Discussion of the 2022 NHT Specialist Exams

This is our post for discussion of the 2022 NHT Specialist Mathematics exams, which have now been posted, here and here. (See also our corresponding 2022 NHT Methods exams post.) We’ve had a quick browse, and found plenty of irritants but no glaringly large errors. We don’t plan to look further or to post on the exams, except to follow up on whatever people find worthy of comment.

UPDATE (23/09/22)

The exam reports are now available, here and here.

61 Replies to “Discussion of the 2022 NHT Specialist Exams”

  1. Some quick thoughts on Exam 1 and with sentiments similar to Marty’s:

    Question 4 (a): Minor niggle – Would have preferred “*static* equilibrium” rather than simply “equilibrium”.

    Question 6: Minor niggle. “.. Find the magnitude of the resultant force, in newtons, after it has acted on the mass for four seconds, …” This sentence suggests that the resultant force is an old friend that has been introduced earlier. It wasn’t.

    Would have preferred that this resultant force was mentioned/introduced earlier:
    “A 2kg mass is moving under the action of a resultant force. Relative to a fixed origin, the position of the mass after t seconds is given by …”

    Question 7: The particle and its speed is approaching infinity after a finite time. I don’t like it. It’s sloppy and makes no physical sense. The time should have been restricted to something like \displaystyle \left[0, \frac{\pi}{3} \right] (But at least it’s not another stunt car with infinite initial acceleration).

    Question 8: Another wordy ‘real-life’ statistics question. I see from part (c) that students are now required to know z = 1.96 rather than z = 2. This is an under-handed change of policy.

    Given the relative irrelevance of the NHT exams, I think the writers get a pass mark for this one. There are no glaringly large errors (but I could argue a case for Question 7 – it could and should have been better with one simple change). But …. the continual shifting of the goal posts for the statistics content of Specialist Maths – given that we have this gratuitous content forced upon us – continues to greatly irritate me.

      1. I don’t have a problem with that.

        The exact answer is 34 and I’d hope that a student would give that answer without the specific instruction. However, what is NOT hilarious is the sort of unsimplified answer a student is likely to give if the instruction is not given …

        Asking for the answer as an integer is perhaps more a sad indictment on the standard of student than the exam.

          1. \displaystyle 15^2 = 225 is something that students should know. So a number a bit bigger than 15 whose square ends in 9 … Hmmm … What could it be …?

            (17 should be obvious, in my opinion. I don’t see the problem with this. Once upon a time students knew perfect squares up to 20^2 = 400, just like the times tables).

            But I agree that it might have been better (albeit introduce a greater chance of dumb arithmetic mistake) if the coefficient of \displaystyle t^2 had been \displaystyle \sqrt{6} rather than 6.

    1. Re: Question 8 and the required value of z. I see on the 2022 NHT Maths Methods Exam 1 Question 3 that the value z = 2 is explicitly given, with an expectation that this value is used. Clearly VCAA cannot make up its mind and establish a consistent expectation across all of its exams.

      1. In fact, I don’t understand why VCAA refuses to include critical values of z on the formula sheet. What is the point in requiring students to memorise 1.96?

    2. Thanks, John. Re Q6, yes, this is the kind of Not Quite Right wording that really gets up my nose. It’s not killingly wrong, but it’s plain bad writing.

  2. Regarding the z=1.96 or z=2 issue…

    1. Yes, it is an issue. Neither are exact for a 95% confidence interval. z=2 is correct to the nearest integer and z=1.96 is correct to 2 decimal places (or 3 significant figures for any IB teachers who may be reading this for ideas) but there is no reason why one should be chosen over the other. It is an arbitrary choice. As is the 95%… why not 94.7% just to be different…?

    2. The formula given for the two different exams is also a bit of an issue. It is the same formula, written differently. Why give two different versions of the same formula in different exams?

    Not an error in either case, but an ongoing frustration.

