Which Method Yields the Correct Mean in Error Analysis?

[Yitzach]

Homework Statement

This comes from the data analysis of a lab on the Michelson Interferometer.
d1={.208 mm, .275 mm, .214 mm, .214 mm, .230 mm, .230 mm}
d2={1.757 mm, 1.505 mm, 1.725 mm, 1.773 mm, 1.640 mm, 1.732 mm}
$$r=5.13945\pm.11981$$
$$\lambda\approx\lambda'=599\pm23.7\ nm$$

Homework Equations

real d=raw d/r
$$\Delta\lambda=\frac{\lambda\lambda'}{2(d_1-d_2)}$$
$$s_f^2=(\frac{\partial f}{\partial x})^2s_x^2+(\frac{\partial f}{\partial y})^2s_y^2\cdots$$

The Attempt at a Solution

$$mean(d1)=.2285\ mm$$
$$mean(d2)=1.6887\ mm$$
$$\bar{d_1}=mean({\frac{d1}{5.13945})=44460\ nm$$
$$\bar{d_2}=mean(\frac{d2}{5.13945})=328570\ nm$$
$$\bar{d_1}=\frac{mean(d1)}{5.13945}=44460\ nm$$
$$\bar{d_2}=\frac{mean(d2)}{5.13945}=328570\ nm$$

Even though at this point the two means are equal to each other, eventually they won't be. I won't bother you with more means of elements versus mean of results here. The question is, "Is the correct mean the one that takes the mean of the raw values and number crunches with that OR the one where the individual data points are number crunched with and then the mean taken of the results?" First set versus second set. The major problem is loss of decimals and accuracy as the exercise progresses. I like to carry more decimals than are significant in case you can't figure that out before the answer.

Feel free to continue through the exercise yourself. My answer is:
$$\Delta\lambda=.63571\pm.06692\ nm$$

 

[NateD]

The correct mean is the one that takes the mean of the raw values and number crunches with that. This is because it results in more accurate values and more precise calculations.