error help

[Sarahisme]

hey i am writing up my prac report i have a question about error caluclation.

 

if i take the average of serveral measurements (instead of drawing a graph), and these measurements are

 

[math]

147 \pm 3

[/math]

[math]

146 \pm 3

[/math]

[math]

143 \pm 3

[/math]

[math]

145 \pm 3

[/math]

[math]

146 \pm 2

[/math]

 

i get ther average to be 145 (to 3 sig. figs.) , now how do i calucalte the error? do i use a 95% confiddence interval or somethign else?

 

I have on method i think might work, but it gives a very big error compared to the ones for the orginal measurements:

this is the method:

[math]

( \delta avg)^{2} = ( \delta m_1)^{2} + ( \delta m_2)^{2} + ( \delta m_3)^{2} + ( \delta m_4)^{2} + ( \delta m_5)^{2}

[/math]

 

where [math] m_1 , m_2, ... [/math] are measurement 1, measurement 2,...

 

so

[math]

(\delta avg) = \sqrt(3^{2} + 3^{2} + 3^{2} + 3^{2} + 2^{2}) = \sqrt(40) = 6

[/math]

 

so avg = [math] 145 \pm 6 [/math]

 

but this error is quite large compared to the others so yeah...

 

any advice would be great!

 

Thanks guys

 

_Sarah

[ydoaPs]

i could calc each individual error, but not the avg.

[Klaynos]

well the formulae for the average is:

 

(x₁+x₂...) / n = a

 

the calculation for combination of errors would be:

 

(I'm using D to signify it's an error, as I'm lazy)

 

(D(x₁+x₂...))² = Dx₁²+Dx₂²...

 

which we'll just call DX

 

so for the final error (Da) you have to take the error on n (Dn) into account:

 

(Da / a)² = (DX / X)² + (Dn / n)²

 

as Dn is 0

 

(Da / a)² = (DX / X)²

 

(Da / 145.4)² = (6.3245 / 727)²

 

which makes Da = 1.2649

 

I think...

 

Or you can use the general equation for an error which is:

 

[math](\Delta z)^2 = \left(\left( \frac {\delta z} {\delta x}\right)\Delta x\right)^2[/math]

[Sarahisme]

ok, but how can the average error be less than all the individual errors? also, what do you guys think of method of getting [math] \pm 6 [/math]

[Sarahisme]

???

[Sarahisme]

nevermind, i'll just graph it and take the 3 lines of bestfit. speaking of which, does anyone know a program which will give you the error in your line of best fit?

[swansont]

ok, but how can the average error be less than all the individual errors? also, what do you guys think of method of getting [math'] \pm 6 [/math]

 

That's why you make multiple measurements - you are zeroing in on the right number, and increasing the confidence that you haven't measured any outlying data points.

[Sarahisme]

is that like the 95% confidence interval thing?

 

should i do something such as giving the mean and the 95% confidence level?