  3. Thoughts on SM1:
    – overall a balanced paper testing mainly outcome 1 and ocassionally outcome 2
    – no apparent conceptual error, although some questions could’ve been phrased with more conciseness and clarity as suggested by JF and RF
    – some questions from old exams were reinvented (i.e. Q7) but requirements should have been tightened compared to the similar one in 2013 SM1. Will be curious to see the assessment report later if students were expected to go on expressing y as a function of x and justifying y must be greater than or equal to zero, rather than just leaving it as a standard hyperbola form x^2/16 – y^2/4 =1
    – the styles and skills being tested look very similar to 2021 November exams (heard somewhere that these papers were developed together) with three vector calculus questions, though one may argue that Q7 is more of parametric differentiation nature rather than vector.
    On a side note I am not sure if the writers have assumed that students are expected to memorize the derivative of sec(x) is sec(x)tan(x), or d(csc(x))/dx = – csc(x)cot(x)…
    From 2013 SM reports I felt that…

    – The choice of 2 + 3 mark allocation for Q10 was again interesting. This makes the life easier for the students compared to Q10 in 2018 (straight 5 marks)
    – Reflection from Q5 in SM1 paper – a similar mcq was there in 2019 NHT SM2, suggesting potentials for tech-active questions to be suitably modified into tech-free questions, vice-versa. This can really test if the students understand and grasp the concepts and skills well rather than button-pressing. Also a warning bell to teachers and students – don’t underestimate the value of solid algebraic skills! If you know your concepts and algebra well, you will do well in Spesh…

    Thoughts on SM2:
    – MCQs are slightly easier compared to last Nov exam. Those who knew their buttons well should have done most questions faster
    – An interesting u-substitution question mcq1
    0. I am guessing if there would be a similar one in SM1 this November. Might be worthy to redo some exercises from Essential textbook – other substitutions.
    – mcq 6: interesting to see some crossovers between polynomial graphs and complex number. Usually SACE exams have questions set in similar manner.

    – Much easier than ERQ1 last year. Decent question.
    – modified based on 2015 SM1 question on complex number. Blended with similar ideas from previous exam questions involving perpendicular bisector
    – Not bad. Benign
    – 1998, 2005, 2011, 2016 all had similar inflow-outflow concentration problems. Every six or seven years, it appears. Are we expecting some logistic equations or population problem or predator-prey models in this November? Hopefully not chemical reaction rate problem…
    Very challenging. Very verbose, inevitably.
    Too many parameters to work with.
    Coefficient of friction is an essential part throughout the question. Some parts favored integration approaches over constant acceleration formulae.
    Interestingly, much easier than last year’s statistics question. The last part assessed binomial distribution, so students who did methods last year need to pick up their probability knowledge again this year before Nov exam…CRV, sample proportion, etc.
    If only students have access to some similar vector calculus questions around 1994-1997…Finally some vector integration questions appeared again. Last two parts were also very challenging. If this question were a November exam question I guess the percentage for students obtaining full marks would tend to be single digit, even decimals or close to zero. Hardly could some similar question to this be found in the mainstream textbooks…Again, this question, especially the last two parts, has proven that physics students are not necessarily advantaged. Specialist maths is far more than that.

    1. Thanks very much, Guardian. I’m sure that will be very helpful to teachers and students. I disagree on 5 versus 2 + 3, re Q10, but that’s a debate for another time.

    1. Hi Reef. I’ll look when home. Do you suspect an issue with the question, or just you’re not sure what is going on?

  4. Hello, All

    I don’t understand why the answer Maximum rate = 30.8 when t = 0.4.

    obviously, d2x/dt2 =0 then t=2.8 and rate =-4.2 gram / L is answer

    1. Hi Jack, I’ve been flat out but I’ll look tomorrow. Since no one has raised this earlier, I suspect you’ve made an error, but I’ll check.

