Predicting Marks from Study Hours with a Lost Register
A coaching centre in Cuttack tracks the weekly hours of self-study (x) and the final marks out of 100 (y) of ten students who appeared for the CHSE examination. The tutor's summary sheet survives but the mark register has been lost. It records n = 10, Σx = 50, Σy = 600, Σx² = 310, Σy² = 40,000 and Σxy = 3,400. Study hours among these students ranged from 2 to 9 per week and marks from 41 to 88. The tutor wants to fit the line that predicts marks from study hours, to tell next year's batch that 'every extra hour of study adds about six and a half marks', and to predict the mark of a student who plans to study 14 hours a week. A colleague has fitted the reverse regression of hours on marks and obtained x = 0.1y − 1, and argues that this line is better because it is 'less steep and so less likely to give impossible predictions'. The centre's accountant, meanwhile, wants to know how much of the variation in marks the study hours actually explain.
1. From the summary statistics, fit the regression of marks on hours, report the correlation, and deal with the tutor's proposed prediction at 14 hours.
Read the worked answer
Means: x̄ = 50/10 = 5, ȳ = 600/10 = 60.
Corrected sums of squares and products: S_xx = 310 − 50²/10 = 310 − 250 = 60 S_yy = 40,000 − 600²/10 = 40,000 − 36,000 = 4,000 S_xy = 3,400 − (50)(600)/10 = 3,400 − 3,000 = 400
Slope b_yx = S_xy/S_xx = 400/60 = 6.667 marks per extra hour; intercept a = ȳ − b·x̄ = 60 − 6.667 × 5 = 26.67.
ŷ = 26.67 + 6.667x.
r = S_xy/√(S_xx·S_yy) = 400/√(60 × 4000) = 400/489.90 = 0.8165; r² = 0.667.
At x = 8 the fitted value is 80.0, well inside the observed range. At x = 14 the fitted value is 26.67 + 93.33 = 120, which is impossible: the maximum mark is 100. Two separate faults are exposed. First, 14 hours lies far outside the observed range of 2–9 hours, so this is extrapolation and the linear relation fitted over that range carries no evidence beyond it. Second, the line is unbounded while marks are not, so even well-meant arithmetic can breach 100; treat the fitted value as an expected mark on a bounded scale, or fit a model respecting the bounds if predictions near 100 are needed. The honest answer to the student is a range/qualification, not '120 marks'.
2. The colleague's reverse line is x = 0.1y − 1. Which regression should the centre use, and what does the contrast between slopes 6.667 and 0.1 reveal?
Read the worked answer
The colleague's line is the regression of x on y: b_xy = S_xy/S_yy = 400/4000 = 0.1, and a = x̄ − 0.1ȳ = 5 − 6 = −1. It answers 'how many hours usually go with a given mark', that is, it minimises the horizontal distances to the points. To predict a mark from hours you minimise vertical distances, which is the y-on-x line, so the tutor's choice is the correct one.
The colleague's argument misfires on two counts. First, inverting x = 0.1y − 1 gives y = 10x + 10, which is steeper than 6.667x + 26.67, and it predicts 90 at x = 8 where the fitted model gives 80, and 150 at x = 14. It is not 'less likely to give impossible predictions'; it is more so. Second, the two lines coincide only when |r| = 1. Here the product of the two slopes is 6.667 × 0.1 = 0.667 = r², exactly as theory requires, and because r² is only 0.667 there are two distinct lines crossing at (5, 60). Neither is 'more accurate' than the other; they answer different questions. If the centre ever wants to predict the hours a student with a given mark typically put in - for example, to ask a student with 55 marks how much study was expected of her - the colleague's line is the right one.
3. Assess the statement 'every extra hour of study adds about six and a half marks' and the claim that hours explain most of the variation in marks.
Read the worked answer
The slope 6.667 is correctly computed, but the sentence is a causal claim from observational data and is not supported. Hours of study were not assigned at random: students choose them, and the choice is entangled with ability, prior achievement, subject difficulty, health, private tuition and home circumstances. A positive slope is equally consistent with 'bright students need fewer hours and still do worse' as a confounder working against the observed relation, or with prior marks driving both. The defensible wording is associational: 'students who studied one hour more per week scored about 6.7 marks higher on average in this sample', with the uncertainty and the range of x stated.
The accountant's question is answered by r² = 0.667: about two-thirds (66.7%) of the sample variation in marks is accounted for by the linear relation with hours, leaving a third to other factors and to scatter. Note the wrong claim that r = 0.8165 explains '82%' - r is the correlation, not a proportion; the proportion is r². Two further cautions belong in any answer. r and the fitted line are sensitive to single points, and with n = 10 one unusual student can move r appreciably - the lost register should be recovered and the scatter plotted before making any public claim. And a per-hour rate estimated over 2–9 hours should not be multiplied out to large changes; returns are likely to be diminishing, so the linear rate is a local description, not a law.