Sig Figs: When Rules Clash

Factual Questions
1

This was my daughter’s homework, but we think the book’s answer has too many sig figs in it. And, I’m not sure what the correct number of sig figs should be! Here’s the situation: Can someone explain step by step what answer you get when applying the rules to sig figs to the following? A football field is 100.0 yards long, how many miles long is it? (1 yard = exactly 3 feet, 1 mile = 5.280 x10^3 feet).

I’ve reviewed a few textbooks that cover the same rules rather well, but I still can’t quite make sense how to apply the rules here.

Please especially answer:
a) How do you readily determine how many sig figs you must display in your final answer?
b) Since this is a two-step problem, do you honor the sig fig rules after each step? Or, do you somehow carry a bunch of numbers until the last step?
c) How many sig figs are in the scientific notation number (given above)?
d) Does an exact measurement* (with many sig figs) take precedence over other rules that would dictate one should default to least number of sig figs?
e) Isn’t there some rule when multiplying two numbers, you add up the sig figs in both numbers? Or, am I just remembering wrong? (It may be a rule for analyzing mathematical error???)
f) Extra credit! Can someone clearly explanation the difference between “precision” vs. “accuracy”? …Or, a precise number vs. an accurate number? I guess the subtle difference eluded me all these years.

I have to look back at the correct answer, but it carried like 6 digits which just seems like too many. OTOH, I am tempted to argue the correct number of sig figs should be one. I’ll re-post the book’s answer tomorrow. It’s late. Thx! (I’ll be back!)

*An exact measurement may be a defined value, like 1 ft = 12 in. I think it can also be an exact count, like if I say there are 5 apples. It is clear there are exactly 5 apples.

2

For the sum of 1+1, 1.99999999999 is a very precise answer, but not very accurate.

3

Precision is repeatability.
Accuracy is how close the average is to the ideal.

The typical example is target shooting - if all you shots fall widely around the bullseye, you are accurate but imprecise. If your shots have a nice, tight group, but miss the bullseye, you are precise but inaccurate.

4

When I learned it when you multiply you should keep all the digits you get not rounding until the last step, then you should keep as many significant digits in factor with the least number of significant digits. However, exact numbers have an infinite number of significant digits. So the 3 in feet/yard and 5280 in feet/mile have an infinite number of significant digits. One might say that a football field is defined as 100 yards so that is exact as well, but in this case you’re to assume it’s measured as 100.0 to tenth of a yard . That’s four significant digits so I’d say the answer should have four.

5

However, in this case I believe rule based methods are wrong. There is only one measurement so you know the football field is between 99.95 and 100.05 yards as any number in that range would round to the given 100.0. Therefore multiplying by 3/5280, which is an exact number, the football field is between 0.05678977… and 0.05684659… miles so we only know 0.0568 none of the rest of the digits are significant. This is three significant digits, as the lead zeros don’t count.

6

I believe the rule is your final answer can have the same number of significant figures as the lowest amount of significant figures in any of the number you derived it from. You derived your answer from three numbers:

3.00000…
1.000 x 10^2
5.280 x 10^3
(The zeroes are significant because they’re the final digit.)

So your final answer can have three digits after the decimal point.

The football field is 100.0 yards long. (1.000 x 10^2) x 3.00000… equals 3.000 x 10^2 to three significant figures.

5.280 x 10^3 feet equal 1 mile. 3.000 x 10^2 / 5.280 x 10^3 equals 5.682 x 10^-2 to three significant figures.

I can’t believe I’m trying to do both significant figures and scientific notation from memory at one o’clock in the morning.

7

What answer did they give? 0.056818? 0.05682?

Using scientific notation allows us to specify how many sig figs easier.

5.681 X 10-2 has four sig figs, which is how I would have answered it, as 100.0 yards has four sig figs.

I majored in EE, but this was sometime in the dark ages, so I’m not as familiar with it as I used to be. I’m sure someone will come along and correct me.

