Statistical Analysis Question: How many measurements to reduce uncertainty?

[katastrophe]

Homework Statement

a. Mike makes six independent measurements of the diameter D of a leap year detector that made its way into the lab, obtaining D = 4.64, 4.78, 4.82, 4.68, 4.80, and 4.95m. What would result would he report in a lab writeup?
b. Alice does the same experiment as Mike, but makes only one measurement. Based on Mike's results, what would you expect for the uncertainty in Alice's single measurement of D?
c. How many measurements would have to be made altogether to reduce the uncertainty in the mean to +/- 0.003 m?

Homework Equations

D = $$\Sigma$$x$$_{i}$$/N
$$\sigma$$D = $$\delta$$D
$$\sigma$$D=$$\sqrt{(1/(N-1))*\Sigma(xi-\overline{x})^{2}}$$

The Attempt at a Solution

So, my real question doesn't come until part C, but I figure it definitely wouldn't hurt to make sure I have the first two parts right to base my work upon:

A. D = $$\Sigma$$x$$_{i}$$/N = (4.64+4.78+4.82+4.68+4.80+4.95)/6 = 4.78 m

B. $$\sigma$$D=$$\sqrt{(1/N-1)*\Sigma(xi-\overline{x})^{2}}$$ = $$\sqrt{(1/6-1)*((4.64-4.78)^{2}+(4.78-4.78)^{2}+(4.82-4.78)^{2}+(4.68-4.78)^{2}+(4.80-4.78)^{2}+(4.95-4.78)^{2})}$$
= 0.11m

C. So here's where my question is. I assume that we would use the same formula as in B, but how can we know what the new x$$_{i}$$ and $$\overline{x}$$ would be? Do you use a different formula to determine how to decrease the uncertainty? I know that each measurement induces a sqrt2 decrease in uncertainty, is this how you calculate it? Thanks so much!

 

[anathema]

Your solutions for parts A and B are correct. For part C, you are on the right track. To decrease the uncertainty to +/- 0.003 m, we need to decrease the standard deviation (sigma) to 0.003 m. This can be achieved by increasing the number of measurements (N) and keeping the same sum of the measurements (Sigma(xi)). In other words, we need to make more measurements without changing the average (x̄).

To calculate the new number of measurements needed, we can rearrange the formula for sigma (standard deviation) to solve for N:

N = (Sigma(xi) / sigma)^2 + 1

Substituting the values from part A and B, we get:

N = (4.78 / 0.003)^2 + 1 = 6422.67

Therefore, to reduce the uncertainty to +/- 0.003 m, we would need to make 6423 measurements in total.

I hope this helps and let me know if you have any further questions. Best of luck with your research!

 

[tomcue]

Based on the given data, Mike's result for the diameter D would be 4.78m, as calculated in part A. For Alice's single measurement, we can expect her uncertainty to be similar to Mike's, around 0.11m as calculated in part B.

To reduce the uncertainty in the mean to +/- 0.003m, we can use the formula for standard deviation, \sigmaD=\sqrt{(1/(N-1))*\Sigma(xi-\overline{x})^{2}}, where N is the total number of measurements. We can rearrange this formula to solve for N, which would give us N = (1/\sigmaD^2)+1. Plugging in the given uncertainty of 0.003m, we get N = (1/(0.003m)^2)+1 = 111,112 measurements. This means that approximately 111,112 measurements would need to be made altogether to reduce the uncertainty in the mean to +/- 0.003m. However, this number may vary depending on the precision and accuracy of the equipment and the experimental setup.