Question TWO [35]

This question concerns the diameters (in inches), heights (in feet), and volumes (in cubic feet) of 31 cherry trees. Minitab calculations concerning this data is attached to this exam.

  1. Which better predicts volume, diameter or height? Supply a sentence explaining your answer. (This item will be graded on the basis of the explanation.)
                                           DIAMETER  HEIGHT
          std of (logarithmic) residuals   0.04993   0.1769
          percentage error                 12%        50%
    On the basis of percentage error, DIAMETER is a better predictor
  2. Convert the relation of log volume to log diameter and log height back to the original (non-logarithmic) coordinates.
    VOLUME = ____ 0.001318 DIAMETER1.98 HEIGHT1.12 (ft3) ___

  3. Based on the standard deviation of the residuals, How well will the formula of (2) predict volume? (Deciding whether the answer should be in cubic feet or percentages is part of this question.)
    Standard deviation of residuals: 0.035
    This corresponds to a 8.4% error.

  4. Convert (2) and (3) to metric units. (1 inch = 2.54 cm)
    VOLUME = 0.181×DIAMETER1.98×HEIGHT1.12 (m3)
                VOL in cubic meters, DIAM and HEIGHT in meters
                (Full marks for (anything)×DIAMETER1.98 HEIGHT1.12
    Prediction accuracy: The percent error is still 8.4%

  5. The descriptive statistics for log volume are not in the Minitab printout. Supply as many as you can. (You will not be able to calculate them all.)
    Calculate: Quantiles as logs of quantiles for volume.
                Mean using any of the regression formulae.
                Std. dev is not possible with material covered in class.
     1.42     ________  1.009   1.28      1.38      1.58    1.89  
     Mean     Std dev  Minimum    Q1     Median      Q3    Maximum
  6. One of the trees has
    version 1:   Diameter=18 inches   Height=80 feet   Volume=51.0 cu feet.
    version 2:   Diameter=11 inches   Height=75 feet   Volume=18.2 cu feet.
    For that tree calculate the residual for the regression of log Volume on log Height and log Diameter.
    version 1:   log(51.0) - (-2.88 + 1.98 log(18) + 1.12 log(80)) = -0.029
    version 2:   log(18.2) - (-2.88 + 1.98 log(11) + 1.12 log(75)) = -0.022

  7. Obtain the sum of squares of residuals for the regression of log volume to log diameter and log height.
    Solve:   Using any of the regressions --
      "s" = sqrt((Sum of squares of residuals)/(N-#coef))
    28x0.035352 = 0.03498