    2. Hi Jack.

      The question asks for “the maximum rate at which the chemical flows \displaystyle from \displaystyle the \displaystyle tank” (my emphasis). That is, the maximum rate of outflow. The outflow rate is NOT dx/dt (dx/dt is the rate of change \displaystyle inside the tank). You’ve used the wrong rate in your calculation.

      1. Just to clarify:

        The rate of outflow is the rate of change of \displaystyle volume, it’s not the same as the rate of change of \displaystyle chemical leaving the tank. It’s a tricky question, containing several different rates that can easily lead to confusion!

    1. Nuts indeed. It’s funny what new crap you discover from an intelligent student who works carefully through a VCAA exam. Some crap (including in Exam Reports) lies dormant for several years. And then VCAA gets it knickers in a knot (or used to, the jury is still out now that there’s a new Maths Manager and Curriculum Manager) when the crap is pointed out.

      But I have issues beyond the picture.

      Questions like 8 are dumb. Not defective, just plain dumb. I’d love to know what the intent of the question is. Most students will simply use their CAS to draw the slope field (and in this case stop at option B). I don’t see the point. Just ask them to draw a slope field in Exam 1 if you want to see if they can. Or give them a slope field and ask about the behaviour of the solutions to the corresponding DE for different initial conditions. Or write a Study design that explicitly contains some worthwhile limit content.

      Students who try to find the correct option by eliminating wrong options with some mathematical analysis (which potentially makes such a question more worthwhile) get screwed over here. Sure, you can reasonably note that dy/dx = 0 when x = 0 and so eliminate A and C. This leaves B, D and E.

      But after this, you need to be very comfortable with limits. Because the only way I see to continue such an analysis is to argue dy/dx –> 0 as x –> -oo (eliminate D) and then dy/dx –> -oo as x –> 0 and y < 0 (eliminate E).

      I don't think many students would approach the question this way. They might eliminate A and C as above, but then would go to the CAS and get lucky with B.

      I consider these questions dumb and pathetic.

      1. I think that the elimination of D is quite easy. I would in fact eliminate E first, as the picture shows that dy/dx depends on both x and y which makes an easy elimination. It does not seem to me that those attempting an analysis strategy should be screwed over by this question any more than any other questions of a similar style.

  5. I realise very few people are probably reading this now that the exam has passed, but I came back to Paper 2 Section B Q5e today and think there is an issue with the final answer given by VCAA.

    I agree that the sample mean required rounds to 0.5026 if we are simply rounding an answer. BUT… in context, I believe that if 0.5026 is used, the p-value remains above 0.05 which is a problem and therefore, the minimum value of the sample mean (assuming the rounded SD value from part a is used) should be 0.5027.

    I accept that very few people are likely to care about this, even less given that it is a NHT paper. The thought bothers me though.

    1. Thanks, RF. It’s stats crap, I don’t do stats crap. But if someone confirms the error, I’ll add it to the list.

      1. I should have spotted it earlier but I was a bit distracted at the time!

        Thanks BiB for confirming my suspicions and I agree with your observation (1). On (2), yes, I am inclined to agree, however there is not a big enough sample size (yet) for me to reject H0 in this case!

        1. Sorry, RF. I made an edit in my original comment that you’re referring to and it turned to crap. I’ll repost it later today.

        2. Yep, RF. This is all your fault.

          Thank you both. Once BiB has repaired his comment, I’ll look and then add to the error list.

        3. RF, I meant to say thankyou as well. I initially thought you were talking about the 2023 NHT exam. I was mistaken but, as a result, discovered that the ‘Reports’ for the 2023 NHT exams are available. I have not checked them for accuracy.

    2. Yes, the answer in the Exam Report is wrong. Well spotted, RF. I care too!

      Let the critical value be a. Pr(Xbar > a) < 0.05 is required.

      To more than four decimal places, a = 0.502601.

      Rounding down to 4 dp: Pr(Xbar > 0.5026) = 0.05004 > 0.05.

      Rounding up to 4 dp: Pr(Xbar > 0.5027) = 0.0438525.

