Errors and significant figures

[Zayan]

Homework Statement

L= 5.7±0.1
B=3.4±0.2
Find area with error limits.

Relevant Equations

∆A/A=∆L/L+ ∆B/B

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as the mean value? Which one?

 

[PeroK]

I'm not sure I know the expected answer, but I'd make the following observation. With that data, the area is between 17.92 and 20.88 units. That's definitely a range of nearly 3 units. It seems valid to say $\pm 1.5$.

 

[Zayan]

I'm not sure I know the expected answer, but I'd make the following observation. With that data, the area is between 17.92 and 20.88 units. That's definitely a range of 3 units. It seems valid to

That's true. But I had studied that we round off to the same no. of significant figures as the data given. I had also studied that we keep decimal places same as the original data. Idk I'm confused lol

 

[PeroK]

That's true. But I had studied that we round off to the same no. of significant figures as the data given. Idk I'm confused lol

My mathematical instinct is satisfied by $19.4 \pm 1.5$.

In this case, reducing the answer to $19$ plus or minus whatever doesn't make much sense.

Say the problem involved buying stuff and you had to have enough. You'd definitely want 21 units and not 20 or 20.5. If 21 is not enough, then the given data was wrong!

That's what error bars are for, IMHO.

 

[PeroK]

PS 20.5 is not an upper limit for the area. It would have to be $19 \pm 2$, assuming the error must be symmetric.

 

[Zayan]

PS 20.5 is not an upper limit for the area. It would have to be $19 \pm 2$, assuming the error must be symmetric.

Then doesn't it add MORE uncertainty than there is? According to that logic shouldn't it be 19.4±1.48 if we're being SUPER precise.

 

[Zayan]

Then doesn't it add MORE uncertainty than there is? According to that logic shouldn't it be 19.4±1.48 if we're being SUPER precise.

Also, I'm not sure whether we should be assuming the mean value to be the arithmetic mean of the upper and lower limits. So rounding off to the original data seems fair.

 

[vela]

The idea behind error propagation is to calculate the uncertainty in the result. Once you know the uncertainty, it tells you where to round to. The rules for sig figs are just general rules of thumb to estimate how the uncertainty propagates without having to do a proper analysis of error propagation. (That means don't use them for this problem.)

Are you sure you're calculating the uncertainty correctly? When you have a formula like $A = LB$, the uncertainty is usually estimated using
$$\left(\frac{\Delta A}{A}\right)^2 = \left(\frac{\Delta L}{L}\right)^2 + \left(\frac{\Delta B}{B}\right)^2.$$

 

[Zayan]

Are you sure you're calculating the uncertainty correctly? When you have a formula like $A = LB$, the uncertainty is usually estimated using
$$\left(\frac{\Delta A}{A}\right)^2 = \left(\frac{\Delta L}{L}\right)^2 + \left(\frac{\Delta B}{B}\right)^2.$$

Area is L*B.
A=L*B
Take natural log on both sides,
lnA= lnL+lnB
Take differential on both sides,
(1/A)dA= (1/L)dL + (1/B)dB.
Thus giving,
∆A/A=∆L/L+ ∆B/B

 

[kuruman]

The way to write the answer formally to $N$ sig figs is using powers of ten in the form $~\text{X.XXX}\times 10^n~$ with $N-1$ digits to the right of the decimal. So 19.38 to two sig figs would be $1.9\times 10^1$.

If you have an uncertainty, you write it as a number to the same power of 10, in this case $0.148\times 10^1.$

When you put the two together, the uncertainty would normally have one sig less than the answer. I say "normally" because there is an exception: if there is a "1" to the left of the decimal the uncertainty is written to the same number of sig-figs as the answer. That's because the dropped sig fig constitutes a considerable percentage of the answer.

So here you would write $(1.9 \pm 0.15)\times 10^1$ which more conventionally is $19\pm 1.5.$

As an additional example, if you changed B to 5.3 keeping everything else the same,
The area is $30.21\rightarrow 3.0\times10^1$
the uncertainty is $1.67\rightarrow 0.167\times10^1$
and put together you get $(3.0\pm 0.2)\times10^1$, conventionally written as $30\pm 2.$

 

[PeroK]

Then doesn't it add MORE uncertainty than there is? According to that logic shouldn't it be 19.4±1.48 if we're being SUPER precise.

It's not a question of being precise. It's a question of the area definitely lying within the range given.

 

[kuruman]

Are you sure you're calculating the uncertainty correctly? When you have a formula like $A = LB$, the uncertainty is usually estimated using
$$\left(\frac{\Delta A}{A}\right)^2 = \left(\frac{\Delta L}{L}\right)^2 + \left(\frac{\Delta B}{B}\right)^2.$$

I had the same question so I looked up some sites and found this
http://www.geol.lsu.edu/jlorenzo/geophysics/uncertainties/Uncertaintiespart2.html
that features both methods.

 

[Zayan]

The way to write the answer formally to $N$ sig figs is using powers of ten in the form $~\text{X.XXX}\times 10^n~$ with $N-1$ digits to the right of the decimal. So 19.38 to two sig figs would be $1.9\times 10^1$.

If you have an uncertainty, you write it as a number to the same power of 10, in this case $0.148\times 10^1.$

When you put the two together, the uncertainty would normally have one sig less than the answer. I say "normally" because there is an exception: if there is a "1" to the left of the decimal the uncertainty is written to the same number of sig-figs as the answer. That's because the dropped sig fig constitutes a considerable percentage of the answer.

