How do you find the uncertainty of an average?

[Student2018]

Homework Statement

I used a spectrometer in class and obtained 6 angles in which and found the angles. I needed to find the average of the angles and account for the uncertainties here is what I did:

Center (A): 179.58°±.26 Center (B): 359.35°±.26
Purple(RA): 194.95°±.66 Purple (RB): 374.32°±.66
Purple(LA): 163.97°±.17 Purple (LB): 343.83°±.17

Homework Equations

Average= (x1+x2+x3+x4)/4
dc=sqrt[(da)^2+(db)^2]

The Attempt at a Solution

1.C(A) - P(RA) = 179.58-194.95 = 15.37±.71
2.C(B) - P(RB) = 359.35-374.32 = 14.97±.71
3.C(A) - P(LA) = 179.58-163.97 = 15.61±.31
4.C(B) - P(LB) = 359.35-343.83 = 15.52±.31

where I got the uncertainties like so:
dc(1)=sqrt[(.26)^2+(.66)^2]= .71

dc(3)=sqrt[(.26)^2+(.17)^2]= .31

I needed to get the average of all the numbers
Average= (15.37+14.97+15.61+15.52)/4 = 15.38

I talked to the professor and he wanted me to use
dt=sqrt[(.71)^2+(.71)^2+(.31)^2+(.31)^2]= 1.0956

My question is, is it 1.0956 or is it (1.0956/4) = .274?
(Sorry if it was long)

 

[haruspex]

dt=sqrt[(.71)^2+(.71)^2+(.31)^2+(.31)^2]= 1.0956

That is clearly wrong since it would say the more measurements you take the less accurate your result.

If you have a number of readings of the same quantity but with different uncertainties on each, the ideal is to use a weighted average to find the mean. The more precise values need to be given greater weight.
The weighting is 1/σ_i². In this case you should get 15.50 rather than 15.38.
The uncertainty in that average goes like σ^-2=Σσ_i^-2. For that I get σ=0.20.

 

[Student2018]

I see, thank you. That uncertainty would make a lot more sense than 1.0956 which is way too high of a range.