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A little bit of fun »

Multiple choice test taking: a professional skill

February 23, 2010 by dberenstein

As I contemplate a set of bubble sheets and a midterm coming up for my class,   I have to think that all those multiple choice tests that I took in my life ended producing a valuable skill. They taught me how to take the tests so that when I design them I can solve them and fix the errors that crop up. I can now make them with great confidence. After all that training, I am a professional.

I also know exactly what are the ‘traps’ that are hidden in these questions: the numbers are not random. Many of them correspond to simple mistakes from unit conversions (factors of 10), dividing instead of multiplying, standard sign errors.

Then there is the pattern theory: that the correct test answers have to have a pattern and not be truly random. That theory is sometimes true. In order to ‘balance’ the set of answers, some professors require that the number of consecutive identical answers is set to zero or at most one sometimes, so that consecutive answers are anti-correlated. If you do this by random numbers,  you can get patterns of four or five consecutive identical answers.

Someone suggested to me that to ensure that the test has a true solution no matter what, the following answer should be appended to the set of choces for each and every question:

None of the above.

I also remember that when I was in high school my chemistry teacher told us of a test with the following instruction at the beginning: if you answer any of the questions in this test, you will fail this test. The whole class failed (because people don’t usually read the instructions).

Finally, this reminds me of a puzzle where there is no question, but the following answers are given (well it was something like this anyhow):

  1. None of the above
  2. All of the above
  3. None of the above
  4. All of the above
  5. None of the above

This fits the pattern that I just described. Which of these is the correct answer?

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Posted in Academia, humor, puzzle | 20 Comments

20 Responses

  1. on February 24, 2010 at 2:53 am excitedstate

    5


  2. on February 24, 2010 at 10:06 am Andy Wood

    I believe the answer is (2). Here’s why:

    If (1) is true, then (2) is true, but (3), (4) and (5) are false.

    If (1) is false, then (2) is false and (3) is true, but (4) and (5) are false.

    (1) is equivalent to the proposition “For all x, x is in the empty set implies false” (since “the above” == “the empty set” in (1)).

    But “For all x, x is in the empty set” == false

    So (1) is equivalent to “For all x, false => false” and we all now that “false => anything” is true.

    So (1) is true, and therefore (2) is true, the rest being false.


  3. on February 24, 2010 at 10:40 am fh

    No, I would say that (1) says “All statements in the empty set are false”. This is, like the statement “All statements in the empty set are blue” vacuously true. Therefore both (1) and (2) are correct answers.

    The following is not justified:

    “For all x, x is in the empty set” == false

    Though it’s a common misconception: “For all x in S” does not imply that there are any x in S.


  4. on February 24, 2010 at 11:00 am Luboš Motl

    I agree with fh but I am afraid that David will not.

    (1) says that “there are no true answers above (1)”. Because there are no answers above (1), and therefore no true answers above (1), the statement (1) is tautologically true.

    Because (1) is true, as we just proved, (2) says that all statements above (2), i.e. the statement (1), is true. But we just proved it’s the case. So (2) is true, too.

    However, statements (3) and (5) say that “none of the answers above them are true”, which is wrong because (1) is true and it is above them. So (3) and (5) are false.

    Consequently, (4) is false, too, because it says that all answers above it, namely (1), (2), (3), should be true. But they’re not: (3) is false as proven previously.

    So only (1) and (2) are true, but both of them are.


  5. on February 24, 2010 at 11:03 am Luboš Motl

    Holy cow, I think that all of us are actually saying exactly the same thing. ;-) Andy just started his answer with a sentence that indicated that (1) was not a correct answer, but then he modified the answer. ;-)


  6. on February 24, 2010 at 11:16 am Andy Wood

    “For all x in S” does not imply that there are any x in S.

    But isn’t the statement “For all x, P(x)” usually shorthand for “For all x in some Universal Set, P(x)”? And I thought that our Universal Set included at least statements (1) – (5) and so is obviously not empty.


  7. on February 24, 2010 at 11:37 am nick

    The answer is C.