So here you would write $(1.9 \pm 0.15)\times 10^1$ which more conventionally is $19\pm 1.5.$

As an additional example, if you changed B to 5.3 keeping everything else the same,
The area is $30.21\rightarrow 3.0\times10^1$
the uncertainty is $1.67\rightarrow 0.167\times10^1$
and put together you get $(3.0\pm 0.2)\times10^1$, conventionally written as $30\pm 2.$

]
Let's say that I had some data with 2 significant figures, and 1 decimal place, eg, 9.2,9.6, etc. Now I calculate the mean value and it comes out to be 9.3. Now since it is the same significant figures, I don't round off. But let's say the error comes out as 0.12. This has 2 significant figures, but it has one extra decimal place. Should I write the answer as 9.3±0.12 OR 9.3±0.1.

 

[PeroK]

The way to write the answer formally to $N$ sig figs is using powers of ten in the form $~\text{X.XXX}\times 10^n~$ with $N-1$ digits to the right of the decimal. So 19.38 to two sig figs would be $1.9\times 10^1$.

If you have an uncertainty, you write it as a number to the same power of 10, in this case $0.148\times 10^1.$

When you put the two together, the uncertainty would normally have one sig less than the answer. I say "normally" because there is an exception: if there is a "1" to the left of the decimal the uncertainty is written to the same number of sig-figs as the answer. That's because the dropped sig fig constitutes a considerable percentage of the answer.

So here you would write $(1.9 \pm 0.15)\times 10^1$ which more conventionally is $19\pm 1.5.$

As an additional example, if you changed B to 5.3 keeping everything else the same,
The area is $30.21\rightarrow 3.0\times10^1$
the uncertainty is $1.67\rightarrow 0.167\times10^1$
and put together you get $(3.0\pm 0.2)\times10^1$, conventionally written as $30\pm 2.$

As I said, my instinct is that the true area must lie within the given range. In this case, area could be 20.8, which is not $19\pm 1.5$.

This is what happened with the blinds for my bedroom window. They were supposed to be whatever $\pm 0.5$ cm. But, they were +1cm more than I specified and they didn't fit and I had to send them back.

It's commercially useless in that context to have an actual size outside the error range.

Don't go into the home furnishings business is my advice.

 

[kuruman]

As I said, my instinct is that the true area must lie within the given range. In this case, area could be 20.8, which is not $19\pm 1.5$.

As you see, the formal way of estimating the error agrees with your instinct and that is why I posted it.

Don't go into the home furnishings business is my advice.

And my advice to your home furnishings supplier is "measure twice, cut once."

 

[kuruman]

]
Let's say that I had some data with 2 significant figures, and 1 decimal place, eg, 9.2,9.6, etc. Now I calculate the mean value and it comes out to be 9.3. Now since it is the same significant figures, I don't round off. But let's say the error comes out as 0.12. This has 2 significant figures, but it has one extra decimal place. Should I write the answer as 9.3±0.12 OR 9.3±0.1.

The mean value of 9.2 and 9.6 is halfway between at 9.4, not 9.3. If the error comes out as 0.12 you should write this as 9.4±0.1. However, if I were making these measurements, I would be concerned with the difference between the two values which is considerably larger than my estimated error and make one more measurement to see where it falls. You should understand, that what follows the $~\pm~$ sign is only an estimate. The idea is that if your measurements follow a normal distribution about the mean, the uncertainty calculated using the standard deviations expression $$\frac{\Delta A}{A}=\sqrt{\left(\frac{\Delta L}{L}\right)^2+\left( \frac{\Delta B}{B}\right)^2}$$ indicates that if you make repeated measurements of the same quantity, about 68% of the values will be within the $~\pm~$ range.

 

[Zayan]

The mean value of 9.2 and 9.6 is halfway between at 9.4, not 9.3.

I also said "etc", implying there are more measurements that I didn't type.

 

[haruspex]

Area is L*B.
A=L*B
Take natural log on both sides,
lnA= lnL+lnB
Take differential on both sides,
(1/A)dA= (1/L)dL + (1/B)dB.
Thus giving,
∆A/A=∆L/L+ ∆B/B

It depends what you mean by the uncertainty.
In many contexts you want the statistical uncertainty, i.e. standard deviation. If X and Y are independent, $(\frac{\sigma(XY)}{E (XY)})^2=(\frac{\sigma X\sigma Y}{E(X)E(Y)})^2+(\frac{\sigma X}{E(X)})^2+(\frac{\sigma Y}{E(Y)})^2$.
For small variances we can ignore the first term on the right, giving @vela's version.

An engineer often cares more about tolerances. If the Deltas are the maximum errors then your formula is the one to use.

 

[Herman Trivilino]

But I had studied that we round off to the same no. of significant figures as the data given.

If that's the convention used in your class then that's what you do.

I had also studied that we keep decimal places same as the original data. Idk I'm confused lol

That is confusing, but when expressing a result in the form $x \pm \Delta x$ both $x$ and $\Delta x$ should be rounded to the same number of decimal places.

So, your instructor would want either $19.4 \pm 1.5$ or $19 \pm 1$.

 

[DaveE]

When I was a practicing EE, none of this ever made sense to me. These problems rarely mention the probability distribution of the data or what the user cares about. If you have a flat density and have to never exceed some value, you'll care about a really different result than if everything is assumed and reported as gaussian, for example.

I think there are many unstated assumptions here, which cloud the reported result. This is a "word problem" not an equation. You do the math on the data set first, then report what the user needs.

Sorry, I know this isn't a helpful response for your problem. Except to say expect confusion in the future. I don't think there are simple rules that are useful IRL.