Fertilizer dose and tomato yield: prediction beyond the trial
A greenhouse trial tested six fertilizer doses from 0 to 24 g per tomato plant, with four plots at each dose (n = 24). The fitted least-squares line is yield = 3.1 + 0.42x, where yield is kg per plot and x is grams of fertilizer per plant. The mean dose was 12 g, Sxx = 480, the residual standard deviation was 1.8 kg, R2 = 0.61, and the standard error of the slope was 0.082. A grower wants to apply 30 g per plant and asks for a predicted yield and a 95% prediction interval. The trial's highest dose was 24 g. A residual plot against fitted values shows a mild U-shape, and one plot at the lowest dose had unusually high yield. The grower says that because R2 is 0.61, "61% of the fertilizer effect is explained," and plans to use the equation as a calibration curve.
1. Interpret the slope, intercept, R2, residual standard deviation, and slope standard error. Correct the grower's statement about R2.
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Slope 0.42 kg per gram: within the tested range, each additional gram of fertilizer is associated with a predicted yield increase of 0.42 kg. Intercept 3.1 kg is the predicted yield at 0 g fertilizer; because 0 g was in the trial, it is an observed-range anchor, though it still depends on the linear model. R2 = 0.61 means 61% of the sample variation in yield around its mean is accounted for by the linear dose relationship. It does not mean "61% of the fertilizer effect" and it does not establish causation, because other greenhouse factors could covary with dose even in an experiment.
Residual SD 1.8 kg is the typical vertical distance of a plot from the fitted line, not the prediction error for a new plot. The slope SE 0.082 gives t = 0.42/0.082 = 5.1 with 22 df, so the positive slope is clearly distinguishable from zero under the model. The grower should treat R2 as a descriptive fit measure, not as a calibration guarantee.
2. Compute the predicted yield at 30 g and a 95% prediction interval. Discuss extrapolation and leverage, and say whether you would present this prediction as a calibration result.
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At x = 30, the predicted yield is 3.1 + 0.42(30) = 15.7 kg. But 30 g is outside the observed 0-24 g range, so this is extrapolation; the linear relationship may not continue. For a 95% prediction interval at x0 = 30: SE_pred = s sqrt[1 + 1/n + (x0 - xbar)^2/Sxx] = 1.8 sqrt[1 + 1/24 + (18)^2/480] = 1.8 sqrt(1.7167) = 2.36 kg. With t(22, 0.975) = 2.074, the interval is 15.7 +/- 4.89, or about (10.8, 20.6) kg.
The leverage is h = 1/24 + 324/480 = 0.717, which is high. That point strongly influences the fit, and the interval is sensitive to model form. I would not present the 30 g prediction as a calibration result without a trial at 24-32 g. At a within-range dose such as 20 g, the prediction interval is narrower and safer.
3. The residual plot shows a mild U-shape, and one low-dose plot has unusually high yield. What do these diagnostics imply for the linear model and for the proposed use of the equation?
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A U-shape in residuals versus fitted values means the linear model systematically under- or over-predicts at the ends: yield may rise with dose but level off or accelerate. That violates the linearity assumption and makes both the slope interpretation and the 30 g prediction suspect. The unusually high yield at the lowest dose may be a legitimate observation, a transcription error, or a local soil or light effect; check the raw data and fit with and without it.
I would add a quadratic term (dose squared) or use a nonlinear plateau model, compare adjusted R2 and residual plots, and validate with new plots. Report the linear fit only as an approximation within the observed range. Do not use R2 alone to choose the model; residual pattern, lack-of-fit tests, and prediction error matter more. A calibration curve needs validation across the range where it will be used.