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Let's say I play a game with where the outcome can be any even number from 10 to 30, with unknown probabilities. After playing this game, I have the following results:

$\bar X=20, s=3$ (with $n=1000$)

If I were to calculate the 95% confidence interval, using the formula:

$\bar X \pm 1.96\cdot s/\sqrt n$

  1. Is this even valid given that the underlying distribution is discrete?
  2. Disregarding #1, am I correct in assuming that the resulting confidence interval tells me that the true mean will lie inside this range 95% of the time?
  3. Similarly, am I correct in assuming that it does not tell me that the outcome value will lie within this interval 95% of the time?
  4. How can I find the 95% confidence interval for the outcome values?
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  • $\begingroup$ The formula you give is for a confidence interval for a mean, (suitable when $X\sim\,\text{iid}$ normal, though it would often also be suitable in very large i.i.d samples). When you ask for an interval for a variable, you'd have to clarify which of several kinds of interval you seek instead (e.g. a prediction interval? a tolerance interval? perhaps something else?), and such a calculation will depend on the underlying distribution even when you collected a large sample. $\endgroup$
    – Glen_b
    Nov 3, 2018 at 7:16

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  1. I see no problem with CI calculated. You have a set of 11 outcomes from 10-30 even nos. (say like a random number selection) and you calculated the mean value to be 20. So yes in the same way you can calculate the CI assuming that the distribution of your outcomes is normal (which is where you are getting your 1.96).

  2. The definition of CI is that if you repeat the experiment a large number of times, say 100 times, the true mean will lie within 95% of the calculated CIs.

  3. Yes, the CI is just one of the intervals in which your true mean may lie. Often people take this to mean that "we are 95% certain that the true mean lies within this interval" which is incorrect

  4. to get a CI for each of the possible outcome probabilities, you would need a binomial CI where you will calculate the probability of each outcome happening (p) and calculate the CI around this. This will give you the probability of each outcome and a CI around the probability.

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  • $\begingroup$ Thank you. In regards to #4, what I meant to ask was: How do I find the values x and y so that I can say: If I repeated this game a large number of times, the outcome (my score) would be between x and y 95% of the time? $\endgroup$
    – Joakim
    Nov 2, 2018 at 11:29
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    $\begingroup$ By the outcome, do you mean, the CI around your mean outcome (20)? or do you mean a CI around all possible outcomes? In case of the former, you have got that done as you described in your question, assuming the outcomes are normally distributed. A 95% CI around 20, say [14, 26], would be interpreted loosely as containing the true mean 95% of the times under repeated experimentation. $\endgroup$
    – EskCargo
    Nov 2, 2018 at 13:41
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    $\begingroup$ If your data is normally distributed, the estimated mean is also the outcome with the highest probability of happening. so the probability that your score is 20 in a random game is also higher than it being another value; but there could be other values with a "high" probability too (not as much as the mean because that is the central tendency value). The probability of another score occurring can be calculated from your sample of 1000 tries. I hope that made sense. it is a conceptual question around probabilities you have asked so you have to think of it that way... Cheers and good luck $\endgroup$
    – EskCargo
    Nov 2, 2018 at 13:41
  • $\begingroup$ I actually meant CI around all outcomes. Sorry, I'm not making this easy for you.. nevertheless, your answers have been super helpful, so thank you :) $\endgroup$
    – Joakim
    Nov 2, 2018 at 14:50

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