Rainfall and paddy yield: correlation, regression and a causal claim
An agricultural co-operative in Salem records, for 10 consecutive years, the seasonal rainfall x (mm) and the average paddy yield y (quintals per acre) on member farms. The summary totals are Σx = 600, Σy = 250, Σx² = 38,000, Σy² = 6,650 and Σxy = 15,600. Rainfall in those years ranged from 42 mm to 91 mm. The co-operative's extension officer wants to use the relationship to predict yield for a coming season and to advise farmers on whether rainfall alone is a reliable planning variable. A local lender has seen a draft note saying 'each additional millimetre of rainfall adds 0.3 quintal per acre, so rainfall causes the increase', and wants the claim checked before it is used in a crop-loan circular. The officer also has a rainfall forecast of 85 mm for the coming season, but a farmer asks what would happen at 120 mm, which is outside the observed range.
1. Compute the correlation coefficient and the coefficient of determination, and interpret both in context.
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First compute the corrected sums: Sxx = 38000 − 600²/10 = 38000 − 36000 = 2000. Syy = 6650 − 250²/10 = 6650 − 6250 = 400. Sxy = 15600 − (600)(250)/10 = 15600 − 15000 = 600.
r = Sxy/√(SxxSyy) = 600/√(2000 × 400) = 600/√800000 = 600/894.427 = 0.6708.
The correlation is positive and moderate to fairly strong: years with higher rainfall tend to have higher average yield. It is not close to 1, so rainfall alone does not determine yield. The coefficient of determination is r² = 0.6708² ≈ 0.450, so about 45% of the observed variation in yield is accounted for by the linear relationship with rainfall; the remaining 55% is associated with other factors such as soil, fertiliser, variety and management.
2. Fit the least-squares regression of yield on rainfall, predict the yield at 85 mm, and interpret the slope.
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The regression of y on x has slope b = Sxy/Sxx = 600/2000 = 0.30 and intercept a = ȳ − b x̄ = 25 − 0.30(60) = 25 − 18 = 7.
So the fitted line is ŷ = 7 + 0.30x, where x is rainfall in mm and ŷ is yield in quintals per acre.
At x = 85 mm, ŷ = 7 + 0.30(85) = 7 + 25.5 = 32.5 quintals per acre.
The slope says that within the observed rainfall range, a 1 mm increase in rainfall is associated with an estimated 0.30 quintal per acre increase in average yield. This is an average association from these 10 years, not a guaranteed response for an individual farm, and it should not be read as a causal effect.
3. Evaluate the lender's causal statement and the farmer's question about 120 mm. What cautions should the extension officer give?
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The lender's statement is too strong. The regression is based on observational data, so an omitted variable such as fertiliser use, soil quality or irrigation could drive both rainfall timing and yield; farmers may also adjust planting dates in wetter years. A slope of 0.30 describes association within the sample, not proof that adding 1 mm of rain would raise yield by 0.30 quintal. A causal claim would require a designed trial or a much richer model.
The 120 mm question is an extrapolation: the observed rainfall range was 42–91 mm, and the linear fit has no support beyond that range. In reality, very heavy rain can waterlog or damage paddy, so yield might fall rather than rise. The officer should restrict predictions to roughly the observed range, report a prediction interval rather than only a point estimate, and warn that the regression of y on x cannot be inverted to predict rainfall from yield. For 85 mm, the point estimate 32.5 quintals per acre is reasonable as a planning figure, but it should be presented with uncertainty and checked against current agronomic advice.