Writing a result from a table and how many digits should the answer have?

[Karagoz]

Homework Statement

[/B]
A 0.02 meters long conductor is placed perpendicular in a homogen magnetic field.

Current through the conductor is changed. And we have measured the data here:

How can we write the magnetic flux density the correct way?

How do I know how many digits my answer should have?

Homework Equations

F=I*l*B

The Attempt at a Solution

After calculating the B (F/(I*0.02)) one by one from the table above (e.g. like: 1,1 / (1,0*0,02) = 0,055 ), I got these numbers:
B (magnetic flux density):
0,055 (min)
0,06
0,0625
0,062
0,056666667
0,062857143 (max)

Difference between min and max divided by 2: (0,0628-0,055)/2 = 0,0039 ≈ 0.004

min + 0.0039 = 0.055 + 0.004 = 0.059 = 59 mT

B = (59 ± 4) mT

 

[kuruman]

Your input values are to 2 significant figures, so your answer should also be to two sig figs which means that 59 is correct. However, I would take the mean or average as opposed to using half the difference between min and max. What if you have a single outlier on the high side, say around 80 mT? It's a good bet that your uncertainties follow a normal distribution; you have enough data points to calculate the standard deviation from the mean as an estimate of your uncertainty.

 

[Karagoz]

Your input values are to 2 significant figures, so your answer should also be to two sig figs which means that 59 is correct. However, I would take the mean or average as opposed to using half the difference between min and max. What if you have a single outlier on the high side, say around 80 mT? It's a good bet that your uncertainties follow a normal distribution; you have enough data points to calculate the standard deviation from the mean as an estimate of your uncertainty.

The means is:
(0,055 + 0,06 + 0,0625 + 0,062 + 0,056666667 + 0,062857143) / 6 = 0.0598373 ≈ 0.060

B = (60 ± c) mT

But how then do I put the correct number as in c?

The correct way is to put the standard deviation as c?

Half the difference between max and min is: (0.062857143 - 0.055) / 2 ≈ 0.004.

But adding 0.004 to 0.060 is 0.064. And 0.064 is +0.001 more than the max value which is ca 0.063.
And subtracting 0.004 from 0.060 is 0.056 which is also 0.001 more than the min value which is ca 0.055.

 

[kuruman]

The correct way is to put the standard deviation as c?

Yes. Be sure to keep only 1 sig fig.

But adding 0.004 to 0.060 is 0.064. And 0.064 is +0.001 more than the max value which is ca 0.063.
And subtracting 0.004 from 0.060 is 0.056 which is also 0.001 more than the min value which is ca 0.055.

So what? The ± uncertainty does not mean that all values will be within a range. It means that statistically, if your measurements are independent of one another and if they obey a normal (Gaussian) distribution, which is very likely in this case, then about 68% of the measurements are expected to be within the ± one standard deviation range.

 

[haruspex]

If you want to be as accurate as possible with this, you should take into account that the same range of error on each datum does not mean the same fractional error. E.g. for 1,0 it is ±,05, so ±5%, but the same range for 3,5 is ±1,4%.
If it were just a matter of averaging such a set of numbers that would not matter, but the division complicates this. The results of the divisions have different absolute ranges.
Thus, one should use a weighted average, reflecting the relative uncertainties.

Edit:
On second thoughts, it looks like the I/A values are set by the experimenter, so are probably known much more accurately than the two sig figs suggest. Since these supply the denominator in the divisions, the division does not really create a complication.

 

[Karagoz]

So when writing (A ± c), the A is the "mean", and c is the standard deviation.

This is what's shown as "solution to a physics problem" to a website of a high-school physics book.

I and F are taken from a table, B values are calculated (length of the conductor is 0.0500 m).

