Chemistry for poets

February 6, 2006 1:28 PM

ChemistryFilter: Help me understand significant figures.

I'm taking an Organic Chem class to fulfill a prerequisite for grad school. I believe I understand what I read in the textbook, but when I have my lab, my professor seems to tell me something completely different. As a result, I'm now confused about something I thought I understood.

I'm trying to understand significant figures and how they relate to measurement. If I'm measuring something with a ruler that is accurate to a centimeter (there are no demarcations in between 1 cm, 2 cm, 3 cm, etc), I should mark my measurement as being exact to the centimeter and estimated to one tenth of a center. Therefore, if I'm measuring something that falls between 2 and 3 cm, closer to 3 cm than 2 cm, I can estimate that the object I'm measuring is 2.7 cm long, given that significant figures are the digits in a measurement that are known with certainty (2 cm) plus one digit that has uncertainty (in this case .7 cm). If my ruler had demarcations between centimeter markings, I would have two certain measurements and one uncertain estimated number - for example, 2.73 cm, where 2.7 is a known number and .03 cm is estimated.

My understanding is that each the numbers in the above examples is significant ... both 2 and 7 are significant figures in my first example and 2, 7, and 3 in my second. My book very clearly states that significant figures are the digits in a measurement known with certainty plus one estimated digit.

Now, when I had my lab, my professor confused me by saying that only the 2 (in my 2.7 cm example) is significant, and that an estimated measurement cannot be included as a significant figure. I've asked another chemist and he said the same thing.

So, should I go with my professor and exlude uncertain numbers when trying to determine significant figures, or should I go with the book? I realize that this is an extremely basic question, but it seems strange when all the printed information I've read says one thing, and the actual human chemists I've asked have said another. At any rate, I'm confused and would greatly appreciate help the wonderfully brainy Mefite community has to offer.

posted by discokitty to science & nature (31 comments total)

So, should I go with my professor and exlude uncertain numbers when trying to determine significant figures, or should I go with the book?

In the example above, your measurement would be 3 cm, not 2.7 cm, because you have no ruler gradations to use that make 0.7 a "significant" measurement.

Another way to think about it is, if you were to give a room filled with students the very same ruler, you can be confident that the room will on the whole measure 3 cm as opposed to 2 cm. There is much less confidence in the collected average of measurements like 2.6, 2.5, 2.8, 2.9 or 2.7 cm, for example.

Go with the professor and chemist. Your book is either wrong or you accidently misinterpreted it.
posted by Rothko at 1:36 PM on February 6, 2006

Well I would probablt disagree with your Proff.
linky

The last digit is always the least accurately known one. If you can estimate 2.7 you know the 2 for sure and the 7 is the best estimate you can come up with knowing its not perfectly accurate.
posted by ozomatli at 1:44 PM on February 6, 2006

Ditto what Rothko said. Significant figures are certain units of measurement -- 3cm in your example. In part of a larger equation, use the smallest *common* unit of measurement that you can. i.e. if you measured the mass of one element to the nearest 3 decimal points, and you measured the mass of another to the nearest 2 points... then you mix them together... you would report your total mass only to the nearest 2 points. Dig?
posted by kaseijin at 1:45 PM on February 6, 2006

On preview: It is true that your last digit is always the least accurately known one, but in none of the chemistry or physics classes that I have taken have we been allowed to eyeball a distance between two demarcations and assign it a value.

To again refer to the case in the question: the unit of mm would be the least precise, purely because you have to round it to the nearest demarcation. Let's convert the unit: 3mm is equal to 0.003m In this case, the 3 is your least accurate value, and you are quite certain about the zeros in there (which are also significant in this case).

At least, that has always been my understanding.
posted by kaseijin at 1:52 PM on February 6, 2006

You're getting into trouble here because the terms are fuzzy. It seems like 'significant figure' and 'error' are being confused.

If you go farther, you'll make much more use of "error" (often coming from the statistical measure standard error), rather than sig. figs.

With a ruler that only has 1 cm markings, the first thing you should do is get a better ruler ;-)

After that, realize that you're really just guessing for any value between 2 and 3. The best bet is to take your measurement, and round down if the value seems to be between x.0 and x.4, and round up on x.5 to x.9. You would then have the 1 sig. fig., like your prof. is suggesting, and you would have an error estimate of +/- 0.5, which is another useful tidbit, as it tells people how good your measurement was (in this case, not very good).