Here’s a web site which explains the rules, and gives examples.

http://www.usca.edu/chemistry/genchem/sigfig.htm

Usually by the least number of sig figs in the given data. The two definitions, 3 ft = 1 yard and 1 mi + 5,280 feet actually have an infinite number of sig figs, since they are definitions and not measured.

Yes, that give the most accurate answer.

For measurements and not definitions, it would be three. This is one reason to use scientific notation is that it allows you to specify the number of sig figs.

Yes, but it’s a definition and not a measurement. Treat it like it has an infinate number of sig figs.

No, you take the least number.

Like **beowulff **says. Here’s a diagramwhich explains it.

8

The rules are simple and easy to find.

the book states the field is 100.0 yards long. Thus, when they measured out this field, they were sure of their accuracy to 1/10 yard, 5 digits. Yes, there are exactly 100 yards in 1 field, but when actually laying out that measure, then accuracy comes down to precision. the chalk lines they laid out for a real-life field are +/- 3.6 inches, measuring from same place (middle of?) each end chalk line.

100.0 yards is 0.05618 miles, or 5.618x10^-2 miles maintaining the same number of significant digits.

Since most work today is done by calculator or computer, rounding after the numbers have crunched is usually the most appropriate way to round.

What you are actually asking about is cumulative errors.
Every measurement ha an associated confidence or error range.
Yes, certain conversion factors -yards/mile, or counts - 5 apples - have an effective error range of zero.
By default, the error is presumed to be half the smallest scale unit of the measurement device; ie. a metric ruler is in mm, so the reading error is 0.5mm; but typically, you measure both ends, so the reading error is double that (1mm.) In other cases, the measuring method determines accuracy.
Actual error in the error in same units - i.e. 100.0 yards, +/- 0.05
Relative error is a percentage or fraction.

When adding or subtracting, add the relative error.
“If I give you $10 give or take a quarter, and then give you $20 give or take a dime, you have $30 give or take 35 cents.”
However, the error can only have 1 significant digit so $30 give or take 40 cents.
The answer cannot be more accurate than the error amount.
You can’t say $42.95 give or take a dollar. the 5 is totally meaningless, and 9 most likely too.

When multiplying, add the relative errors.
“I’m giving you $10 give or take a quarter. I’ll do this 20 times, give or take a time.”
So 10.00+/-0.25 x 20+/-1

Relative errors
1/40+1/20=3/40

10x20=200

200x3/40=15
So I end up with $200+/- $15.

Calculating resultant error from initial inputs is the most important first term lesson in the practical physics lab I used to be a tutor for.

9

If you have Length x Width, and each measurement is 5% error, then error in area can be 10%. You add the error margins together.
Thats the answer to F as well. There’s the accuracy you are are writing a value, which is the result of the calculaton but there is ALSO the error margin that results from the method of measurement and/or calculation … which is a seperate calculation.

In your simple case, the conversion from yard to feet to miles introduces no further error, except if you use too few decimal places along the way, so the original error margin remains… well it wasn’t specified nor asked for.

University level science (and some engineering) may ask for error margins to be calculated.

10

Absent other indicators, the error is implied in the number of significant digits supplied.

If the football field (a measured distance, not an abstract) is specified as 100.0 yards, the implied error is +/- 0.05 yards, since this is the number given. If the rounding had been for a measured 100.05 or more, the measurement would be specified as 100.1; 99.949 or less, the result would have been 99.9; since 4 digits accuracy is what is supplied, your answer should have 4 significant digits.

This is where scientific notation is nice, since without a decimal, trailing zeroes may not be significant. Without the decimal point, we would have no idea if the field was measured with a precise tape measure (accurate to the nearest quarter inch, say) or by pacing it off (10% error, say). A field of 1.0x10^2 and a field of 1.00x10^2 imply different precisions.

As mentioned, conversions and constants have relatively zero error, since you can use them with more precision than any other imprecise number in your calculation. When working with 4 digits, use Pi=3.14156265 and the error in your Pi specification will be of no significance. (This is what your calculator does). integer specifications (5280 feet in a mile) have zero error.

