Simple Significant Figure Problem

[BillhB]

Homework Statement

Convert $120. \frac{km}{h}$ to $\frac{m}{s}$

Homework Equations

$1 km=10^3 m$
$1 hour = 3600 seconds$

The Attempt at a Solution

Not so much concerned with the solution, it's easy, as to what happened after the solution was found. So you get $33.33\overline{33}$, after which the professor rounded to 33.3 citing significant figures. That makes no sense though, if we go back in the other direction we would see this new value is actually $119.88\frac{km}{h}$ which is more precise than the original value. If you used this value in another calculation it seems like it would defeat the entire purpose of doing the whole silly significant figure thing in the first place.

Now you could just cite significant figures again and get back to 120., but if you wanted to convert the value into use again at a later date you'd have rounding error. The chemistry lecture professor basically brushed me off when I brought it up. I was arguing that the entire $33\frac{1}{3}$ was significant.

Who's right? Any insights here?

 

[haruspex]

I believe the decimal point shown in "120." says the 0 is a significant figure (not sure how it works if it had been 1200 with three sig figs). This means you should take the exact value as being anywhere in the range 119.5 to 120.5. On dividing by 3.6 that gives 33.2 to almost 33.5, so even quoting 33.3 (implying 33.25 to 33.35) is overexact. But hey, there's only so much you can do without quoting specific bounds.
Give your prof a break.

 

[David Lewis]

...if we go back in the other direction we would see this new value is actually 119.88 km/h which is more precise than the original value.

I believe you add one or two more significant figures when you run the computation backwards.
(e.g. 33.333) That prevents imprecision from compounding.
But you still round the final answer off to 3 significant figures when you're done.

 

[BillhB]

I believe you add one or two more significant figures when you run the computation backwards.
(e.g. 33.333) That prevents imprecision from compounding.
But you still round the final answer off to 3 significant figures when you're done.

This ain't really helpful.

I believe the decimal point shown in "120." says the 0 is a significant figure (not sure how it works if it had been 1200 with three sig figs).

Probably use scientific notation? I would guess.

This means you should take the exact value as being anywhere in the range 119.5 to 120.5.

I should have said so, my bad, but the class generally assigns the last digit as the uncertain one and puts the error bounds at $\pm 1$, so 119 to 121 km. That's another convention that doesn't seem very logical to me. If I were to use a measuring device with increments in which I would guess at a whole kilometer, I don't see how I could possibly be an entire kilometer off in either direction. This seems like it would true for any measurement that's around everyday units. $\pm .5$ makes a lot more sense. End rant.

On dividing by 3.6 that gives 33.2 to almost 33.5, so even quoting 33.3 (implying 33.25 to 33.35) is overexact. But hey, there's only so much you can do without quoting specific bounds.

So on dividing by 3.6 we would get $33+\frac{1}{18}$ to $33 + \frac{11}{18}$, so I just think we should keep the $33 + \frac{1}{3}$ and specify a new uncertainty of $\pm\frac{5}{18}$ to propagate both the precision and uncertainty correctly of the original measurement.

Would this be right?

Give your prof a break.

It would be easier if the professor didn't say other things which made no sense, like accuracy of measuring devices isn't important, only precision is. She went on to say they pay 2000 dollars for each digit after the decimal, and I told her I would sell her a scale that had $10^{whatever}$ precision for 2 grand a digit, but wasn't accurate at all if that's what she thought. Maybe she brushed me off because I came across like a dick, but I only wanted to illustrate the silliness of the statement.

Last semester the professor in the same department said similar things, but was actually worse. He gave me a F on one exam because I didn't round after each step in a calculation. I tried to explain that's not how it works, and eventually had to go to the dean. He hated me the rest of the course, possibly because I made him look like an idiot in front of the dean.

I don't think he was an idiot, he had a Ph.D in chemistry, I just don't feel like they care or think we as students are too dumb to learn things the correct way. Maybe we are, but try me first before preaching something that seems error ridden.

I just want to learn things the correct way, it's frustrating when none of my science courses can agree on how to handle the math in science. At least the physics courses generally treat this stuff better, it makes me question my desire to be a doctor and continue with this chemistry department. end second rant.

 

[haruspex]

He gave me a F on one exam because I didn't round after each step in a calculation.

That prof gets an F from me. I regularly have occasion on these forums to point out that rounding prematurely leads to accumulation of errors. By the time you're done you might have no precision left.

keep the $33+\frac{1}{3}$ and specify a new uncertainty of $\pm\frac{5}{18}$

You should keep it in decimals. Taking the original to be +/-1 (ugh) that gives 33.3+/-0.3.

 

[BillhB]

You should keep it in decimals. Taking the original to be +/-1 (ugh) that gives 33.3+/-0.3.

Got it, thank you so much for the help and for letting me vent.

 

[lychette]

That prof gets an F from me. I regularly have occasion on these forums to point out that rounding prematurely leads to accumulation of errors. By the time you're done you might have no precision left.
This is a rather rash statement ! Can you quote some of your regular occasions when you have pointed this out (save me searching)
Do you mean precision or accuracy?...is it better to be precise or accurate?
'Insignificant figures never become significant figures'

 

[SammyS]

I believe you add one or two more significant figures when you run the computation backwards.
(e.g. 33.333) That prevents imprecision from compounding.
But you still round the final answer off to 3 significant figures when you're done.

This looks to me like it would be very helpful.

 

[BillhB]

This looks to me like it would be very helpful.

It just wasn't useful to the question that I was asking. I had already said in the post that you can get back to 120. using significant figures again.

The question was more along the lines as to what the proper answer would be that maintained the precision of the original measurement. The rounding to the significance of the original measurement bothered me for that reason, because it was technically more precise than the original measurement with identical error bounds... $\pm1$ That's why I thought it best to just give the fractional answer with fractional uncertainty that covered the entire range and maybe with a caveat that either gave the original significance of the measurement or the original measurement itself.

Practically, if I were going to use the value in the future in another calculation which would include going back to the original measurement I would just use the entire fractional value. Rounding at the end.

In the end, if the question is just a conversion type problem, I'll keep it in decimals, rounded to the significance of the original measurement, with the best new uncertainty as possible.

 

[haruspex]

Practically, if I were going to use the value in the future in another calculation which would include going back to the original measurement I would just use the entire fractional value. Rounding at the end.

Yes, that is the right approach - or, at least, to keep a few extra digits of precision until the end.
You should also be aware that when intermediates are irrational you may be forced to round them, and this might still lead to less precision than you think. E.g. a careless attack on π-22/7.