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I'm looking at a problem where the number of samples, $N$, is 2 million. Each sample represents a profit or loss. Let $X_n$ be the dollar payoff to the n-th game. $E[X_n] = 0$. Let the total payoff of $D(N)$ be $D(N)=\sum_{n=1}^N X_n$, then $E[D(n)] = 0$. The standard deviation of the D(N) is 7141. Then it states that, there is roughly a 5% chance that the gain/loss exceeds 1.96 standard deviations (i.e., a 5% chance of gains or losses beyond +/- 1.96*7141=13,997).

I plugged in these numbers on https://www.socscistatistics.com/confidenceinterval/default3.aspx, and got the numbers in the below image.

Can someone explain to me why the confidence interval here was computed using the $S_M$ (standard error) instead of the standard deviation?

enter image description here

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Using the standard error allows us to tighten up the bounds as the sample size increases.

If all we used were the standard deviation, we would not gain confidence in our estimate as we increased the sample size. That’s not how it should work. If you flip a coin four times and get HTHH, maybe the coin is in fair, but the evidence is modest. If you flip 100 times and get 25 T and 75 H, you’re more confident that the coin favors H.

We should get more confident in our estimates as we have a larger sample size.

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  • $\begingroup$ Right I think I understand this. In this particular example that I saw, the standard deviation of $D(N)$ is proportional to $\sqrt{N}$ so it actually grows as the sample size gets larger. It's not intuitive to me why the example used the standard deviation to compute the confidence interval here $\endgroup$
    – roulette01
    Apr 5, 2020 at 14:56
  • $\begingroup$ Standard deviation and standard error are not synonymous. You mean standard error, not standard deviation. $\endgroup$
    – Dave
    Apr 5, 2020 at 15:02
  • $\begingroup$ No, I'm talking about standard deviation in this comment. The standard deviation that the example derived is $\sqrt{N\frac{51}{2}}$ $\endgroup$
    – roulette01
    Apr 5, 2020 at 15:04

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