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Using ratios to calculate weights in a weighted average (or not)


New member
May 25, 2018
As a 'non-mathematician' (somebody who sucks at maths) I've been enjoying reading up on a lot of the stuff that flew over my head in maths class many years ago. I've encountered some examples of weighted averages and I've been puzzling over how one might derive the weights in the following scenario.

Let's say I have a score of effectiveness for wheat fertiliser products (such as a growth rate), along with measurements of various associated behaviours in the crop (improved absorption levels, conversion rates etc.) Observations are made at various farms at various times of the year so conditions vary, hence the need for averaging over many observations in an analysis of the products. Also, notably, the effectiveness of a product and its behaviour profile often vary a lot from one wheat variety to the next with some products working better on variety A, others on variety B etc.

Therefore, if I want averages for the various readings of a particular product on variety A, I could simply exclude all observations made on other varieties from the analysis. However, let's assume the amount of data is limited so that excluding those observations would be too costly. In that case, I wouldn't want to treat all of the observations for the product in question with equal relevance since observations on non-A varieties would carry less 'weight'. Taking weighted averages would of course address this obstacle. My query here (or confusion) concerns the weighting method - how to go about calculating the weights in this case?

My first idea would be to simply use the ratio of a given product's average effectiveness score between variety pairs. For example, if a product's effectiveness score is 50 units on variety A and 40 units on variety B, and I want averages of the behaviour readings of this product on variety A, could the weight assigned to observations on variety B be calculated as follows: 40 / 50 x 100 = 80% (or a weight of 0.8, where observations on variety A are assigned a weight of 1)? But then what if the scores were reversed (40 units on variety A and 50 units on variety B) - would the following work:

50 / 40 x 100 = 125
125 - 100 = 25
100 - 25 = 75 (i.e. a weight of 0.75)?

Apologies if all these ramblings belie some major rookie errors (or just plain don't make sense!) Any pointers would be appreciated.