You're trouble here is that you're "making up" one of the numbers in 2.7 cm, so it's better not to do that. Call it 3 cm, +/- 0.5 cm. Then you get the requisite 1 sig. fig. If you do want to call it 2.7 cm, it sort of defeats the purpose of keeping track of sig. figs., as one of your number is a guess, and hence not significant. (By the freshman lab definition of significant, which really won't get you very far. In a real lab, you'll be using much better measures of accuracy).

Now, you're eye is probably good enough to get somewhere close to the 1/10th cm unit, without markings, so you could use them (with an indication of your suspected error, which might be +/- 1 or 2 or 3 10ths, but your prof. is most likely just being pedantic to illustrate the importance of sig. figs.)
posted by teece at 1:54 PM on February 6, 2006

Rothko writes "In the example above, your measurement would be 3 cm, not 2.7 cm, because you have no ruler gradations to use that make 0.7 a 'significant' measurement."

I'm pretty sure you're wrong about this, man.

It's been awhile since I've had to think about these things formally, but in practice, I always estimate the last digit of this kind of measurement. And when a measurement of this sort is reported with an uncertainty, the uncertainty is always of the same order of magnitude as the last reported digit in the measurement: in other words, the last significant figure is uncertain. for example, the measurement we're discussing here would be reported as 2.7±0.1 cm.

Plus, every resource I can find on the web agrees with discokitty's textbook. Overwhelmingly so.

That said, you should probably go with what your professor says, or bring it up again with some resources to back you up.

kaseijin writes "It is true that your last digit is always the least accurately known one, but in none of the chemistry or physics classes that I have taken have we been allowed to eyeball a distance between two demarcations and assign it a value."

You sure as hell are allowed to do so; keep in mind that there's some uncertainty associated with the process, though.

teece writes "You're trouble here is that you're 'making up' one of the numbers in 2.7 cm, so it's better not to do that."

Just to reiterate, I'm pretty sure you're wrong about this...
posted by mr_roboto at 2:10 PM on February 6, 2006

I'm gonna have to go ahead and disagree with Rothko et al here. My understanding of significant figures matches what is written in your book: you are allowed to estimate within tenths of the smallest increment on your instrument, and significant figures may contain one estimated digit. The Wikipedia entry on the subject supports this (see the first paragraph of the "Measuring with significant figures" section).

Apparently there is some controversy here, but I believe you and your book are technically correct. If I were you, I'd approach your professor with the relevant passage from the textbook, and ask him/her to explain the discrepancy. But in the end, you should probably do whatever your professor says, since that's where the grade is coming from.

(On preview, what mr_roboto said.)
posted by purplemonkie at 2:13 PM on February 6, 2006

Oh; and for a real-world example...

When I would do this most often would be in measuring a fluid in a pipette or a graduated cylinder. In this case, I'll try and measure at one of the demarcations (e.g. at the 35 mL line). However, I wouldn't trust my measurement to be absolutely precise, so I would record it as 35.0±0.1 mL. In other words, one more significant digit than the device demarcations allow me to measure precisely.
posted by mr_roboto at 2:13 PM on February 6, 2006

And not to be too verbose about this...

The logic of doing it this way is that I certainly wouldn't report my measurement as 35±1 mL, since I know that I'm less that 1 mL off of the 35 mL line (assuming there are also lines marked at 34 and 36 mL). So my uncertainty must be at the next order of magnitude down; hence the reported measurement of 35.0±0.1 mL and three significant figures.
posted by mr_roboto at 2:17 PM on February 6, 2006

My last post is incoherent.

Short advice: don't dwell on this.

Longer: 2.7 could indeed be considered a measurement with 2 sig. figs., given the method you outlined. You'd need to make sure to include error estimates, so that people knew your measurement of the last digit was not all that precise.

Your prof. is telling you that you only have 1 sig. fig. to illustrate a point, I suspect, and make the significance of the exercise clearer.

The only really important thing to remember about sig. figs. relates to derived values. If you take many measurements at 1 sig. fig., and get an average, your calculator might spit out an average of 3.8020986464. You need to know that your actual answer should be 4 with 1 sig. fig., 3.8 with 2, etc.

Sig. figs. are meant to disabuse you of the notion of saying "my widgets are 3.8020986464 cm long" when you only used a simple, 1cm-marked ruler to measure. But the real meat of measurement uncertainty is in error, which sig. figs. only scratch the surface of.