They first find the mean, which is (27.00 + 31.0 + 29.3 + 32.0 + 27.6 + 29.0 + 30.6) / 7 = 29.5 ≈ 30

Then they just find the (1/2)*(B_max-B_min) which is: 0.5*(32.0 - 27.0) = 2.5 ≈ 3

They write: B = (30 ± 3) mT

Is it correct to calculate that way? Or should we use standard deviation?

But when taking the standard deviation, what I got is:

√(((29.5-27.00)^2+ (29.5-31.0)^2 + (29.5-29.3)^2 + (29.5-32.0)^2 + (29.5-27.6)^2 + (29.5-29.0)^2 + (29.5-30.6)^2) / 7) = 1.68 ≈ 2

so B = (30 ± 2) mTMy other question is: it's only in the final result we count the significant digits?

One of the values of B (as seen on the table above) is 29.3,
and is calculated that way: B = mF/(I*l) = 0.88 / (0.60*0.0500) mT = 29.3 mT

The I has only 2 significant digits: 0.60

So shouldn't the value of B also have only two significiant digits, which would be just: 29 mT?

 

[haruspex]

Is it correct to calculate that way? Or should we use standard deviation?

Neither.
Even if the same scatter persists, the more datapoints you have the more precisely you know the true value. Taking max - min is especially wrong since the more samples you have the greater that difference is likely to be.
If your sample of n datapoints has a standard deviation σ then the standard error of the mean is approximately σ/√n.
See https://en.m.wikipedia.org/wiki/Standard_error

 

[kuruman]

If your sample of ##n## datapoints has a standard deviation σ then the standard error of the mean is approximately σ/√n.
See https://en.m.wikipedia.org/wiki/Standard_error

Here is where I get confused. The definition for the standard deviation of ##n## measurements is $$\sigma= \sqrt{\sum_{i=1}^{n}\frac{(x_i- \bar x)^2}{n-1}}.$$ The wikipedia article linked in #7 states "The sampling distribution of a population mean is generated by repeated sampling and recording of the means obtained. This forms a distribution of different means, and this distribution has its own mean and variance." Here, OP has sampled and recorded only one mean, so shouldn't ##\sigma## as calculated above be the standard error of the mean? In other words, in the equation SEM ≈ σ/√n, n = 1 because there is only one observation of the mean.

 

[haruspex]

Here is where I get confused. The definition for the standard deviation of ##n## measurements is $$\sigma= \sqrt{\sum_{i=1}^{n}\frac{(x_i- \bar x)^2}{n-1}}.$$ The wikipedia article linked in #7 states "The sampling distribution of a population mean is generated by repeated sampling and recording of the means obtained. This forms a distribution of different means, and this distribution has its own mean and variance." Here, OP has sampled and recorded only one mean, so shouldn't ##\sigma## as calculated above be the standard error of the mean?

The formula you quote is the unbiased estimator of the variance (σ²) of the population. (It is not actually an unbiased estimator of the standard deviation, but it is hard to do any better.)
In the context of the thread, it tells you what spread (as standard dev) to expect if you were to do vast numbers of measurements.

What the wikipedia article is saying is that if you were to take M batches of N measurements (samples), you could generate M estimates of the mean, one from each batch. If you were then to look at the variance in these estimates you would find that it is approximately σ²/N, where σ² is the variance of the underlying population. Note that M does not feature. But that is just a thought experiment for calculating the standard error in the mean as the square root of that, i.e. σ/√N. You do not actually need to take M batches of N.
So we use the formula in post #8 to approximate σ, then divide by √N to gauge the error in our estimate of μ.

In other words, in the equation SEM ≈ σ/√n, n = 1 because there is only one observation of the mean.

Every measurement is an observation of the mean.

 

[Karagoz]

Neither.
Even if the same scatter persists, the more datapoints you have the more precisely you know the true value. Taking max - min is especially wrong since the more samples you have the greater that difference is likely to be.
If your sample of n datapoints has a standard deviation σ then the standard error of the mean is approximately σ/√n.
See https://en.m.wikipedia.org/wiki/Standard_error

So we should write that way:

(Mean ± (σ/√n)) ?

