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gonnis

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- Thread starter gonnis
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gonnis

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I think it also depends on the stated precision of the answer relative to the accuracy of the measurements.

BUT ... take all of that with a grain of salt since I don't really remember much of my measurement theory course that I took about 50 years ago and I'm perfectly willing to just make stuff up

- #3

gonnis

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- #4

Stephen Tashi

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if the weighing scale is known to be accurate to say, +/- .01g, then does the mean of repeated measurements = more accuracy?

You've described a situation, but the description is not specific enough to be a mathematical question. To be a mathematical question, you'd first have to define what you mean by "more accuracy".

For example, if we are considering errors to be "random" in some sense then there are no mathematical guarantees about accuracy as a deterministic quantity. Applying mathematics would only produce statements about the

- #5

nsaspook

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http://www.ncsu.edu/labwrite/Experimental Design/accuracyprecision.htm

http://meettechniek.info/measurement/accuracy.html

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I would think about this in the following sense. There are two kinds of errors associated with measurements (I) random errors and (II) biases.

Random errors can influence your result in either direction and when the measurement is repeated, they will take on different values.

A bias is an error that, on repeated measurements will always influence the result in the same direction with the same magnitude. This might be the case if, in your example, the scale is incorrectly calibrated and always results in a measurement that is 2 g high of the true value.

You probably already know that if you measure a value many times and plot your results in a histogram, that histogram will likely have a bell-shape to it (or a normal distribution). This is because for the case of many, small potential random errors (as is commonly encountered in the real world) some will push the result up and others will push it down. In some rare, extreme cases all the random errors will point in the same direction putting you out on the tail of the distribution. More often, most of the errors will cancel out leaving you with a result that's closer to the true value.

Mathematically you can show that the mean value of this distribution is actually your best estimate for the real value of the parameter you are measuring. In fact, if you were to make multiple sets of N measurements, that would give you a set of mean values, those mean values will be normally distributed with a width that is characterized by the standard deviation over the root of those N measurements. (This is probably the answer that you're looking for - that the uncertainty in your best estimate of the true value is inversely proportional to the root of N - therefore more measurements improves your answer).

But it's important to remember that this distribution will not resolve any bias in your apparatus or measurement technique. In that sense, your best estimate will never be perfect, because it cannot eliminate bias.

- #7

gonnis

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If the tip of the bell in a bell-shaped histogram is the best estimate of the true value, then doesn't the accuracy (the difference between measured value and actual value) of the average of all your measurements increase as the data points increase? So that if you had infinite amount of data points it would tend toward some value, which (excluding biases) would be the actual value? thanks for any more helpMathematically you can show that the mean value of this distribution is actually your best estimate for the real value of the parameter you are measuring. In fact, if you were to make multiple sets of N measurements, that would give you a set of mean values, those mean values will be normally distributed with a width that is characterized by the standard deviation over the root of those N measurements. (This is probably the answer that you're looking for - that the uncertainty in your best estimate of the true value is inversely proportional to the root of N - therefore more measurements improves your answer).

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- #9

gonnis

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- #10

nsaspook

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90% would be extremely bad accuracy.

http://en.wikipedia.org/wiki/Calibration#Standards_required_and_accuracy

- #11

homeomorphic

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Assuming independent trials. This might not always be the case in the real world because maybe you are causing wear and tear on your measuring device. Of course, that's why their are companies and institutions that deal with calibration, standards, and so on.

- #12

Svein

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Well, that is really a long lecture. Let us just say that as long as you are doing

- The standard deviation gives you an estimate of how precise your measurements are (if you have a large standard deviation, your mean can be far off the "true value").
- If you do repeated measurements of things that change over time (like temperature), the calculated mean does
*not*give you a better estimate of the temperature at a given point in time.

- #13

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- The standard deviation gives you an estimate of how precise your measurements are (if you have a large standard deviation, your mean can be far off the "true value").

- #14

gonnis

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thanks... If I may use a different example then.. an automated blood pressure cuff. The repeated measurements are all automated so biases are less. If just one automated measurement is 99% accurate to the true blood pressure, then wouldn't the mean of 60 automated measurements over 60mins beLet us just say that as long as you are doingrepeated measurements of the same thing, themeanwill give you a better estimate of the "true measurement".

also phinds I get the distinction b/w accuracy and precision (if I read your comment correctly) thanks

- #15

Svein

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No, you can have a very small standard deviation around your measurements and be WAY off in accuracy if you measuring device is precise but inaccurate. A different device making the same measurements could give a larger standard deviation but be much more accurate if that device is accurate but less precise.

As I said - calculating the mean is just a mathematical exercise. In the real world it may not even make sense.

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