#### Buzz Bloom

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- Summary
- I used a spreadsheet (PNG image in main body section) to analyze 12 values with +/- error ranges which I assume for the purpose of this calculation to be standard deviation values. I am seeking conformation or corrections regarding the validity of the method I used.

Below is the spreadsheet image.

The source of the data in the two columns, Ho and +/-, is the following article

The Ho column contains the 12 values, H

The +/- column contains for each H

The Source column specifies where in the article the particular 12 values for H

The Wt=... column contains values, W

The Wt*Ho column contains weighted Ho values

I feel confident that this result is OK for AH. I am less confident that my calculation below of AE is OK. I will present my analysis, and explain my main concern about AE, in what follows.

The D2^2=... column contains values, D2

I now define Aσ as square root of a quotient: the weighted sum of squared differences and the sum of weights. Using [3] and [10]

I do not know how to adjust this for weighted samples.

Another minor concern is that my assumtion to use 1/E

The source of the data in the two columns, Ho and +/-, is the following article

Let AHo be the conjoined weighted average values for Ho and the corresponding error ranges.[1] AHo = AH +/- AE = 67.65 +/- 0.22.

The Ho column contains the 12 values, H

_{i}, for the Ho variable, i being the index (of a variable value) ranging from 1 to 12.The +/- column contains for each H

_{i}value the corresponding value of an error range, E_{i}, which I assume to be of 1 standard deviation σ_{i}.The Source column specifies where in the article the particular 12 values for H

_{i}and E_{i}are found.The Wt=... column contains values, W

_{i}, for the weights for calculating weighted sums. I assume that the appropriate choice of weight are the squares of the reciprocals of the E_{i}values.[2] W

The sum of this column is_{i}= 1/E_{i}^{2}[3] ΣW = Σ[i=1 to 12] W

_{i}= 25.8982 .The Wt*Ho column contains weighted Ho values

[4] WH

The sum of this column is_{i}= W_{i}H_{i}.[5] ΣWH = Σ[i=1 to 12] WH

_{i}= 1751.9631 .

The weighted average, AH, is[6] AH = ΣWH / ΣW = 67.65.

I feel confident that this result is OK for AH. I am less confident that my calculation below of AE is OK. I will present my analysis, and explain my main concern about AE, in what follows.

The D2^2=... column contains values, D2

_{i}, of the squares of the 12 differences between AH and a value of Ho.[7] D2

Althought not part of the calculations, it will later be useful to also have defined the sum of the differences squared._{i}= (Ho-H_{i})^{2}[8] ΣD2 = Σ[i=1 to 12] (D2

_{i})The Wt*D2 column contains values, WD2

_{i}, of the product of a weight and a squared difference.[9] WD2

_{i}= W_{i}D2_{i}[10] ΣWD2 = Σ[i=1 to 12] (WD2

_{i}) = 1.2992I now define Aσ as square root of a quotient: the weighted sum of squared differences and the sum of weights. Using [3] and [10]

[11] Aσ = (WD2/ΣW)

^{1/2}= 1.2992/25.8982 = 0.22

The main reason I am uncertain is that __if__all of the E_{i}values were the same, say 1/s, then then the sum of the weights, ΣW, would be[12] ΣW = (12/s)

Then the sum of the weighted squares of differences, ΣWD, would be^{2}.[`13] ΣWD2 = (1/s

^{2}) Σ[i=1 to 12] (D2_{i})= (1/s)

and Aσ would be^{2})ΣD2[14] Aσ = [(1/12)ΣD2]

Ordinarily, since there are 12 samples I would expect^{1/2}^{1/2}.[15] Aσ = [(1/11) ΣD2]

Equation [15] is based on the article^{1/2}^{1/2}.### Standard deviation - Wikipedia

en.wikipedia.org

I do not know how to adjust this for weighted samples.

Another minor concern is that my assumtion to use 1/E

_{i}^{2}as weights may not be the right approach. However, I think I have a good rational for doing this, but it is rather complicated, so I will not include it in this post. If anyone would like to see this rationale, I will post it later.
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