Question on the error of the estimated mean.

[NATURE.M]

So I took a series of measurements, where the reading error for the measurements are either ±0.01 mm or ±0.02 mm or ±0.03 mm or ±0.04 mm. Now, I can calculate the estimated mean of the measurements by x_est=Ʃ x_i / N, where x_i is the i th measurement. Then I can propagate the errors in the x_i quantities by Δx = √((Δx_1)^2 +...+(Δx_n)^2)
Then the error is estimated mean would be Δ(x_est) = Δx/N.
So, if I did everything correctly would the Δ(x_est) be the standard error?
And would each individual measurement X_i have to have the same reading error if the same device was used to measure them (ruler), then the above equation would become Δ(x_est) =Δx/√N?

 

[mathman]

Your presentation is a little confusing. The variance involves the error squares and the standard deviation is its square root. When combining independent measurement errors, the divisor is N for the variance and √N for the standard deviation.

 

[K^2]

mathman, OP has a situation where each measurement can have its own uncertainty. This happens if you use different instruments for measurements, for example. Say, estimating velocity by taking measurements from GPS and accelerometers. In the simple example where each measurement is a direct measurment of a quantity and you estimate the mean, OP's equation is exactly right. Estimated error is the square root of the sum of the squares.

In case where all \Delta x are the same, you get \Delta x/\sqrt{N} for the answer, as you'd expect.

 

[mathman]

Then the error is estimated mean would be Δ(x_est) = Δx/N.

then the above equation would become Δ(x_est) =Δx/√N?

The above statements look confusing.