σ: standard deviation
n: number if samples

Maybe ½(max - min) used in the high school book to keep it simple and easy for that level.

 

[haruspex]

So we should write that way:

(Mean ± (σ/√n)) ?

σ: standard deviation
n: number if samples

Yes.

Maybe ½(max - min) used in the high school book to keep it simple and easy for that level.

That is a crude way of estimating the standard deviation in the (theoretical) population of infinitely many samples, but it is completely wrong for estimating the error in the calculated mean from n samples.
As I noted, it would tend to increase the more samples you took!
½(max - min) /√(n-1) might be ok. The -1 is needed. Consider taking only one sample. ½(max - min) = 0, whereas you really have no idea what the error is. ½(max - min) /√(n-1) = 0/0, which correctly reflects the utter uncertainty.

 

[haruspex]

Something else that needs clarifying.

There are two sources of error here. There is the experimental error in taking the measurements, from whatever cause, and the rounding error in expressing the readings to only two significant digits.
(I thought there was a further major complication from the division, but see my edit to post #5.)

My post #9 is in respect of handling a batch of readings suffering only from experimental error. The theory assumes the underlying distribution of these numbers is roughly Gaussian and the numbers are independent and identically distributed ("iid").

If, in fact, the F/mN readings (to unlimited precision) are Gaussian, iid then the rounding to two digits changes the distribution.

If the underlying Gaussian has a std dev much less than 0.05 then the rounding wipes that out and replaces it with a uniform distribution [-0.05,+0.05]. At the other extreme, if the Gaussian std dev is a lot more than 0.05 then the rounding has little effect. In between the two extremes it gets complicated.
If you were to plot the data on a graph you would use the rounding error, ±0.05, to set the error bars.

The division is still troublesome in this regard: the division also divides the uncertainty, so now each ratio has a different error range.
To cope with this, you should use regression analysis to find the slope rather than averaging the individual ratios. But you would want to fix (0,0) as a required intercept for the graph. I.e. you calculate the slope as ##m=\frac{\Sigma y_ix_i}{\Sigma x_i^2}##. This gives a mean of 60.65, not 59.

The standard error in the slope is ##\sqrt{\frac{\Sigma(y_i-mx_i)^2}{(\Sigma x_i^2)(n-2)}}##, which, after the division by 0.02, is about 1.37.
See e.g. http://www.statisticshowto.com/find-standard-error-regression-slope/ and https://en.wikipedia.org/wiki/Simple_linear_regression#Normality_assumption.

Note that the standard deviation of the errors y_i-mx_i is around 0.32, which dwarfs the ±0.05 rounding error. This means the 2-sig figs rounding error is irrelevant.

Putting all this together, you would be justified in quoting a result of 60.6±1.4.

 

[kuruman]

Every measurement is an observation of the mean.

That's what I've come to realize and where I had a blind spot. Thanks for pointing it out. I have another question regarding the following statement.

But you would want to fix (0,0) as a required intercept for the graph.

Would it not be better to do two calculations, one with floating intercept and one with zero intercept and see if it makes a difference? Clearly, the theory predicts zero force for zero current, but there could be a systematic error that the non-zero intercept would be able to pick up.

 

[haruspex]

Clearly, the theory predicts zero force for zero current, but there could be a systematic error that the non-zero intercept would be able to pick up

In general, yes, but here I rather assumed it would have been evident if there had been a force before any current was applied.

 

[kuruman]

In general, yes, but here I rather assumed it would have been evident if there had been a force before any current was applied.

Without knowledge of the experimental design, I was thinking more along the lines of opposition to the magnetic force (friction?) in the low current range. In any case, I agree that there is no evidence for an intercept here which is easy to see that

 

[haruspex]

I was thinking more along the lines of opposition to the magnetic force (friction?) in the low current range

Yes, it depends how the force is measured.