# Basic statistics calibration question

1. Jul 31, 2013

### dislect

Hi guys,

My friend was asked the following question in an interview:

There are 3 weight scales and a 10kg weight.
You weigh the 10kg on each scale 5 times (total of 15 measurements for the 3 scales), each measurement is a bit off the 10kg mark (lets say the first scale shows: 11kg, 9.4kg, 12.1kg, 10.6kg. 10.2kg).

You are suppose to choose the criteria of which scales is more accurate.
Why is it better to use standard deviation instead of the medium or average?
How do you create a calibration formula for the scale?

Thanks!

2. Jul 31, 2013

### verty

The question is wrong. It is not better to use the standard deviation. This was a trick question, I think.

I won't say more but think about why the standard deviation might be bad to use.

3. Jul 31, 2013

### dislect

I wish you would say more though :)
Standard deviation measures the variation from the average. If the standard deviation for one of the scales is smaller the of the others then the measurements are closer to the average... yet, the average could be someting like 20kg which is very far away from the true 10kg. Is that why you ment its not a good indication for accuracy?
So is using the average of measurements for each scale and measuring how far the average is from the 10kg the bets way to measure the accuracy and choose the "best scale"?
How about using the standard deviation formula with the mean value as 10kg instead of the true average of that scale?

Plus, any clarifications about a calibration formula would be great.

thanks

4. Jul 31, 2013

### verty

Yes, my point was that scale A could be less accurate even though it has a smaller standard deviation, if it has a crazy average. Without paying attention to the average, you couldn't calibrate the scale. For me it is more important.

I think by calibration formula they mean a list of steps to follow to calibrate one of the scales as much as possible. I don't you need many hints here, think of Hooke's law and how you would calibrate a scale at home.

5. Jul 31, 2013

### Mandelbroth

Consider the following set of data:

$$\begin{matrix}500 & 0 & 4922 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4003 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}$$

What's its average? What's its standard deviation? :tongue:

This is something of a ridiculous example, but it shows that in the context of the problem, standard deviation IS the best indicator.

The problem says that each one is "a bit off" the actual value of 10, so the average will be about 10.

Thus, we use our formula for a sample standard deviation, replacing $\bar{x}=10$, to obtain $\displaystyle s=\sqrt{\frac{1}{n-1}\sum_{i=1}^{n}\left[(x_i-10)^2\right]}$. The scale with the smallest value of $s$ is the most accurate.

6. Aug 1, 2013

### dislect

Ok, great demonstration !
Follow up questions: Why 1/(n-1) and not 1/n?
How would you calibrate the following results: 11kg, 9.4kg, 12.1kg, 10.6kg. 10.2kg?

7. Aug 1, 2013

### chiro

With regards to n-1 its because n-1 provides an unbiased estimate of the population value.

In statistics we use random variables to estimate population parameters (like mu or sigma). An unbiased estimator means that E[theta_hat] = theta where theta is the true population value and theta_hat is an estimator that represents the estimator distribution of theta.

8. Aug 1, 2013

### Mandelbroth

There are $n-1$ degrees of freedom.

For the data you gave, I would calculate $s=\sqrt{\frac{(11-10)^2+(9.4-10)^2+(12.1-10)^2+(10.6-10)^2+(10.2-10)^2}{5-1}}=\sqrt{\frac{1+.36+4.41+.36+.04}{4}}=\frac{1}{2}\sqrt{6.17}\approx 1.241975 \text{ kg}$.

Note that the calculation assumes the "average" is 10, so it is not a standard form of standard deviation as much as a measure of the deviation from the "wanted" mean. The average of the data is not 10, but it should be fairly close.

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