Margin of error for standard deviation

[CarlosMarti12]

Hello, I am trying to calculate what the margin of error for the standard deviation would be if each data point has a margin of error E. The standard deviation (for 3 data points) is defined as

$σ = \sqrt{\frac{(x_{1}-μ)^{2}+(x_{2}-μ)^{2}+(x_{3}-μ)^{2}}{3}}$

Where

$μ = \frac{x_{1}+x_{2}+x_{3}}{3}$

In other words, $x_{n}$ for n = 1, 2, or 3 has a margin of error of ±E. Is there a way to find what the maximum margin of error would be for the standard deviation, given the margin of error for each individual data point? (Notice that the margin of error would affect the mean value of μ as well.)

Any help would be greatly appreciated. Thanks!

 

[Stephen Tashi]

The standard deviation (for 3 data points) is defined as

$σ = \sqrt{\frac{(x_{1}-μ)^{2}+(x_{2}-μ)^{2}+(x_{3}-μ)^{2}}{3}}$

Where

$μ = \frac{x_{1}+x_{2}+x_{3}}{3}$

This is apparently a "sample standard deviation" rather than the standard deviation of a probability distribution. You can define the sample standard deviation of a sample of 3 things that way, but some texts would say divide by 2 instead of by 3 in your first equation. Let's say we define it your way.

In other words, $x_{n}$ for n = 1, 2, or 3 has a margin of error of ±E. Is there a way to find what the maximum margin of error would be for the standard deviation, given the margin of error for each individual data point? (Notice that the margin of error would affect the mean value of μ as well.)

If you are talking about "margin of error" in the sense of the rules of thumb used in lab measurements, then I don't know the answer. These rules of thumb seem to vary from textbook to textbook.

If you are talking about a real math problem --- well, now your talking!

Let the errors in $x_1, x_2,x_3$ be respectively $e_1, e_2, e_3$.


Define

$$\mu_e = \frac{ (x_1 + e_1) + (x_2 + e_2) + (x_3 + e_3)}{3}$$

$$\sigma_e = \sqrt{ \frac{(x_1 + e_1 - \mu_e)^2 + (x_2 + e_2 - \mu_e)^2 + (x3 + e_3 - \mu_e)^2 }{3}}$$

$$F(e_1,e_2,e_3) = \sigma_e - \sigma$$

Regarding $x_1, x_2, x_3$ as constants the problem is:

Given

$$-E \le e_1 \le E$$
$$- E \le e_2 \le E$$
$$-E \le e_3 \le E$$

Find the tightest bounds $L, U$ such that

$$L \le F(e_1,e_2,e_3) \le U$$

If we have to, we can approach this as multi-variable caclulus problem:

Find the max and min values of $F(e_1,e_2,e_3)$ subject to the above constraints on the $e_i$.

That would involve taking partial derivative of $F$ with respect to the $e_i$ and setting them equal to zero. If would also involve checking all the "boundary" cases. Before we worry about that, let's see if the above is a correct statement of you question.

In a probabilistic setting for errors this formulation gives a worst case analysis, which may be an unlikely case.

 

[viraltux]

Hello, I am trying to calculate what the margin of error for the standard deviation would be if each data point has a margin of error E. The standard deviation (for 3 data points) is defined as

$σ = \sqrt{\frac{(x_{1}-μ)^{2}+(x_{2}-μ)^{2}+(x_{3}-μ)^{2}}{3}}$

Where

$μ = \frac{x_{1}+x_{2}+x_{3}}{3}$

In other words, $x_{n}$ for n = 1, 2, or 3 has a margin of error of ±E. Is there a way to find what the maximum margin of error would be for the standard deviation, given the margin of error for each individual data point? (Notice that the margin of error would affect the mean value of μ as well.)

Any help would be greatly appreciated. Thanks!

By "margin of error ±E" you mean a confidence interval? I am not sure what you mean.

 

[haruspex]

CarlosMarti12, you ask for the "maximum margin of error". Taking that literally, think of it geometrically. The feasible 'true' values for the triple x1, x2, x3 form a cube of side 2E. The point where they all equal the observed mean may be inside or outside the cube. Sigma is (proportional to) the distance from that point to a point in the cube.
So you can just evaluate the distance from the 'mean point' to each of the eight vertices to find the max sigma.
The min sigma is more complicated. It can be 0, or the distance to the nearest face of the cube, or the distance to the nearest edge of the cube, or the distance to the nearest vertex of the cube. This will be according to whether the mean lies inside the +/-E range of all, 2, 1 or none of the measurements.

That said, I would consider the standard deviation of sigma more interesting. For that, it would be best to reinterpret E as some number of standard deviations of the measurements.

 

[blue_raver22]

Hello,

Thank you for your question. The margin of error for the standard deviation can be calculated using the formula:

ME = t * \frac{s}{\sqrt{n}}

Where ME is the margin of error, t is the critical value from the t-distribution based on the desired confidence level, s is the sample standard deviation, and n is the sample size.

In your case, since you have 3 data points, n = 3. Also, the sample standard deviation can be calculated using the formula you provided:

s = \sqrt{\frac{(x_{1}-μ)^{2}+(x_{2}-μ)^{2}+(x_{3}-μ)^{2}}{3}}

Therefore, the margin of error for the standard deviation would be:

ME = t * \frac{\sqrt{\frac{(x_{1}-μ)^{2}+(x_{2}-μ)^{2}+(x_{3}-μ)^{2}}{3}}}{\sqrt{3}}

We can see that the margin of error will be affected by the individual data point's margin of error, as well as the sample size and the confidence level. The maximum margin of error for the standard deviation would depend on the maximum margin of error for each individual data point.

I hope this helps. Let me know if you have any further questions. Thank you.