  8. on February 24, 2010 at 1:34 pm Luboš Motl

    OK, this new comment by Andy made it clear that even if I agreed about some of his results, I probably don’t agree with his methodology. What kind of a “universal set” is being discussed here?

    fh was manifestly talking about the propositions of the form “for all statements in the set S, something is satisfied” where “S” were all statements above a given statement “u” – and this set is empty for “u=1″.

    In his newest comment, Andy completely omitted the words “in S” that fh wrote previously, and silently replaced “in S” by “in some universal set” (even though S was a clear, well-defined, and actually empty set), so it directly follows that Andy’s newest comment about “universal sets”, whatever they are, cannot possibly have any relevance for fh’s arguments where the words “in S” were essential. ;-)


  9. on February 24, 2010 at 2:35 pm Andy Wood

    Andy completely omitted the words “in S” that fh wrote previously,…

    There is a very good reason for that. In my original post I said:

    “For all x, x is in the empty set” == false

    which fh says cannot be justified on the grounds that a proposition beginning “For all x in S,…” does not imply that S is non-empty. That is, fh inserted “in S” into my original statement.

    I was simply stating that “For all x” (with “in S” missing) implicitly assumes that “S” is some universal set (ie a superset of all the sets we are currently interested in), which, in the case in hand is indeed non-empty.

    So my phrasing of statement (1) should be expanded to “For all statements x, if x is one of the statements above (ie in the empty set) then false”, although, I now think it’s better to replace the “false” with “x is false”. That should make it equivalent to fh’s version.


  10. on February 24, 2010 at 3:30 pm Luboš Motl

    Dear Andy, fh didn’t insert “in S” into your original statement because he wanted to misinterpret what you’re saying – but because he was answering David’s puzzle, and David’s puzzle definitely does say “in S”.

    The word “above” in “nothing above” plays the role of “in S”, and the set “S” is defined as the set of propositions that are above a particular copy of the answer “nothing above”. ;-)

    So it is you who omitted “in S”, either when you were answering David’s puzzle at the beginning, or reacting to fh’s and my answers later. Also, I have understood why you talk about “universal sets”. Well, “universal set” in your comments seems to be an object such that “x belongs to the universal set” is a tautology (which, by the way, can’t be consistently defined due to Russel’s paradox) for every “x”.

    So it doesn’t hurt at all if you omit “in the universal set” everywhere. And the real trick of yours is not that you include “in the universal set” but that you omit “in S” which you shouldn’t! ;-) To say something relevant to David’s puzzle, you need “in S”, and “in S” is extremely well defined – instead of the philosophical diatribes in your comments, sorry. And for the case of the answer (1), “S” actually *is* empty.

    fh’s very correct point was that the proposition “for all x in S, P(x)” doesn’t include the claim that “S” is non-empty. In fact, the proposition “for all x in S, P(x)” is not only compatible for an empty “S”: it is tautologically true for an empty “S” and any property P(x)!


  11. on February 24, 2010 at 11:01 pm fh

    Ah I’m starting to see you logic now Andy, and I think it’s consistent, with what Lubos and I have been saying. Sorry to misunderatnd you. To rephrase your argument in maybe more standard language:

    (1) says forall x in S, x is false

    You want to repharse this as:

    (1) Forall x, if (x in S) then (x is false)

    This statement is true as the antecedent is always wrong.

    To see this consider that to prove a statement “if A then B” we can assume not-B and show not-A. In our case that means we assume that x is true and want to show that it follows that x is not in S. As S is the empty set this is trivially the case and statement (1) is indeed proven true.


  12. on February 25, 2010 at 12:18 am Haelfix

    I wouldn’t circle any answer, and simply take the whole paragraph as a proposition and scribble ‘true’ on the side.

    Two can play at this game


  13. on February 25, 2010 at 9:30 am Luboš Motl

    Dear fh, sorry if I am losing you. But while I agree with everything you wrote, I don’t understand why you write it.