With your ruler, I'd certainly note that my millimeter measurements were imprecise (with error bars), but I would not simply throw away all millimeter information. But in an intro. exercise about sig. figs., a prof. very well might want you to do that.

Thus I'd say 2.7, with two sig. figs, and an error of +/- 2.5 mm. But if it bugs a prof. (which it only will in freshman labs), shrug and round the number off.

mr_roboto: in my experience, when they're teaching you about sig. figs., uncertainty in the form of error bars is only very lightly touched upon, if at all. Hence, the prof. is trying to hammer home the point that the measurement is uncertain, I suspect. Since the last digit is rather uncertain with such a ruler, he's ruling it insignificant. That's completely kosher -- it's just not how you'd do it in a real lab.

It's never wrong to make a measurement less precise, and if you're very conservative that's exactly what you'll do with a ruler like the one mentioned here. Sure, it's possible to be more precise with such a tool, but there is no requirement to be.
posted by teece at 2:17 PM on February 6, 2006

teece writes "The only really important thing to remember about sig. figs. relates to derived values."

This is a lab class, in which the students are making actual measurements on actual devices. So "derived values" are not the only point of concern: the significant figures on the measured values need to be right, too.

I think you're giving the professor lots of credit in framing this all as a pedagogical exercise. My guess is she just forgot how beginners are taught significant figures. Not too surprising, if you've known many professors....
posted by mr_roboto at 2:23 PM on February 6, 2006

My prof isn't a native english speaker, so I think he's struggling to communicate himself clearly and I don't always "get" everything he's trying to say.

I probably am overthinking things, I'm just glad that I'm understanding the concept properly.

Thanks everyone.
posted by discokitty at 2:25 PM on February 6, 2006

I was taught (book and lab) the method described in your book, and therefore disagree with Rothko et al. If you have a flask with 100mL gradations, reporting "140mL" for an amount of liquid in the flask seems perfectly correct. (And that's what I was taught in Chemistry, Physics, etc.) The Wikipedia entry seems correct to me.

The *important* thing they're trying to teach you is not to report results with absurd precision just because that's what your calculator says. You multiply "X.XXX" by "Y.YYY" and your calculator says "Z.ZZZZZZZZZZZ", stop right there and realize the calculator is giving you a false concept of the precision of the result. If you get that, you're 9/10 of the way to understanding significant figures.
posted by jellicle at 2:30 PM on February 6, 2006

the significant figures on the measured values need to be right, too.

Sure, but what constitutes significant is quite open for debate. The profs. seem to take the stance that the error of measuring with such a ruler is around 1 cm. If you make that assumption, then you measure 2.7, use 2.7 for any intermediate results, and report 3.0 +/- 1 cm. (per Taylor, An Intro. to Error Analysis).

The real quibble seems to be that we all know you can indeed be a bit more accurate with such a ruler than merely getting to the nearest centimeter.

But if the profs. are making that assertion (as they seem to be), then it is indeed the case that you only have 1 sig. fig. in a measurement such as 2.7 cm. With only one sig. fig., you report such a measurement as 3 cm.

The profs. may be confusing here, but they're not wrong unless they agree that 2.7 is more accurate that +/- 1 cm, but still feel it only has one sig. fig.

I don't know which is the case.
posted by teece at 2:36 PM on February 6, 2006

I was taught (book and lab) the method described in your book, and therefore disagree with Rothko et al. If you have a flask with 100mL gradations, reporting "140mL" for an amount of liquid in the flask seems perfectly correct. (And that's what I was taught in Chemistry, Physics, etc.) The Wikipedia entry seems correct to me.
posted by juv3nal at 2:42 PM on February 6, 2006

teece writes "The profs. seem to take the stance that the error of measuring with such a ruler is around 1 cm."

Which is a ridiculous stance: the ruler is marked at every 1 cm, so the error must be less than 1 cm. Conventionally, the measurement error is considered to be an order of magnitude smaller than the value of the smallest demarcation.

The professor is just wrong. Which, again, is not entirely surprising.
posted by mr_roboto at 2:44 PM on February 6, 2006

ack hit the wrong button what I meant to say:


I was taught (book and lab) the method described in your book, and therefore disagree with Rothko et al. If you have a flask with 100mL gradations, reporting "140mL" for an amount of liquid in the flask seems perfectly correct. (And that's what I was taught in Chemistry, Physics, etc.) The Wikipedia entry seems correct to me.

