thoroughly deserved whack, Dennis was not referring to blatantly ill-formed problems of the type above, but rather to a systemic and much more worrying issue. Dennis noted that NAPLAN doesn’t offer a mathematics test or an arithmetic test, but rather a numeracy test. Numeracy is pedagogical garbage and in the true spirit of numeracy, NAPLAN’s tests include no meaningful evaluation of arithmetic or algebraic skills. And, since we’re doing the Peru thing, it seems worth noting that numeracy is undoubtedly a first world disease. It is difficult to imagine a poorer country, one which must weigh every educational dollar and every educational hour, spending much time on numeracy bullshit.

Finally, a general note about this blog. It would be simple to write amusing little posts about this or that bit of nonsense in, um, rural Peru or wherever. That, however, is not the purpose of this blog. We have no intention of making easy fun of people or institutions honestly struggling in difficult circumstances; that includes the vast majority of Australian teachers, who have to tolerate and attempt to make sense of all manner of nonsense flung at them from on high. Our purpose is to point out the specific idiocies of arrogant, well-funded educational authorities that have no excuse for screwing up in the manner in which they so often do.

The Median is the Message

Our first post concerns an error in the 2016 Mathematical Methods Exam 2 (year 12 in Victoria, Australia). It is not close to the silliest mathematics we’ve come across, and not even the silliest error to occur in a Methods exam. Indeed, most Methods exams are riddled with nonsense. For several reasons, however, whacking this particular error is a good way to begin: the error occurs in a recent and important exam; the error is pretty dumb; it took a special effort to make the error; and the subsequent handling of the error demonstrates the fundamental (lack of) character of the Victorian Curriculum and Assessment Authority.

The problem, first pointed out to us by teacher and friend John Kermond, is in Section B of the exam and concerns Question 3(h)(ii). This question relates to a probability distribution with “probability density function”

    \[  f(x) =   \left\{\aligned &\frac{(210-x)e^{\frac{x-210}{20}}}{400} \qquad && 0\leqslant x \leqslant 210,\\ &0 && \text{elsewhere.} \endaligned\right.}\]

Now, anyone with a good nose for calculus is going to be thinking “uh-oh”. It is a fundamental property of a PDF that the total integral (underlying area) should equal 1. But how are all those integrated powers of e going to cancel out? Well, they don’t. What has been defined is only approximately a PDF,  with a total area of 1 - 23/2e^{21/2} \approx 0.9997. (It is easy to calculate the area exactly using integration by parts.)

Below we’ll discuss the absurdity of handing students a non-PDF, but back to the exam question. 3(h)(ii) asks the students to find the median of the “probability distribution”, correct to two decimal places. Since the question makes no sense for a non-PDF, of course the VCAA have shot themself in the foot. However, we can still attempt to make some sense of the question, which is when we discover that the VCAA has also shot themself in the other foot.

The median m of a probability distribution is the half-way point. So, in the integration context here we want the m for which

a)      \phantom{\quad}  \int\limits_0^m f(x)\,{\rm d}x = \dfrac12.

As such, this question was intended to be just another CAS exercise, and so both trivial and pointless: push the button, write down the answer and on to the next question. The problem is, the median can also be determined by the equation

b)     \phantom{\quad}  \int\limits_m^{210} f(x)\,{\rm d}x = \dfrac12,

or by the equation

c)     \phantom{\quad} \int\limits_0^m f(x)\,{\rm d}x = \int\limits_m^{210} f(x)\,{\rm d}x.

And, since our function is only approximately a PDF, these three equations necessarily give three different answers: to the demanded two decimal places the answers are respectively 176.45, 176.43 and 176.44. Doh!

What to make of this? There are two obvious questions.

1. How did the VCAA end up with a PDF which isn’t a PDF?

It would be astonishing if all of the exam’s writers and checkers failed to notice the integral was not 1. It is even more astonishing if all the writers-checkers recognised and were comfortable with a non-PDF. Especially since the VCAA can be notoriously, absurdly fussy about the form and precision of answers (see below).

2. How was the error in 3(h)(ii) not detected?

It should have been routine for this mistake to have been detected and corrected with any decent vetting. Yes, we all make mistakes. Mistakes in very important exams, however, should not be so common, and the VCAA seems to make a habit of it.

OK, so the VCAA stuffed up. It happens. What happened next? That’s where the VCAA’s arrogance and cowardice shine bright for all to see. The one and only sentence in the Examiners’ Report that remotely addresses the error is:

“As [the] function f  is a close approximation of the [???] probability density function, answers to the nearest integer were accepted”. 

The wording is clumsy, and no concession has been made that the best (and uniquely correct) answer is “The question is stuffed up”, but it seems that solutions to all of a), b) and c) above were accepted. The problem, however, isn’t with the grading of the question.

It is perhaps too much to expect an insufferably arrogant VCAA to apologise, to express anything approximating regret for yet another error. But how could the VCAA fail to understand the necessity of a clear and explicit acknowledgement of the error? Apart from demonstrating total gutlessness, it is fundamentally unprofessional. How are students and teachers, especially new teachers, supposed to read the exam question and report? How are students and teachers supposed to approach such questions in the future? Are they still expected to employ the precise definitions that they have learned? Or, are they supposed to now presume that near enough is good enough?

For a pompous finale, the Examiners’ Report follows up by snarking that, in writing the integral for the PDF, “The dx was often missing from students’ working”. One would have thought that the examiners might have dispensed with their finely honed prissiness for that one paragraph. But no. For some clowns it’s never the wrong time to whine about a missing dx.

UPDATE (16 June): In the comments below, Terry Mills has made the excellent point that the prior question on the exam is similarly problematic. 3(h)(i) asks students to calculate the mean of the probability distribution, which would normally be calculated as \int xf(x)\,{\rm d}x. For our non-PDF, however, we should should normalise by dividing by \int f(x)\,{\rm d}x. To the demanded two decimal places, that changes the answer from the Examiners’ Report’s 170.01 to 170.06.