Sum of squared residuals and R² Problem set A Answers

\(R^2\) = 0.883

\(R^2\) = 0.969

\(R^2\) = 0.888

1. (a) \(R^2\) = 0.677
2. (b) 67.7%

1. (a) 0.2, 0.3, 0.2, 0.3, −0.2
2. (b) 0.3

1. (a) \(T = -0.0121x^3 + 0.489x^2 + 1.75x - 0.500\)
2. (b) 140
3. (c) x = 35 is outside the range of the data (0.6 to 10), so this is extrapolation and is not reliable. The cubic has a negative x³ coefficient: it reaches a maximum at x = 28.6 and then falls, but a moon further from the planet would not take less time to orbit it.

1. (a) 0.962
2. (b) 0.992
3. (c) The cubic has the larger \(R^2\) (0.992 against 0.962), so it explains more of the variation and fits better.

1. (a) 3.19
2. (b) 0.35
3. (c) Model 2: its \(SS_{res}\) (0.35) is smaller than the other (3.19)

1. (a) \(y=0.533t+14.7\), \(R^2\) = 0.991
2. (b) 46.7
3. (c) 60 is far outside the data range (0 to 14), so the prediction is extrapolation. A town's population does not usually keep growing by the same amount for 60 years: housing, jobs and land all limit it. The high \(R^2\) only describes the fit inside the data.