That reporting of 140mL isn't right regardless of whether you have 1 or 2 sig figs because not doing it in scientific notation ambiguates the position of the significant figure. you either mean 1.4X10^2 mL or 1.40X10^2 mL; reporting 140mL specifies neither.
posted by juv3nal at 2:46 PM on February 6, 2006

The profs. seem to take the stance that the error of measuring with such a ruler is around 1 cm.

I find it hard to believe any scientist would make such a nutty assumption. Unless you're using a ruler that was drawn freehand on construction paper by a third grader, how can the error be equal to its smallest increment? It seems far more likely that the professor and chemist have simply forgotten the sig fig rules. It happens, especially with stuff like this; I'm in the process of getting a Ph.D. in biochemistry and I bet most of my colleagues would be hard-pressed to explain sig figs correctly. Lots of people just don't give it much thought these days, since the need for estimation has been reduced by the availability of accurate and/or electronic instruments.

The profs. may be confusing here, but they're not wrong unless they agree that 2.7 is more accurate that +/- 1 cm, but still feel it only has one sig. fig.

According to the real definition and rules of significant figures, they are indeed wrong. As I said above, you are allowed to estimate within tenths of the smallest increment on your instrument, and significant figures may contain one estimated digit. But again, when all is said and done it's probably best that discokitty do as the professor says for the purposes of this class.
posted by purplemonkie at 2:57 PM on February 6, 2006

I have no idea what you're talking about with that last digit. Your accuracy is to the centimeter, and you have one significant digit, '3'.

Your example is a little skewed because you're using the ruler, but you're actually trying to use another device, your eyes, with greater precision. If you had a ruler with half-millimeter markings, or Vermeer calipers, you wouldn't be able to estimate anything beyond what the ruler told you either.

In a lab setting, you'll be using equipment, like a scale, or calipers, or a spectrometer that won't give you any non-significant digits, and will provide you with no means to estimate beyond what it gives you on the readout.

If you used the same ruler to mesure something 42cm long, then you would have two digits of precision.

Finaly, If you really wanted too you could write something like 2.7 ± 2 or something like that, but if you're going for digits of precision, you would go with just '3'.
posted by delmoi at 3:17 PM on February 6, 2006

First of all, if you're eyeballing you're eyeballing literally, using your eyeball to measure the difference in angle between the two points to your eye, and measuring that against a reference angle on the ruler, which will be more accurate if the angular difference is enough for you to see. Remember, the actual measurements are not being done by the ruler, or beaker, but your eyes.

Second of all you're "allowed" to do whatever you want, but if you are being graded you should do what the professor asks of you, and if you can't figure that out, you should follow the standards and hope for the best. Measuring real world values is fuzzy and you could do this in a number of different ways and still be doing valid science, depending on how subtle (or obvious) the results are.
posted by delmoi at 3:29 PM on February 6, 2006

From your example, yes, only the number 2 is "significant." The 7, you pulled out of the air.

The examples used in this thread are a little absurd. If you have a container intended to measure fluids that is marked in 100mL intervals, the intended purpose would be to measure volumes of at least 1L (or more like, 10's or 100's of L).

Example: you measure out 1.50mL (0.00150x10^3 mL) of liquid in a small pipette where the smallest gradation is 0.01mL (0.00001x10^3 mL). You measure 1.2L (1.20x10^3 mL)of liquid in a much larger pipette where the smallest gradation is 10mL. You add the two liquids togather.

You would report it as 1.20x10^3 mL, not 1.2015x10^3 mL.
posted by PurplePorpoise at 3:37 PM on February 6, 2006

juv3nal: while using scientific notation is fine (and usually preferred), 140mL isn't unclear or wrong. If you wanted to indicate 140mL with three significant figures, you could write either "1.40 x 10^2 mL" or "140. mL".

This whole question is confusing two different things: measurement uncertainty and significant figures. Measurement uncertainty describes the limits of your measurement tools. As above, I think most people are taught to measure one place past the gradations when using tools amenable to that. If the tool isn't amenable to that, then you use whatever the tool tells you.

Significant figures is a ROUGH concept of the total uncertainty in a calculation (measurement uncertainty is only one of many different types of uncertainty). It keeps you from giving ludicrously precise answers to questions when you didn't measure your input values with great precision.

