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Question from a Statistics textbook written in 1978 (ISBN 0-393-09076-0):

It is claimed that a vitamin supplement helps kangaroos to learn to run a special maze. To test whether this is true, sixteen kangaroos are divided up into eight pairs. In each pair, one kangaroo is selected at random to receive the vitamin suppplement; the other is fed a normal diet. The kangaroos are then timed as they learn to run the maze. In six of the eight pairs, the treated kangaroo learns to run the maze more quickly than its untreated partner. If in fact the vitamin supplement has no effect, so that each animal of the pair is equally likely to be the quicker, what is the probability that six or more of the treated animals would learn the maze more quickly than their untreated partners, just by chance?

So, my thinking is that the pairs are a red herring and we only need to look at the 8 "fast" kangaroos. if the vitamin has no effect, then each of these 8 kangaroos have a 0.5 chance of being "fast" or "slow" ("fast" defined as "learning the maze quicker than its counterpart"). So this is a binomial distribution, the cumulative probability for 5 out of 8 at P=0.5 is 0.855, so the chance of 6 or more kangaroos being "fast" out of 8, if the vitamin has no effect, is 1-0.855=0.145. Is this the right solution, or am I missing something with the pairs of kangaroos?

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Yes, binomial is the correct interpretation here. Note that, without a calculator, it could be easier to calculate $\sum_{k=6}^8P(X=k)$ directly (Let $X$ be the number of pairs that treated kangaroo is faster). Note also that, due to the symmetry of the problem, this is also equal to the case where untreated kangoroo is faster in more than 5 trials.

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  • $\begingroup$ Yay, so I got it right! :) Thank you very much, @gunes! Sadly I don't understand the part about doing it without a calculator :( but that's obviously my lack of knowledge. I also didn't realise that it was symmetrical (but I understand that part). $\endgroup$
    – Reader 123
    Commented May 8, 2020 at 20:54
  • $\begingroup$ I meant calculating it by hand. Certainly, you'd prefer the way where there are less number of operations to perform. $\endgroup$
    – gunes
    Commented May 8, 2020 at 22:09
  • $\begingroup$ I know you meant calculating by hand - I just don't understand how... :| $\endgroup$
    – Reader 123
    Commented May 9, 2020 at 16:16
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    $\begingroup$ $$P(X=k)={8 \choose k}\frac{1}{2^8}$$ $\endgroup$
    – gunes
    Commented May 9, 2020 at 16:35
  • $\begingroup$ Wow, so the whole thing boils down to 37:256 = 0.1445 ! :-o. Thanks, @gunes, I would have never thought of that! $\endgroup$
    – Reader 123
    Commented May 9, 2020 at 19:47

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