Ice Cream Sales and Drownings
A council analyst finds a correlation of r = 0.86 between daily ice cream sales (in pounds) and daily incidents at a beach over 90 summer days. A regression line predicting incidents from sales has slope 0.0021 incidents per pound. A councillor proposes restricting ice cream sales to reduce incidents. A colleague points out that air temperature was also recorded and correlates strongly with both variables. The analyst also notes that on the two hottest days, sales were unusually high and incidents were lower than the trend predicted.
1. Interpret r = 0.86 and the slope of 0.0021 in context, and explain why the councillor's proposal does not follow.
Read the worked answer
r = 0.86 means a strong positive linear association: days with higher ice cream sales tend to be days with more beach incidents. r² = 0.74, so about 74% of the variation in incidents is associated with the linear relationship with sales - but association is not causation.
Slope 0.0021 means that, on average, each additional £1 of daily sales is associated with 0.0021 more incidents - roughly one extra incident per £476 of sales. This is a description of the fitted line, not a mechanism.
Why the proposal fails: the credible explanation is a lurking (confounding) variable - hot weather. Hot days raise ice cream sales and bring more people to the beach, which raises the number of incidents. Ice cream does not cause incidents; both are consequences of temperature. Restricting sales would not change the weather, so incidents would be largely unaffected. The regression cannot distinguish these explanations, because it only measures how two variables co-vary, not what drives what.
2. The two hottest days fall below the trend line. Explain what this tells you and how to investigate the confounding explanation.
Read the worked answer
Points below the line at high sales are influential-ish outliers in the sense that they contradict the fitted pattern: at the extreme of the predictor, the observed response is lower than predicted. This is a warning that the relationship may not be linear across the whole range, or that a third variable is doing the work.
How to investigate:
- Partial correlation or multiple regression. Fit incidents on both sales and temperature. If the coefficient on sales shrinks towards zero once temperature is included, the association was largely explained by temperature.
- Stratify. Group days into temperature bands (e.g. under 18°C, 18–24°C, over 24°C) and compute the sales–incidents correlation within each band. If it drops sharply, confounding is confirmed.
- Check the outliers. Verify the two hottest days are not recording errors, and consider whether a curved term is needed.
- Look for a natural experiment - e.g. a period when the kiosk was closed on hot days - to see whether incidents fell.
Only after temperature is controlled for could any causal claim about sales be entertained, and even then it would need stronger evidence than a correlation.
3. The analyst wants to predict incidents for a day with £5,000 of sales. State two cautions about using the regression line here.
Read the worked answer
Caution 1: extrapolation. £5,000 may lie outside the range of sales observed in the 90 days. If the data covered, say, £400–£3,000, then £5,000 is extrapolation, and there is no evidence the linear relationship continues beyond the observed range. Predictions there are unreliable and the standard error widens rapidly.
Caution 2: the model is confounded and the prediction is not causal. Because temperature drives both variables, a £5,000 day is probably a very hot day, and the prediction mixes the effect of heat with the sales figure. The line also ignores the two outliers below trend, suggesting the fit at high sales is poor. Additionally, a point prediction should be given as an interval: with r² = 0.74, roughly a quarter of variation is unexplained, so the prediction interval will be wide.
The honest answer is a predicted value with a clearly stated interval, flagged as conditional on the observed range and on temperature being typical for such a day.