Rainfall and Paddy Yield: Reading Regression Output Before Trusting It
The agriculture officer of a panchayat has ten years of records for a paddy variety. For each year she notes x, the rainfall in centimetres during the growing season, and y, the yield in quintals per hectare. The summary she hands to a Plus Two project team is: n = 10, mean of x = 85 cm, mean of y = 42 quintals, sum of squared deviations of x = 800, sum of squared deviations of y = 500, and sum of the cross-products of deviations = 560. Rainfall in those ten years never fell below 70 cm or rose above 95 cm. One student submits a write-up claiming that the regression coefficient of y on x is 1.30, that the regression coefficient of x on y is 0.95, and that r = 0.95. The officer is more interested in a practical question: the neighbouring panchayat recorded 100 cm of rain this season, so what yield should the department expect? She also wants to know how much of the year-to-year variation in yield the rainfall actually explains.
1. Compute r and both regression coefficients from the summaries, and check the student's three claimed values.
Read the worked answer
The three quantities come from the same two sums:
$$r=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^2\sum(y-\bar{y})^2}}=\frac{560}{\sqrt{800\times500}}=\frac{560}{632.46}=0.885$$
$$b_{yx}=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2}=\frac{560}{800}=0.70$$
$$b_{xy}=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(y-\bar{y})^2}=\frac{560}{500}=1.12$$
Consistency checks: $b_{yx}\times b_{xy}=0.70\times1.12=0.784=r^2$, and $\sqrt{0.784}=0.885=r$. Both hold, so the numbers agree.
The student's write-up is impossible. A regression coefficient may exceed 1 (as $b_{xy}=1.12$ does), but the product of the two coefficients must equal $r^2\le1$. The claimed product is $1.30\times0.95=1.235>1$, which would force $r=\sqrt{1.235}>1$ - a mathematical impossibility. Also, both coefficients must carry the sign of r; here both are positive, which is consistent.
2. Write the regression line of yield on rainfall, predict the yield at 100 cm, and state how far that prediction can be trusted.
Read the worked answer
Line of y on x:
$$y-\bar{y}=b_{yx}(x-\bar{x})\ \Rightarrow\ y-42=0.70(x-85)$$ $$y=0.70x-17.5$$
At x = 100 cm: $y=0.70(100)-17.5=52.5$ quintals per hectare. Report this as the expected average yield, not a guaranteed one.
Trust:
- Extrapolation. Observed rainfall ran only from 70 to 95 cm; 100 cm is outside that range, so the straight-line pattern is being extended into unobserved territory. If yield levels off or falls in very wet years (waterlogging, lodging), the true value will be lower.
- Explained variation. $r^2=0.784$, so about 78.4% of the variation in yield goes with rainfall; the remaining 21.6% is due to seed quality, fertiliser, pests, soil and so on. A single prediction therefore has a real margin of error around 52.5.
- Correlation is not causation. The slope describes association; it does not prove that extra rain alone raises yield.
A good answer states the point estimate, then names the extrapolation risk and the residual scatter.
3. Interpret b(y on x) = 0.70 in context, and explain why there are two different regression lines rather than one.
Read the worked answer
Interpretation: on average, in the observed range, one additional centimetre of seasonal rainfall is associated with an increase of 0.70 quintal per hectare in yield. It is an average rate over the ten years, not a promise for any single field.
Why two lines: the line of y on x is fitted by minimising vertical deviations, and answers 'what yield do we expect for a given rainfall?' The line of x on y minimises horizontal deviations and answers the reverse question, 'what rainfall would typically go with a given yield?' These are different minimisation problems, so they give different slopes (0.70 and 1.12 here) and different lines.
Properties worth stating:
- Both lines pass through $(\bar{x},\bar{y})=(85,42)$.
- They coincide only when $|r|=1$; the weaker the correlation, the wider the angle between them.
- $b_{yx}\cdot b_{xy}=r^2$, so neither coefficient can be read as r itself.
Since the officer wants to predict yield from rainfall, the correct line is $y$ on $x$, giving $y=0.70x-17.5$.