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

[steveistumped]

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.

 

[jvicens]

Thank you for bringing up this question about weighted averages and how to calculate the weights in a scenario involving different varieties of wheat and their effectiveness with fertiliser products. I can understand your confusion and appreciate your efforts to find a solution.

Firstly, let me assure you that your approach of using the ratio of a product's average effectiveness score between variety pairs to calculate the weights is a valid one. This method takes into account the varying effectiveness of the product on different wheat varieties and assigns a higher weight to observations on the more effective variety. However, there are a few points to consider when using this method:

1. The weights should be calculated based on the effectiveness score for the specific variety that you are interested in. In your example, if you want to calculate the weights for observations on variety A, then the scores for variety A (50 units) and variety B (40 units) should be used. Similarly, if you want to calculate the weights for observations on variety B, you should use the scores for variety B (40 units) and variety A (50 units).

2. The weights should be calculated separately for each behaviour reading. For example, if you want to calculate the weighted average for improved absorption levels, you should use the effectiveness scores for that specific behaviour and calculate the weights accordingly.

3. The weights should be normalized to add up to 1. In your example, the weights for observations on variety A (1) and variety B (0.8) add up to 1. Similarly, if the scores were reversed, the weights for observations on variety A (0.8) and variety B (1) should add up to 1.

I hope this clarifies your confusion and helps you in your analysis. Keep up the good work of exploring and learning about topics that interest you, even if they may have seemed daunting in the past. Science is all about curiosity and continuous learning.