What the Practice-Hours Data Can and Cannot Promise a Parent
The director of a competitive-exam coaching centre in Mysuru holds records for 50 students on weekly practice hours (x) and score in a 100-mark mock test (y). The observed practice hours ranged from 4 to 18. The summary values are x̄ = 12, ȳ = 64, σx = 4, σy = 9 and r = 0.60. A parent whose son practises 20 hours a week wants a predicted score. Another parent, whose daughter practises 30 hours, cites the same prediction formula and expects a proportionately higher score. A faculty member tells both parents that 'r = 0.60 means 60 per cent of the marks are explained by practice, so the prediction can be trusted almost exactly', and adds that whichever variable is taken as y, the slope of the line remains 1.35. The director wants a written reply that separates what the data support from what they do not, and a realistic statement of the error a parent should expect.
1. Obtain both regression coefficients and both regression lines, and check that they intersect at the point of means.
Read the worked answer
Regression of y on x (score predicted from hours): byx = r·σy/σx = 0.60 × 9/4 = 1.35 y − 64 = 1.35(x − 12), i.e. y = 1.35x + 47.8
Regression of x on y (hours implied by a score): bxy = r·σx/σy = 0.60 × 4/9 = 0.2667 x − 12 = 0.2667(y − 64), i.e. x = 0.2667y − 5.07
Intersection check: at x = 12, y = 1.35(12) + 47.8 = 16.2 + 47.8 = 64. At y = 64, x = 0.2667(64) − 5.07 = 17.07 − 5.07 = 12. Both lines pass through (12, 64), as they must.
Consistency check: byx × bxy = 1.35 × 0.2667 = 0.36 = r². Both coefficients carry the sign of r, here positive. The two slopes are not equal and are not interchangeable - 1.35 is a marks-per-hour figure, 0.2667 is hours-per-mark.
2. Predict the score of the student who practises 20 hours, and draft the reply to the parent who expects 30 hours to give a proportionately higher score. Support the reply with a measure of prediction error.
Read the worked answer
At x = 20: y = 1.35(20) + 47.8 = 27 + 47.8 = 74.8 marks.
Error to quote: standard error of estimate syx = σy√(1 − r²) = 9 × √0.64 = 9 × 0.8 = 7.2 marks. So 74.8 is a conditional average, and an individual student with 20 hours could plausibly score roughly 75 ± 7.
For 30 hours the arithmetic gives 1.35(30) + 47.8 = 88.3, but 30 lies far outside the observed range of 4 to 18 hours. The data say nothing about the shape of the relationship up there; scores are also capped at 100, so the linear formula must eventually break. Percentage logic ('50% more hours, 50% more marks') is not what the regression says either - each added hour is associated with 1.35 marks on average, and that average itself has a margin of error.
Suggested reply: use the formula inside 4–18 hours only, report predictions as ranges, and never treat a prediction as a guarantee.
3. Evaluate the faculty member's two statements: that r = 0.60 explains 60 per cent of the marks, and that the slope stays 1.35 whichever way the regression is run.
Read the worked answer
Both statements are wrong, and the errors are of different kinds.
First: r = 0.60 explains r² = 0.36, i.e. 36 per cent of the variation in mock scores, not 60 per cent. The remaining 64 per cent is down to everything not in the model - prior preparation, subject difficulty, test-day condition, attendance. r is a measure of linear association; its square is the share of variance accounted for. A parent hearing '60 per cent explained' would badly over-trust the formula.
Second: the slope depends on which variable is predicted. Predicting score from hours gives byx = 1.35; predicting hours from score gives bxy = 0.2667. The two lines are distinct unless r = ±1, and the choice must follow the purpose: here we predict marks, so y on x is correct. Only the product byx·bxy = r² is symmetric. The faculty member has confused a regression coefficient with a correlation coefficient.