      Therefore the answer, correct to four decimal places, is a = 0.5027. Rounding UP is required, counter-intuitive as that seems.

      We make two observations:

      1) To avoid counter-intuitive answers, one should choose values so that the correct answer corresponds to the usual mathematical rounding. Then this sort of 'issue' does not occur.

      2) Stats is not done well in VCAA exams.

      1. Thanks very much, BiB. I’ll figure out what the hell you guys are talking about and then add to the error list.

      2. In hindsight, I don’t think a more unfortunate set of values could have been chosen for this question. I can almost understand both answers being accepted. (Almost).
        I think it could make for a very interesting classroom discussion. In reality, how reasonable would it be to round down rather than up? (Terry, this is your cue …)

      3. RF refers to using the rounded SD from (a). Does this matter? If one uses a more precise SD does that change the answer (to four decimal places) of (e)?

        1. VCAA’s answer is wrong whether you use the exact sd or the approx sd (0.0016) from part (a). (I should have been clear that I used the exact sd) in my earlier calculations). The only difference is that the probabilities calculated in each case make rounding up less counter-intuitive. Using the approx sd of 0.0016:

          a = 0.5026 give the probability as 0.0521, whereas

          a = 0.5027 gives the probability as 0.0458.

          Your question is very pertinent because VCAA has consistently said that unless there’s a “Hence” somewhere in the later question (implying approx values found earlier should be used), more accuracy than what the answer requires must be used during the calculation to avoid rounding error. It doesn’t matter here, but I can think of situations where it would.

      4. OK, I read through it and understand the issue. Except I’m not sure why it is counterintuitive to round up. You’re effectively finding the value of a so that P(Xbar > a) = 0.05. Isn’t the need to round up obvious to ensure we still have P < 0.05?

        1. You’re talking like a mathematician, Marty! Of course it’s obvious if you understand that you’re simply looking to satisfy an inequality at a given level of accuracy. But from experience I know it’s counter-intuitive to students (and I’d guess to many teachers as well). Have you read Judgement Under Uncertainty:

          Click to access TverskyKahneman1974.pdf

          Highly recommended once the ‘obvious’ is under discussion.

          1. Ok, thanks. But you seem to suggest this a trickier or more misleading situation than normal. You were suggesting maybe both rounded answers should be accepted: you rejected the idea, but it didn’t seem to be so clear-cut in your mind. So, I’m just trying to think why this is, at least arguably, a more subtle situation that usual. You indicate VCAA could have chosen the values better. How? Just to not have the precise answer so slightly above 0.5026?

            1. Although I think it’s important that students understand how to round to get the correct critical value, for a 1 mark (yes, 1 mark) question I would choose values such that the correct answer is obtained using the usual ‘mathematical rounding’.
              You have to ask yourself a question: What’s being tested and how many marks is it worth? I think checking rounding up and rounding down and then justifying the final answer demonstrates understanding that is worth more than 1 mark.

              Personally, I think it would be a very courageous student who gave, correctly, a = 0.5027 as the answer. I advise my students that in these situations, even if it’s only 1 mark, they should always justify their rounding. Because it’s lamentable that VCAA cannot be trusted to use correct answers in their marking scheme.

              Re: “I can almost understand both answers being accepted. (Almost).”
              I can understand something but still disagree with it. I only raised it because of the unfortunate precedent set by the Exam Report for 2108 Exam 2 Question 6 part (e). The real tragedy is that students read these Exam Reports, cannot make sense of what they read and so – quite reasonably – ask their teacher what they should do.

              1. OK, I understand, and fair enough. I’ve added the question to the error list. Thanks again, to both you and RF, for raising and clarifying the problem.

                1. I’m going to check through some Methods exams again now, because I have had a lot of questions from Methods students (I don’t teach Methods this year, but that doesn’t stop anyone…) about this rounding question.

                  To be fair, a lot of those were about confidence intervals which is less of a big deal if you round the wrong way.

