Math · Introductory statistics · Worked example

Fit a least-squares line and interpret r²

Five students report hours studied and exam scores: (1, 55), (2, 62), (3, 64), (4, 71) and (5, 78). Find the least-squares line, r and r², and predict the score after 3.5 hours.

Find the means

\begin{array}{l} \frac{1 + 2 + 3 + 4 + 5}{5} = 3 \\ \frac{55 + 62 + 64 + 71 + 78}{5} = 66 \end{array}

Find the slope

Multiply each pair of deviations and add: (−2)(−11) + (−1)(−4) + 0(−2) + 1(5) + 2(12) = 55. The squared x-deviations add to 10.

\frac{55}{10} = 5.5

Find the intercept

The line passes through (x̄, ȳ) = (3, 66).

66 - 5.5 \cdot 3 = 49.5

Find r and r²

The squared y-deviations add to 310.

\begin{array}{l} \frac{55}{\sqrt{10 \cdot 310}} \approx 0.988 \\ {0.988}^{2} \approx 0.976 \end{array}

Predict

Substitute x = 3.5, which lies inside the data.

49.5 + 5.5 \cdot 3.5 = 68.75

Result

ŷ = 49.5 + 5.5x; r ≈ 0.988 and r² ≈ 0.976, so the line accounts for about 97.6% of the variation in scores. At 3.5 hours it predicts 68.75.

Your turn

The line ŷ = 49.5 + 5.5x predicts 71.5 at x = 4, where the observed score was 71. What is the residual?

Show the answer and explanation

−0.5.

Residual = observed − predicted = 71 − 71.5 = −0.5, so the point sits just below the line.

71 - (49.5 + 5.5 \cdot 4) = - 0.5

Keep exploring

In Statistics, Regression shows the same equation, r and R², and the residuals 0, 1.5, −2, −0.5 and 1. The line would predict 115.5 for 12 hours, an impossible score: extrapolation.

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