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Consider the following question:

enter image description here

I have calculated:

  • margin of error for the first 50 jumps is 0.127,
  • margin of error for the last 10 jumps is 0.248,
  • margin of error for all 60 jumps is 0.156.

The correct answer is supposed to be (D), but it seems to me like the narrowest confidence interval is the one for the first 50 jumps (B). Strangely, answer choice (B) is wrong according to College Board. Any insight? Thanks!

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Sorry to say it but you've just got a calculation error on your last line where you calculate the standard error for the "all 60 trials". I can sympathise!

When calculating q (not having the event, which here is $\dfrac{23}{60}$) you've double-dipped and taken the complement of q. In other words you've written the numerator of the fraction within the square root as $p*p$ rather than $p*q$

What you wrote in your calculator window:

$$ \sqrt{\dfrac{\dfrac{37}{60}*(1-\dfrac{23}{60})}{60}} $$

Whereas what you wanted to do:

$$ \sqrt{\dfrac{\dfrac{37}{60}*(1-\dfrac{37}{60})}{60}} $$

or equivalently

$$ \sqrt{\dfrac{\dfrac{37}{60}*\dfrac{23}{60}}{60}} $$

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  • $\begingroup$ As a further comment your hand-written note (in green) was the correct formula so it's just how you've specified in the calculator. $\endgroup$
    – James Stanley
    Commented Feb 28, 2022 at 3:06
  • $\begingroup$ Thank you! This time, I got a smaller margin of error (0.123). I appreciate you finding and pointing out the mistake. Just to recap / follow up: is it generally true that the margin of error should decrease when you add more data, because n (the denominator) gets bigger? Or is it possible for p and (1-p) to change so much that the margin of error increases despite having more data? $\endgroup$
    – EthanAlvaree
    Commented Feb 28, 2022 at 17:38
  • $\begingroup$ In some ways that's a different question as well, and there will undoubtedly be other people here on CV who will be able to provide better answers than I could (from several different angles) so if you're keen to ask it then I'd say go for it. $\endgroup$
    – James Stanley
    Commented Feb 28, 2022 at 20:31
  • $\begingroup$ (update to a deleted comment) The answer is quite complex: for a fixed denominator, standard errors are smaller when $p$ sits closer to 0 or 1, and larger when $p$ sits at 0.5. So the actual distribution of standard errors with a fixed increment to the denominator (but no increment to the numerator) depends on the value of $p$ you start from as well as the denominator. $\endgroup$
    – James Stanley
    Commented Feb 28, 2022 at 20:54

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