Least Squares Regression Line Math Example 2

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Example 2

hard
The LSRL for predicting weight (yy, kg) from height (xx, cm) is y^=−100+0.8x\hat{y} = -100 + 0.8x. Interpret the slope and intercept, predict weight for height=175 cm, and explain why extrapolating to height=50 cm is problematic.

Solution

  1. 1
    Slope: for each 1 cm increase in height, weight increases by 0.8 kg on average
  2. 2
    Intercept (−100-100): predicted weight at height=0 cm; clearly nonsensical (no person has height=0); the intercept serves as a mathematical anchor
  3. 3
    Prediction at x=175: y^=−100+0.8(175)=−100+140=40\hat{y} = -100 + 0.8(175) = -100 + 140 = 40 kg
  4. 4
    Extrapolation at x=50: y^=−100+0.8(50)=−100+40=−60\hat{y} = -100 + 0.8(50) = -100 + 40 = -60 kg — impossible (negative weight!); linear model only valid in observed height range

Answer

Slope=0.8 kg/cm; intercept is meaningless extrapolation. Prediction at 175 cm = 40 kg. Extrapolating to 50 cm gives nonsensical -60 kg.
Always interpret slope in context: 'for each 1-unit increase in x, y increases by b on average.' The intercept is often contextually meaningless (x=0 may be outside observed range). Never extrapolate beyond the observed x range — the linear relationship may not hold.

About Least Squares Regression Line

The unique straight line y^=a+bx\hat{y} = a + bx that minimizes the sum of squared vertical distances (residuals) between the observed data points and the line.

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More Least Squares Regression Line Examples

Example 1 medium

Find the least-squares regression line for: [formula]: [formula]. Use [formula] and [formula].

Example 3 easy

Given [formula]: (a) predict [formula] when [formula], (b) interpret the slope, (c) does the line pa

Example 4 hard

The LSRL has the property of minimizing [formula]. Explain why minimizing squared residuals (rather

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