Talk:Significant figure

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Regarding the significant digits of square roots: consider the square root of 54478, which has 5 significant digits. The root is 233.405227; rounding this to 5 significant digits gives 233.41. But to state the answer as "233.41 with 5 significant digits" is wrong: after all, the true initial value could have been 54477.5 whose square root is 203.404155, and that is not captured by 233.41. The correct answer is "233.4 with 4 significant digits". AxelBoldt 01:17 Jan 9, 2003 (UTC)

Regarding significant figures of products: I was taught that the significant figures of a product of two numbers with m and n significant digits, respectively, was simply the lesser of m or n, NOT one less than that lesser figure. In the specific case presented, which is to multiply 23.2 by 146.5, we get an answer of 3398.8 and need to estimate its precision. Now, by writing "23.2" and understanding "3 significant digits", we are saying we have very high confidence that the truth lies somewhere between 23.15000 and 23.249999999... That is, we estimate the outer bounds of error to be ± 0.05. Similarly, the error maximum for 146.5 also is ± 0.05. I was taught that, when one muliplies two numbers together, the error estimate in the resulant product is given by ADDING the two RELATIVE errors. The relative errors are the error estimates divided by their respective values. So the relative error of 23.2 ± 0.05 is .05/23.2 = ± 0.00215517 (to an unjustified number of digits!). Similarly, the relative error for 146.5 ± 0.05 is ± 0.00034130. So in the product we expect a relative error of the sum of these, or 0.00249647. Multiplying this by the product itself yields an error estimate for the product of ± 8.4850. So we are confident the product must fall somewhere betweeen 3390.315 and 3407.285. Note that, were we to apply the error estimates to the original input numbers and calculate, we would decide the answer must lie between 23.15 x 146.45 = 3390.3175, and 23.25 x 146.55 = 3407.2875. Now, suppose we agree that the product must have 3 significant digits (because that is the least significant input, 23.2). Then we will write our product in exponential form as 3.40 x 10^3 (sorry no superscipts here), which we should interpret as: "the answer must fall between 3395 and 3405". As we see, that does not cover the whole range created by using the maximum possible error in the input values. The real problem here touches on statistical arguments. When we say that we estimate the error on 23.2 as ± 0.05, a probabilitic view is that there is a certain precisely-known confidence that the truth lies betweeen 23.15000 and 23.24999. Very often that confidence interval is chosen as 99.7% Condfidence, given by ± 3 Standard Deviations of the population of repeated measurements of the varaible whose mean value is 23.2. Although we don't actually have any statement of Confidence Interval or Standard Deviation here, that does not affect the rest of the argument. The point is that, for each of the input values, there is a (presumed Normal) distribution of observed values with a Confidence Interval which may be used to express an "Error Estimate". But when it comes to the product of the two input numbers, the distribution of observed values has its own mean and Variance, and we find that the Confidence Interval for it is certainly less than the simple sum of relative errors of the inputs. Viewed another way, the probability that BOTH input values are truly at the maximum expected deviation from their stated values SIMULTANEOUSLY is quite small. A statistician, given all the information he/she would normally expect to have, would calculate a "pooled estimate of Variance" to apply to the product, and it would be less than the classically-derived maximum Error Estimate derived above by summing the relative errors. In that case, to say that the product DOES actually "fall between 3395 and 3405" would correspond well to what statisctical tools predict, and we would be considered correct to state that the product has 3 significant digits. User:Ken_McLellan 16:52 June 12, 2003 (UTC)

I have some concerns about how much we stress significant figures to students (undergrad and h.s.). It can seem like one of those rules that makes chemistry so distant and arcane to students. Instead, waiting until students are engaged in research where sig figs actually have meaning and consequence encourages a more "real world" approach. In the meantime, they could perhaps just agree to use 3 sig figs! Janet Russell