## I. Combining Experimental Errors

After measuring the length and width of the block last week, you were asked the following question: "Since your measurements of the length and width are subject to random error, how should you report the area of the block including a 90% confidence interval?" Remember what we mean by "confidence interval". We expect the mean of a small number of measurements, plus or minus one standard error, to overlap the true mean approximately 6 times out of 10. Thus, an interval equal to +/- the standard error in our measurements is referred to as a 60% confidence interval. Similarly, an interval of +/- 2 times the standard error in our measurements is referred to as a 90% confidence interval.

Assume that your length and width have been measured to 90% confidence with the following results:

L = 17.63 +/- 0.04 cm,

W = 7.598 +/- 0.004 cm.

To find the area to 90% confidence, you might be tempted to reason as follows:

if the mean area A is 17.63 cm X 7.598 cm = 133.95 cm2,

and the maximum area A is 17.67 cm X 7.602 cm = 134.33 cm2,

then the 90% confidence interval in A is = 134.33 cm2 - 133.95 cm2 = 0.38 cm2.

This seemingly reasonable way of determining your uncertainty is not correct. The reason it is wrong follows from an examination of probabilities.  Since there is only one chance in ten that our estimate of the length does not overlap the mean of an infinite number of measurements, there is only one chance in twenty that the true mean lies more than 0.04 cm above our mean L and similarly only one chance in twenty that the true mean for the width lies more than 0.004 cm above our mean W.  The probability that we are too high for both measurements is thus only one in 400, not one in 10.  Thus, an uncertainty of +/-0.38 cm2 actually represents a much higher level of confidence than is warranted by our measurements.

If we wish to report the area of the plate to the same level of confidence we report for L and W, we must allow for the possibility that errors in independent measurements might offset one another. The correct procedure is as follows:

First, calculate the uncertainty in A caused by the uncertainty in L only, using the mean value of W in this calculation.  Call the result (DA)L:

(DA)L = A(L + DL, W) - A(L, W).

In our example this result is

(DA)L = (17.67 cm X 7.598 cm) - (17.63 cm X 7.598 cm) = 0.304 cm2.

Next, calculate the uncertainty in A caused by the uncertainty in W only, using the mean value of L in the calculation.  Call this result (DA)W:

(DA)W = A(L, W + DW) - A(L, W)

= (17.63 cm X 7.602 cm) - (17.63 cm X 7.598 cm) = 0.071 cm2.

Now, if we think of the area A as a function of two independent variables L and W, then small variations in L and W will cause A to change by

or

The interval +/-DA is the 90% confidence interval for the area. (It is customary to retain only one significant figure when reporting confidence intervals; two significant digits were shown here in order to make the example clearer. In other words, the reported result for the area should be 134.0 +/- 0.3 cm2.)

This procedure is a generally valid one. There can be one, two, three or more variables in your calculation and the calculation can involve any function or arithmetic operation. Note, however, that all measured quantities must be known to the same level of confidence, and for our purposes this means that each experimental parameter must be measured the same number of times.  A slightly more complicated calculation would allow us to remove this restriction.

Use your measured values of the length and width of the block from last week's lab to determine the area of your block including a 90% confidence interval.