    Don’t you agree that the following two propositions of yours,

    (1) says forall x in S, x is false
    (1′) Forall x, if (x in S) then (x is false)

    are exactly equivalent? Why is it so important to formulate them (it) in both ways?

    Healfix: funny. One may view it as instructions to add checks to the lines. After 1), one doesn’t add any. After reading 2), one adds a check after 1). After reading 3), he erases it. After reading 4), he adds checks after 1,2,3). After reading five, he deletes all of them again. Problem solved. :-)


  14. on February 25, 2010 at 9:40 am fh

    Oh yes, they are completely equivalent, I just thought the second was closer to the notions Andy had been trying to express in his line:

    “For all x, x is in the empty set” == false

    I think the mistake is merely that this should be phrased as:

    For all x: “x is in the empty set” == false

    and that one should distinguish between (“x in empty set” == false) and (x == false) more clearly.

    I now think this is what he was trying to say, in which case the underlying reasoning wasn’t wrong after all, so my original criticism was possibly indeed ill founded.

    Just trying to communicate.


  15. on February 25, 2010 at 6:39 pm Luboš Motl

    Dear fh,

    thanks for your clarification. But I am still equally puzzled.

    You want to “distinguish between (“x in empty set” == false) and (x == false) more clearly”.

    Well, I distinguish them quite clearly because the first one has “in empty set” in it while the second doesn’t. ;-) Is this difference I use to distinguish them just an illusion? Or are we forbidden to use this difference to distinguish them?

    Or do you want to design an algorithm that allows one to distinguish any two propositions even after four words are randomly dropped from one of them? That can’t work, can it? :-) For example, “two multiplied by two equals one plus one plus one plus one” couldn’t be distinguished from “two multiplied by to equals one plus one”. ;-)

    What would seem sensible would be if you wanted to distinguish propositions with the same words which are “differently bracketed”. But the difference between two of yours seems to be more than just different brackets – you want to erase “in empty set” in one of them, don’t you? This is the actual difference, not the brackets.

    If I am annoying by persistent questions, you don’t have to answer again. ;-)

    Lubos


  16. on February 26, 2010 at 10:05 am fh

    I meant Andy should have distinguished more clearly. That is look at this minor editing of his original post:

    ————–
    (1) is equivalent to the proposition “For all x, (x is in the empty set) implies (x is false)” (since “the above” == “the empty set” in (1)).

    But for all x, (x is in the empty set) == false

    So (1) is equivalent to “For all x, false => (x is false)” and we all now that “false => anything” is true.
    ————–

    And it suddenly makes sense.

    Either way, since he doesn’t seem to be reading anymore I think there is little point in continuing this thread now :D


  17. on February 27, 2010 at 8:48 pm Adam Solomon

    The answer is 3, right?

    The free parameter is 1. If 1 is correct, then 2 is also correct. Then 3 is wrong, as are 4 and 5. If 1 is wrong, then so is 2. Therefore 3 (“none of the above” – that is, “1 and 2 are both false”) is right. 4 and 5 then are not.


  18. on February 27, 2010 at 8:53 pm Adam Solomon

    Unless (as Lubos et al. did above) you take 1) to be tautologically true. In which case I’m not the first to point out that there are two true answers. So I discount that possibility as an idiosyncracy of the problem :)


    • on March 3, 2010 at 11:45 pm dberenstein

      But why discount it?

      Nobody said the test had just one correct answer although it was somewhat implied. On the other hand, it shows that a test can not always be fixed with a `none of the above’ clause.


      • on March 5, 2010 at 5:03 pm Luboš Motl

        Oh, I see, that was your moral message. ;-) But the person who talked to you probably meant that if one wants at least one answer to be correct, there should be “none of the above” at the bottom.

        I don’t think you have falsified the statement. From this viewpoint, your example with 1,2,3,4 is just another test where the question 5 is not needed for the existence of a correct answer, and is therefore wrong. ;-)

        The options 1,3 look just like 5 and 2,4 is morally similar, too, but this fact doesn’t seem to create no “new” problems.



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