Footfall and Sales: Correlation, Regression and Extrapolation
A grocery owner in Guwahati records monthly footfall (x, in hundreds) and sales (y, in ₹ thousand) for 12 months. The summary values are: mean x = 420, mean y = 38.5, Σ(x−x̄)(y−ȳ) = 9,240, Σ(x−x̄)^2 = 16,800, and Σ(y−ȳ)^2 = 8,400. The owner wants to forecast sales for a month with 500 hundred footfall and also for a festival month with 700 hundred footfall. He has heard that a high correlation proves that footfall causes sales. He also fitted a line and found the intercept negative. The observed footfall ranged from 310 to 480 hundred. A nearby shop with 500 hundred footfall reported sales of ₹95 thousand, which the owner thinks his model should match. He asks a statistics student to use the summary values rather than the raw data, because the original register is incomplete.
1. Compute the correlation coefficient and interpret it in context. Does it prove causation?
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[ r=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^2\sum(y-\bar{y})^2}} =\frac{9240}{\sqrt{16800\times8400}} ] [ \sqrt{141{,}120{,}000}\approx11879.4,\qquad r\approx0.778 ]
There is a moderately strong positive linear association: months with higher footfall tend to have higher sales. The coefficient of determination is (r^2\approx0.605), so about 60.5% of the variation in sales is associated with the linear relationship with footfall. Correlation does not prove causation; both variables could be driven by season, festivals, promotions, or location. The most that can be said is that the linear association is useful for prediction within the observed range.
2. Fit the least-squares regression line of sales on footfall and predict sales at 500 hundred footfall.
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The slope is
[ b=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2} =\frac{9240}{16800}=0.55 ]
The intercept is
[ a=\bar{y}-b\bar{x}=38.5-0.55(420)=38.5-231=-192.5 ]
So the fitted line is
[ \hat{y}=-192.5+0.55x ]
where y is in ₹ thousand and x is in hundreds of customers. At x = 500:
[ \hat{y}=-192.5+0.55(500)=82.5 ]
The predicted sales are about ₹82.5 thousand. The slope means that, on average, sales rise by ₹0.55 thousand (₹550) for each additional hundred customers in the observed range.
3. Comment on the negative intercept, the prediction at 700, and the nearby shop's sales of ₹95 thousand.
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The negative intercept (−192.5) is a mathematical consequence of fitting a straight line; it has no sensible interpretation because zero footfall would not produce negative sales. It mainly signals that the linear model should not be used far outside the data range. A prediction at x = 700 is an extrapolation well beyond the observed maximum of 480, so it is unreliable; sales may level off because of store capacity, staffing or stock limits. The prediction at 500 is also slightly outside the observed range and should be treated cautiously. The nearby shop's ₹95 thousand versus the model's ₹82.5 thousand is one residual of about ₹12.5 thousand; one observation does not disprove the model, because other factors and random variation matter. Report the prediction with its uncertainty rather than as a guarantee.