                2. If you wanted to picky (and I’m in the mood), you could also say that the notation for the answer to part (a) is wrong. It uses the symbol s for the standard deviation of the sample means. This symbol conventionally denotes the standard deviation of a sample. The appropriate symbol to use is \displaystyle \sigma_{\bar{X}}. In my view it’s an error albeit a minor one.

                  Correct use of notation is also something the VCAA is pedantic about and pontificates upon.

                    1. Yes.

                      There are several different types of standard deviations (and indeed different types of means) that students meet when learning this stuff:

                      s (sample standard deviation, which can sometimes be used to approximate the population standard deviation).

                      \displaystyle \sigma (population standard deviation).

                      \displaystyle \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} (standard deviation of the sampling mean).

                      It is essential that students understand the conceptual difference between each of these. The Report makes an error (*) in using s and this error has the potential to confuse students.

                      I’m happy for you to wait and see what others have to say.

                      (*) A *ahem* standard error …

      5. My bad but I blame RF for leading me on. The question asks:

        “Find the largest sample mean of 10 ball bearings that could be observed for the null hypothesis \displaystyle not to be rejected at the 5% level of significance” (my emphasis).
        So the required inequality is > 0.05 not < 0.05.
        I misread the question (twice!!) The VCAA answer is correct.

    1. Hi Don.

      A quick check, I think the given answer is correct. (And I imagine if there were an error, someone would have alerted me to it earlier.) I don’t know how you approached the question, but here is how I looked at it.

      You have \boldsymbol{z_0 = \sqrt3 + i}, and you want \boldsymbol{z_3} so that the line through the origin and \boldsymbol{z_0} is given by \boldsymbol{|z-i| = |z-z_3|}. That means the given line is the perpendicular bisector of the line connecting \boldsymbol{i} and \boldsymbol{z_3}. (Draw a picture.)

      Now \boldsymbol{z_0} makes an angle of 30 degrees with the origin, and \boldsymbol{i} makes an angle of 90 degrees. That means \boldsymbol{z_3} should be 60 degrees in the other direction from \boldsymbol{z_0}. (Draw on the picture.)

      So, \boldsymbol{z_3} should make an angle of MINUS 30 degrees from the positive real axis. Also, \boldsymbol{z_3} should have modulus 1. That gives \boldsymbol{z_3} as suggested in the report.

      1. I agree with the answer to this one, but I did it with geometry of angles and then simple trigonometry to determine the x and y.

    1. Again, quickly.


      STUFF = x

      VOLUME = volume of water = 15 + 15t

      The 15 comes from starting with 15L in the tank.

      The 15t comes from the overall flow of water into the tank (30L – 15L each minute).

      1. There is some weird wording in the latter part of Q3 on this paper that was raised earlier in this thread which is worth reading if you are a student trying to understand the VCAA answers to this specific question. Commenter JF gave a very clear explanation, which does not excuse VCAA’s original choice of wording.

  6. Hi All,
    I would really appreciate if someone could help me with Q6) part d and e.
    I thought that the aircraft being at an altitude of 460 m would mean equating
    500-2t = 460 –> t = 20 then r(20) = 1200i + 460j. So the package would fall directly down since there is no air resistance and then 1800 – 1200 = 600. Could someone explain why a constant acceleration formula was used? Thanks

    1. Hi Tubes.

      There are several ways of solving it but drawing a picture of what’s happening in each part is always important. I’ve attached a brief solution that uses only the straight line motion formulae for constant acceleration. The solution is only meant as a guide and is not exemplary.

      NB 1: When the package falls, it still has horizontal motion.
      NB 2: Position is relative to O.
      NB 3: The acceleration is constant in each direction: 9.8 m/s^2 down for the vertical motion and 0 m/s^2 to the right for the horizontal motion (since air resistance is negligible). This is obvious from the components of the double derivative of the position vector wrt time.

      Best wishes for Exam 2.

      2022 Specialist NHT Exam 2 Q6 parts d-e

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