However, if you are working with 4 or 5 digits, using Pi=3.14 (or 22/7) means the calculation now only has 3 significant digits.

11

I agree with your number of significant digits but I think you transposed the last two digits. I get



     __
0.05681

rounded to 0.05682.

12

If you’re using a calculator, you let it keep however many digits it has, and only round at the end.

If you’re doing this by hand, it’s not always practical to keep all the digits. Multiplying five four-digit numbers would give you twenty digits before rounding. You can instead keep a “guard digit” or two. In your example with four significant digits, you would keep five or six digits of accuracy in the intermediate calculations, and round down to four at the end.

Wikipedia has an article on Guard Digits, but it kind of sucks (it does warn that it’s a stub).

13

Only OldGuy did it the same way as I did and I agree with his answer of 0.0568. But I woiuld like to make a point here. How many significant digits are there in 100.0? Well it looks like 4 but the number of sig digs in 99.9 is only three. But they are practically the same number. Well sig digs is only an approximation. Let me just say that 99.9 has a robust 3 sig digs, while 100.0 has a very shaky 4. In any case, the final answer to the question has only 3 as both OldGuy and I found.

The only robust way to do this kind of problem is to use error intervals and something called interval arithmetic. I don’t go into detail, but in this case it was greatly simplified by the fact that 2 of the numbers were exact by definition.

14

I remember this rule from my chemistry class, which almost never comes up: The only numbers that are “rounded from” are 1-9.; 1-4 are rounded down. If you always round 5-9 up, then you are rounding up 55% of the time, which skews reported numbers upward. So if the last digit is a 5 you always round to the even number – that is, 99.95 and 100.05 both round to 100.0, while 101.05 rounds to 101.1.

15

Of course, any rule concerning rounding fives is very rarely relevant, anyway, since you usually have more digits. You can debate how to round 0.5, but there’s never any question about how to round 0.5137 or whatever.

16

I would argue that it is both precise and accurate. A better example for 1+1 might be 97.2937579101. That answer is very precise and not very accurate.

For the OP: Accuracy measures how far your answer is from the right one. Precision measures how well you know the answer you have (regardless of how right or wrong it is).

I could say I’m 80 inches tall. That is more accurate than saying I’m 30 inches tall. But saying I’m 30.7845 inches tall is more precise than saying I’m 80.2 inches tall.

17

Rounding in mathematics or sciences is an epistemology issue - what *do *we know, and what *can *we know?

But my field is engineering, which I teach at the university level. In engineering, we are generally dealing with real measurements which can be reasonably obtained with real engineering tools, and with manufacturing methods which can produce products to real tolerances. And our rounding rules are based on the physical constraints of those realistic technologies. Naturally, these are different than the epistomological rules of pure sciences and math.

Most of our current textbooks use these recommendations: For numbers starting with a 1, assume that there are four significant digits. For numbers NOT starting with a 1, assume that there are three significant digits. So 100 meters will be assumed to mean 100.0 meters, and 23,000 Newtons will be assumed to mean 2.30x10^4 Newtons.

I also teach my students to NEVER round down until they are ready to present their absolutely final answer. I wasn’t familiar with the term “guard digit” before reading this thread, but it perfectly describes what I try to instill in my students. Only I tell them to use as many guard digits as their calculators will hold, until they are ready to present their final answer.

Oh, and anyone using 22/7 for Pi automatically loses five points! That may have saved time in Isaac Newton’s day, but nowadays it is completely unsupportable.

18

355/113 for the win!

19

Really, there’s no reason to use any fraction for pi any more. It’s easier to just type in digits, and to include as many digits as you need. Or better yet, just hit the pi button on your calculator and get about ten digits with almost no effort.

Alternately, if you’re just estimating something in your head, just use 3.

20

For a fractional approximation of Pi, I just always used 2*pi/2. When that didn’t work, I decided to use C/(2r).