# Tag Info

1

Just a guess, but perhaps he actually said (or meant) "1 cubic centimeter". Increasing the rod length by 10% is not the same as adding 1 cubic centimeter (because the rod is circular, not square, in the other dimensions). Also assuming the diameter of the rod is 1cm.

4

If the 10 cm was a measured value with a corresponding error estimate, then the results would be different. Suppose the bar was $10 \text{ cm} \pm 0.10 \text{ cm}$. Adding exactly $1 \text{ cm}$ to the length of the bar would yield $11 \text{ cm} \pm 0.10 \text{ cm}$, whereas increasing the length of the bar by 10% yields $11 \text{ cm} \pm 0.11 \text{ cm}$. ...

3

$10\%$ of $10\text{ cm}$ is $1\text{ cm}$. $10\text{ cm}\times\frac{10}{100}=1\text{ cm}$. I really don't see how this is a 'nit-pick', or 'relative' vs 'absolute', as suggested in the comments. Are you absolutely, $100\%$, or at least relatively, sure you got the quote right? (There are very close variations of it, where indeed it would be rather stupid ...

2

One thing would be that it's not practical to use '%' in that example. I can easily imagine metal worker going nuts if whole 'tech spec' is in ratios. Aftervards you can't just decrease by 10% and get the same original length. Or maybe because he didn't specify that it's 10% of original length. I have this 'dejavu' feeling that I have heard something ...

2

A chi-squared test is used to compare binned data (e.g. a histogram) with another set of binned data or the predictions of a model binned in the same way. A K-S test is applied to unbinned data to compare the cumulative frequency of two distributions or compare a cumulative frequency against a model prediction of a cumulative frequency. Both chi-squared ...

1

Do the calculation at the computer/calculator precision then quote result with the appropriate number of significant figures. Yes that is correct procedure. Otherwise you could introduce rounding error.

1

Letting $\mathbf{F}$ and $\mathbf{F}^{-1}$ be the forward and inverse discrete Fourier transform, the cyclic autocorrelation of a signal $A$ is given by $$S(A)=\mathbf{F}^{-1}\left[\mathbf{F}(A)\mathbf{F}(A)^*\right].$$ Let the low-passed signal $A_L$ be $$A_L=\mathbf{F}^{-1}\left[\mathbf{F}(A)\mathbf{F}(L)\right]$$ where $L$ is the low-pass filter in the ...

0

I remember reading somewhere that the problem of exact time-keeping on ships could have been solved a lot earlier than it was if somebody would have had the idea of keeping time with a whole array of imprecise clocks - taking the average of clock-times would have given a precise time. Define precise . Precise for timekeeping in ships would be accuracy ...

0

You can get a more accurate measurement from multiple measurements. But each measurement has room to damage the target, and you would not get enough measurements to get to atomic standards. Most likely, you could get to micron measures over 1000000 people, assuming each measurement is to the nearest mill. But as likely, most readers would report 1234.12 ...

2

This is highly unlikely. It comes down to bias and variance. Individual people of course will estimate with limited accuracy, whether just guessing, eyeballing, or using latest and greatest measuring technology. By itself that would not be a problem if people were unbiased estimators and their estimates were independent. All errors would then be variance ...

4

This will not work. I'm going to use the standard error of the mean as the measure of the precision: $\mathrm{SEM} = \sigma_x / \sqrt{N}$. $N$ is the number of people you have make estimates of he length, and $\sigma_x$ is the standard deviation of the estimates that everyone makes of the length. The standard deviation of the sample is given by the square ...

0

I believe the answer is no. Let's simplify the question a little by limiting the number of persons doing the measurements to 1. Of course if you show the same stick over and over again and if the person knows she is shown the same stick, she will be making only one measurement. This can be avoided by sampling different length sticks, but the problem will ...

5

No, of course not. Yes, some people will overestimate and others will underestimate. Averaging would cancel out the bias to some extent, but there's no reason to expect it to cancel out the bias perfectly. We all have similar eyes and brains. We are all deceived by the same optical illusions, in the same way. We all have a shared cultural understanding of ...

0

The people making measurements don't know what the precise answer is so that they can make 'imprecise' measurements. I think you need to understand what 'error' exactly means. Read this explanation: A measurement may be made of a quantity which has an accepted value which can be looked up in a handbook (e.g.. the density of brass). The difference between ...

0

I guess this depends on what exactly we mean by "estimate". If estimate means making up some number that's one thing. But if it implies some kind of [visual] measurement that's another thing. It may be difficult to think about a human being producing any good "measurement" visually, so let's ask a question instead: Can an accurate measurement of an object be ...

6

I believe you are thinking of the Central Limit Theorem. The mean and variance of the averages of many measurements are better estimates of the precision of your measuring rule, but don't tell you anything about the accuracy of your measuring rule. Your measuring rule may be biased. The Central Limit Theorem is a part of mathematics. IMO you should also ...

7

No because none of them know the actual answer. The averaging process you describe only works if each estimate is of the exact answer plus noise. Otherwise it is known as the "Emperor's nose" problem. Nobody can see the Chinese emperor's face so they ask a million peasants how long his nose is, they average the results, and since they have such a large 'N' ...

0

For the Benfords' law to apply, one needs a set of unrelated numbers that span many many orders of magnitude. The "natural" comparison scale for these very different numbers turns out to be the logarithmic scale. If the numbers are really unrelated, then you can use Pascal's principle of indifference to say that the probability to have a given number in the ...

0

Notice that the Fermi-Dirac integral comes from the use of the Fermi-Dirac distribution. Another way to define a system degenerate or non degenerate is to compare the Fermi temperature $T_{F}$ to the actual temperature $T$ of the fermionic system under the study.

1

The reason is the following. If you write down the formula for Fermi-Dirac distribution, you can see that when x is negative, expanding the Fermi-Dirac distribution function using Maclaurin series, you will get the Boltzmann distribution which holds for a classical gas. So it means that your gas is nondegenerate.

0

I think you would underestimate the error if you used that calculation. For example, if you had a thermometer that the manufacturer said was accurate within 1 degree, and you took a reading every second for 100 years to calculate the average temperature in Chicago, the uncertainty in the average could still be about 1 degree, because the thermometer could ...

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