How to calculate error for squared data?

How to calculate standard error?

• given the data set 50.6, 59.8, 50.9, 51.3, 51.5, 51.6, 51.8, 52.0 how do you calculate the standard error? the answer is found to be (along with average wave speed) (51.3 +/- 0.2)m/s the 0.2 is the standard error, but i have no idea how u get it. any help would be appreciated, thanks.

• Answer:

  The easiest way to get the mean and standard deviation of a set of numbers like this is to calculate the sum of each number, then to square each number and calculate the sum of the squares. (Incidentally, the second number should be 50.8 to get the reported average.) Once you have done that, your mean (average) will be sum / n, where sum is the sum of the numbers and n is how many numbers you have. If sumsq is the sum of the squares, the standard deviation will be stddev = sqrt ((sumsq - sum^2 / n) / (n - 1)) and the standard error will be stddev / (sqrt (n)) I subtracted 50 from each number before doing the sums and sums of squares to make the numbers easier to work with and to reduce the loss of significance that would have otherwise resulted when subtracting two nearly equal numbers during the computation of sumsq - sum^2 / n. Subtracting 50 from each entry reduces the mean by 50 and has no effect at all on the numerical result of the standard deviation. For these quantities, I got these results: sum = 10.5 sumsq = 15.55 mean = 10.5 / 8 = 1.31 When adding back the 50 to the mean that was subtracted from each datum, the resulting mean is 51.31, retaining one additional significant figure than is really justified to combat roundoff error. stddev = sqrt ((15.55 - 10.5^2 / 8) / (8 - 1)) = sqrt ((15.55 - 13.78125) / 7) sqrt (1.76875 / 7) = sqrt (0.253) = 0.503 stderr = 0.503/sqrt(8) = 0.18. The result is therefore 51.31 +/- 0.18 m/s, or, to report to the correct number of significant figures, 51.3 +/- 0.2 m/s. It is probable that the formula you were given for standard deviation is stddev = sqrt (sum ((x - mean)^2) / (n - 1)) but this can be shown to be mathematically equivalent to the formula for stddev I gave earlier.