Uncertainty is a more advanced version of significant figures. Calculations using uncertainty keep track of both the value and the uncertainty associated with it separately, and carry both forward to the end, and then figure out how uncertain the result of the calculation is. Same CONCEPT as significant figures, more accurate, harder to calculate.
posted by jellicle at 3:44 PM on February 6, 2006

Significant figures are just poor-man's error bars. If you write down a result with n significant figures, without any other information about the range of possible error in that result, you're asserting that if you round the true value to n digits, the result will be what you've written down.

For SI units, the number of significant figures is independent of the units chosen (which is where I think kasejin has it a bit wrong). For example: if a result is written down as 3mm, that's one significant figure regardless of units; expressing it in metres just makes it 3 x 10^-3m. The fact that this can also be expressed as 0.003m doesn't mean that the result is now correct to three significant figures.

Measurements that are exact multiples of ten units can be confusing. If your measurement is 3cm but you're more comfortable working in mm, it's tempting to write the result down as 30mm - which looks like you have more significant figures than you do. Consistently using scientific notation is your friend here. Even though 3 x 10mm looks kind of strange, it still clearly shows that your result is only good to one significant figure - clearly differentiating it from 3.0 x 10mm.

A distinction that most people don't think about much is that significant figures are a precision thing, not an accuracy thing. Accuracy is a measure of how confident you are that your measurement reflects reality; precision is about reproducibility. A pistol shooter who makes a nice tight group of holes in the target is very precise, even if the holes are nowhere near the bullseye.

In an ideal world, nobody would ever quote results more precise than the accuracy of their measurement process can justify.

If your eye is experienced enough that you can reliably measure levels between 10-unit gradations to a precision of +/- 0.5 units, you're justified in quoting those measurements to 1-unit precision rather than rounding to the nearest 10-unit gradation. But you're unlikely to convince a grad school prof you're that good.
posted by flabdablet at 3:50 PM on February 6, 2006

2.7 is THE correct answer there are 2 significant digits. The last digit is the uncertain one. As I said in the beginnng: your professor is mistaken. Mr Robot and purplemonkey are indeed correct.
posted by ozomatli at 3:52 PM on February 6, 2006

Man, I'm surprised at the confidence with which some of the incorrect answers in this thread have been delivered. These are both WRONG:

PurplePorpoise writes "From your example, yes, only the number 2 is 'significant.' The 7, you pulled out of the air."

delmoi writes "I have no idea what you're talking about with that last digit. Your accuracy is to the centimeter, and you have one significant digit, '3'. "

Where are you getting these answers from, guys? Every resource on the web--plus both chemistry textbooks I can get my hands on right now (Harris' Quantitative Chemical Analysis and Skoog, West, and Holler's Fundamentals of Analytical Chemistry)--agrees with discokitty's textbook. The logic behind this interpretation has been thoroughly explained in the answers on this thread. Discokitty's understanding of the concept, as explained in discokitty's textbook, is correct. The professor is wrong.

If you don't know the answer to the question, don't just make something up.
posted by mr_roboto at 4:03 PM on February 6, 2006

People do it differently. 2 is definitely significant, but you certainly treat the uncertain digit as significant when dealing with multiple measurements. You add the error in quadrature. Certainly.

Make your measurement. It's between 2 and 3 cm, you call it 2.7 +/- 0.5 cm (you get to pick the +/-, just use your head).

20 other people do the same. They get 2.5, 2.9, 2.3, etc... They estimate their error to be 0.5, 0.3, 0.8, 1, etc...

Take the average of the measurements as normal. Add the error in quadrature:

sigma_tot = Sqrt[sigma1^2 + sigma2^2 + sigma3^2] / (number of measurements)

In the end, you should have something that looks like

2.6 +/- 0.3 cm

Not 2.65, not 2.63, only 2.6 (note you can't get hundreths when adding only tenths, treat the tenths place as the last significant digit while dividing). Because your lowest sigma value was greater than or equal to 0.1 cm, your sigma can't be less than 0.1 cm.

Next step: buy a book on experimental methods and/or statistics, and stop taking organic chemistry.
posted by dsword at 4:08 PM on February 6, 2006

What mr_roboto said. This is getting highly ridiculous. And furthermore:

"...the significant figures method is that, when measuring using a non-electronic instrument, the observer should estimate within the nearest tenth of a division marked on the instrument. For example, if a graduated cylinder were marked off at every millilitre (ml), the observer should measure the amount of volume contained in the cylinder to the nearest tenth of a millilitre."

