# Use of statistics in experiment

• Astudious
In summary, the best estimate for the random error σ(X) in a single measurement is given by σ(X)2 ≈ 1/(n-1) * ∑((xi-μ)2) where the sum is over all i. However, this equation pertains to multiple measurements rather than a single measurement. Additionally, it corresponds to the sample variance and may not accurately estimate the value of X's random error for the population as a whole. To predict the value for a new or unknown single data point, the equation should use the sample average of the existing data divided by n.

#### Astudious

I have seen that "the best estimate for the random error σ(X) in a single measurement is given by

σ(X)2 ≈ 1/(n-1) * ∑((xi-μ)2) where the sum is over all i"

I have two questions about this: firstly, how can this pertain to a "single measurement" if it requires the data from multiple measurements (x1, x2, x3, ... xi)? Secondly, this seems to correspond to the sample variance - wouldn't it be a more accurate estimate of the value of X's random error to convert to the variance of the population of X as a whole?

Astudious said:
I have seen that "the best estimate for the random error σ(X) in a single measurement is given by

Where did you see this?

If you look at your equation and plug in n = 1, is the variance defined?

Where did you see this?

If you look at your equation and plug in n = 1, is the variance defined?

http://en.wikipedia.org/wiki/Unbiased_estimation_of_standard_deviation

No, but the variance need not (and perhaps should not) be defined for n=1 - a single measurement by definition cannot have a "spread".

I don't see where it says anything about the variance determined from a single measurement in that article. Where did you see that?

If you want to use several data points that you already have to predict what will happen for a new or unknown single data point, that is the equation you should use.

PS, The correct equation uses the sample average of the existing data in place of μ. If some how you know μ, you can use it, but divide by n rather than n-1.