I've read in Bill Haywood's book that SD = square root of the sample size divided by 2. In Arnold Synder's book BB in BJ however, it is the square root of the sample size. Which is correct?
For example, If I was to use 100 coin flips is the SD 5 or 10.
Thank you
> I've read in Bill Haywood's book that SD =
> square root of the sample size divided by 2.
> In Arnold Synder's book BB in BJ however, it
> is the square root of the sample size. Which
> is correct?
> For example, If I was to use 100 coin flips
> is the SD 5 or 10.
Yours is a common question, and the answer is: It depends!
If you flip a coin 100 times, saying the standard deviation is 10 is interpreted as I did on pp. 151-2 of BJA. Start at a particular spot and move one to the right for every head and one to the left for every tail. Your expectation is to wind up where you began, but the standard deviation is the square root of 100, or 10, so there is roughly 16% probability that you will land 10 to the left or 10 to the right of where you began when the 100 tosses have been completed.
Contrast this with saying that, for 100 tosses, the mean is 50. Now, after we complete the 100 tosses, see how far we are to the right or to left of that mean. Here the SD is the square root of npq, where n is the number of tosses, p is the prob. of heads (1/2), q is the prob. of tails (also 1/2), and so (npq)^.5 = sqrt of 25 = 5.
We now say that there is a 68.3% prob. that we will land within a one-standard-deviation bandwidth around the mean, falling, therefore, between 45 and 55.
The above is actually incorrect for such a small sample because of the rather large prob. that we could land precisely on 45 or 55, and so a correction factor is used to get a more accurate answer for this binomial distribution. But, when the sample is very large, that correction becomes inconsequential and the methodology gives excellent statements of probability.
In any event, the formula for SD does not change.
Don
> Yours is a common question, and the answer
> is: It depends!
Hmmm... I'm afraid that the statement above is somewhat misleading--I'm not sure why you said "it depends." The formula for standard deviation (SD) is invariate (unless you want to consider 'grouped' or 'classed' data).
"Standard deviation" is defined precisely as the sqaure root of the arithmetic mean of the squared deviations from the mean. I can't write the formula here, but, in words: one takes the deviation of each score (i.e., heads or tails) from the mean, squares the difference of each score from the mean, sums the results, divide by the number of cases (in this example, N=100), then takes the square root. [Note that some statisticians, for theoretical reasons, divide by N-1, or, in this case, 99.]
Where the confusion lies may be that this formula is more technically referred to as the 'sample' standard deviation; as a result, one can't determine what the standard deviation is for the problem presented above--because we don't know the arithmetic mean. In two samples of 100 coin flips, for example, the mean (and, therefore, the standard deviation, is likely to be different. You might have 60 heads in the first sample of 100 flips, and 55 tails in second sample. If so, the mean--and, thus, the SD, would be different for each sample.
Maybe that's what Don meant when he said "it depends"; i.e., the SD for your 100 coin flips depends on what the mean for those 100 flips is.
In any case, Haywood's (and I'm not sure who that is) formula is not correct.
I'm sorry for the confusion. Let's forget about blackjack and talk about a binomial coin flip.
If we flip 100 times, then the formula for standard deviation, as deviation from the mean, is precisely as I gave it: (npq)^.5, or 1 SD = 5.
The mean number of heads for 100 tosses is 50, and one standard deviation is 5 in either direction from the mean.
The 10 comes in when we take the square root of 100, and it can be interpreted as one standard deviation of the NET result of EXCESS heads over tails. So, for the same experiment, above, if the breakdown is 55-45, we have ten extra heads, and that represents a one standard-deviation result from the mean result of 50-50, or no excess at all.
The confusion stems from how we express the results. If we express the RAW number of heads, it's possible to have 100, and the SD is 10. If we express the deviation from the mean, the maximum deviation can be only 50, and one standard deviation is 5.
I hope this is clear to everyone.
Don
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