"The number of significant figures in a measurement, such as 2.531, is equal to the number of digits that are known with some degree of confidence (2, 5, and 3) plus the last digit (1), which is an estimate or approximation."

"We let the number of digits reported for a measurement tells us about the certainty of a measurement. The rule is that we know all of the digits with certainty up to the last digit, which is estimated."

"The last digit given for any measurement is the uncertain or estimated digit. It is uncertain because it is estimated. The last digit given for a measurement is always assumed to be estimated and it is significant."

"The number of significant figures is the number of digits believed to be correct by the person doing the measuring. It includes one estimated digit."

"Significant digits or significant figures are digits read from the measuring instrument plus one doubtful digit estimated by the observer."
posted by purplemonkie at 4:09 PM on February 6, 2006

Wait! I meant the divide by (number of measurements) for the averaging... NOT for adding in quadrature. Think of that like adding perpendicular vectors in a space of many dimensions and finding the distance between them (like the pythagorean theorem).
posted by dsword at 4:14 PM on February 6, 2006

Keep in mind we are talking about tracking errors using significant digits. This is not a great way of doing things, but useful for teaching the concept of systematic errors. In real life there would be error bars on every piece of data (where the error bars were generated via a more exact way of tracking uncertainty). You would get a data point like 2.7 (+- .4) or something like that. The idea of significant digits is that 2.7 is the estimate, thus you have TWO significant digits. Becuase we use a base ten system we trach students to estimate to the nearest tenth of the most certain degree of measument. That way you get a nice clean estimate like 2.7 (+- 0.1). The last digit is uncertain (to +- 0.1 in this case but we throughout the real length of the error bars so it could be +- 0.7 we don't know or care. Students are taught that with significant digits the last digit is suspect.) When all of the different measurements are averaged together it is the LAST digit wich will converge to the real precise measurement.
posted by ozomatli at 4:25 PM on February 6, 2006

If you don't know the answer to the question, don't just make something up.

This isn't about 'right' or 'wrong'. Either way would be just as useful in a lab, as long as you have some consistent standard, and as I said, there is no way to estimate using measuring devices more precise then your eyes.

---

Anyway, I have an interesting story about this. When a friend of mine was in high school they did semester projects in Chemistry, and the teacher told them to use 2 digits of precision, or else he would give them an F. My friend worked with a friend of his who's father was a chemistry teacher at the local collage, and so they had access to much more accurate measuring equipment. The son of the professor wrote the lab report, and used 3 digits of precision because his equipment was capable of that.

The Teacher gave them both an F, and almost ruined my friends 4.0 GPA (I guess they appealed to the school board).

What's the point of this story? To me the point is that that real scientists are not pendants about significant digits, only idiotic Highschool chemistry teachers. The other point is that you should do what your teachers tell you to do, because they're the ones doing the grading.

There's no wrong or right about this, only standards. If the effect that you're measuring is so sensitive that you have to get close to your errors, then you'll use error bars, or probability distributions or something, not significant digits. If the effect your measuring isn't so sensitive then it doesn't really matter, as long as you use a sensible standard.
posted by delmoi at 7:00 PM on February 6, 2006

delmoi: There's no wrong or right about this...

Actually, there is. For example, this statement of yours -- Your accuracy is to the centimeter, and you have one significant digit, '3' -- is flat-out wrong.

discokitty asked how to determine the number of significant figures when making measurements with a ruler. There is a set of rules about how to do this, which means that there is indeed a right and a wrong way of going about it. Regarding this particular question, the book was right, and you and the professor (as well as a bunch of other people) were wrong. This isn't just my opinion, or mr_roboto's opinion, or ozomatli's opinion; it's a fact, according to the definition and rules of significant figures. If you doubt this, check out any of the multitude of links I posted above, or pull out any chemistry textbook. Now I'm not suggesting that discokitty go tell the professor to shove it; I agree (and have said twice already) that it's generally a good idea to do things the way your professor wants. Still, there's something to be said for understanding concepts correctly for one's own sake.

You're right that error bars would be a better way of indicating error than significant figures but I really don't see how that's relevant to the particular question discokitty asked.
posted by purplemonkie at 8:16 PM on February 6, 2006