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V 1.0 Carbon on the stump: how forest carbon inventories are verified and why it is important Western Mensurationists’ Meeting, June 2011 Banff, Alberta,

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Presentation on theme: "V 1.0 Carbon on the stump: how forest carbon inventories are verified and why it is important Western Mensurationists’ Meeting, June 2011 Banff, Alberta,"— Presentation transcript:

1 V 1.0 Carbon on the stump: how forest carbon inventories are verified and why it is important Western Mensurationists’ Meeting, June 2011 Banff, Alberta, Canada Zane Haxton Verification Forester Scientific Certification Systems

2 V 1.0 The agenda  What is different about “verification cruising”?  How do we verify inventories?  Some metrics and statistical techniques for inventory verification

3 V 1.0 What is different about “verification cruising”? Conventional check cruising  Conducted during inventory effort to suggest corrections that could be made  Conducted by someone internal to the organization, who is checking the work of internal cruisers or contractors “Verification cruising”  Conducted after inventory effort for verification purposes  Conducted by outside auditors

4 V 1.0 What is different about “verification cruising”? Conventional check cruising  Lower stakes – negative findings may result in corrective action (if checking employees) or docked payment (if checking contractors) “Verification cruising”  Higher stakes – negative findings may result in a repeat of the entire inventory

5 V 1.0 Forest inventory verification  A sample-based process  Select random plots of plots to re-measure  Re-measure the plots using the same sampling protocol and type of equipment used by the forest owner

6 V 1.0 In the beginning: percent error  General formula:  Can be calculated at the plot level and averaged, or calculated across the entire sample, depending on subtle semantic differences in the protocol used.

7 V 1.0 An improvement: the paired t-test (two-sided) One method is to conduct a statistically valid sample of original measurement plots… you can then employ a Paired t-Test to compare the check volumes against the original. Using a t-test makes it clear that:  You have a sample; and that  The confidence of your conclusions depends on sample size among other things. (Paraphrased from John Bell & Associates Inventory & Cruising Newsletter, “Check cruising”, January 1999, available at http://www.proaxis.com/~johnbell/regular/regular45a.htm)

8 V 1.0 An improvement: the paired t-test (two-sided)  Null hypothesis: there is no (really important) difference between verifier and forest owner measurements.  Alternative hypothesis: there is a (really important) difference between verifier and forest owner measurements.

9 V 1.0 An improvement: the paired t-test (two-sided)  General formula: Estimated bias of the forest owner’s inventory, as determined by comparison with verifier measurements (in units). Maximum allowable inventory bias (in same units) Standard error of the estimated bias (in same units).

10 V 1.0 An improvement: the paired t-test (two-sided)  General formula: Estimated bias of the forest owner’s inventory, as determined by comparison with verifier measurements (in units). Maximum allowable inventory bias (in same units) Standard error of the estimated bias (in same units).

11 V 1.0 An improvement: the paired t-test (two-sided)  General formula: Estimated bias of the forest owner’s inventory, as determined by comparison with verifier measurements (in units). Maximum allowable inventory bias (in same units) Standard error of the estimated bias (in same units).

12 V 1.0 An improvement: the paired t-test (two-sided)  General formula: Estimated bias of the forest owner’s inventory, as determined by comparison with verifier measurements (in units). Maximum allowable inventory bias (in same units) Standard error of the estimated bias (in same units).

13 V 1.0 A problem: sample size The sample size needed The Z-value for the given value of α. The Z-value for the given value of β The expected standard deviation of the bias estimate The smallest degree of bias we want to be able to detect

14 V 1.0 A problem: sample size The sample size needed The Z-value for the given value of α. The Z-value for the given value of β The expected standard deviation of the bias estimate The smallest degree of bias we want to be able to detect

15 V 1.0 A problem: sample size The sample size needed The Z-value for the given value of α. The Z-value for the given value of β The expected standard deviation of the bias estimate The smallest degree of bias we want to be able to detect

16 V 1.0 A problem: sample size The sample size needed The Z-value for the given value of α. The Z-value for the given value of β The expected standard deviation of the bias estimate The smallest degree of bias we want to be able to detect

17 V 1.0 A problem: sample size The sample size needed The Z-value for the given value of α. The Z-value for the given value of β The expected standard deviation of the bias estimate The smallest degree of bias we want to be able to detect

18 V 1.0 (Possible) solution: sequential sampling  Null hypothesis: there is no difference between verifier and forest owner measurements.  Alternative hypothesis: there is a difference between verifier and forest owner measurements.

19 V 1.0 (Possible) solution: sequential sampling  Re-measure at least two plots.  Compute the necessary sample size:

20 V 1.0 (Possible) solution: sequential sampling  If the necessary sample size has been attained, stop. Otherwise, re- measure another plot and re-evaluate.  If stopped, compute K:

21 V 1.0 (Possible) solution: sequential sampling  If the necessary sample size has been attained, stop. Otherwise, re- measure another plot and re-evaluate.  If stopped, compute K:

22 V 1.0 (Possible) solution: sequential sampling  If the necessary sample size has been attained, stop. Otherwise, re- measure another plot and re-evaluate.  If stopped, compute K:

23 V 1.0 (Possible) solution: sequential sampling  If the necessary sample size has been attained, stop. Otherwise, re- measure another plot and re-evaluate.  If stopped, compute K:

24 V 1.0 (Possible) solution: sequential sampling  If, accept null hypothesis  If, reject null hypothesis

25 V 1.0 Possible refinements:  Sample in batches of plots  Sample normally for some minimum sample size, then adopt sequential sampling  Require the null hypothesis to be accepted across a minimum number of samples before verifying the inventory

26 V 1.0 Conclusions:  This is an exciting time to be doing forest inventory work!  Advances in forest inventory verification may also be applicable to check cruising.  We’ll let you know!

27 V 1.0 Special thanks to:  Tim Robards, Spatial Informatics Group  John Nickerson, Climate Action Reserve  Ryan Anderson, Scientific Certification Systems

28 V 1.0 Another idea: equivalence testing*  In t-testing, the null hypothesis is one of no difference, so an inventory is accepted unless statistically significant evidence indicates that the forest owner’s measurements are defective.  Equivalence testing, originally developed for model validation, puts the burden of proof on the forest owner by reversing the null hypothesis. *See Robinson and Froese, 2004, “Model validation using equivalence tests”, Ecological Modelling 176:349-358

29 V 1.0 Another idea: equivalence testing*  In t-testing, the null hypothesis is one of no difference, so an inventory is accepted unless statistically significant evidence indicates that the forest owner’s measurements are defective.  Equivalence testing, originally developed for model validation, puts the burden of proof on the forest owner by reversing the null hypothesis. *See Robinson and Froese, 2004, “Model validation using equivalence tests”, Ecological Modelling 176:349-358

30 V 1.0 Another idea: equivalence testing Procedure:  Beforehand, construct a “region of indifference” (e.g. ±15% of the forest owner’s estimated carbon stocks) and select an α-level.  Calculate a “special confidence interval”, equal to two one-sided confidence intervals of size α, around the bias estimate.  If the special confidence interval falls outside the region of indifference, the null hypothesis is not rejected. Otherwise, it is rejected.

31 V 1.0 Another idea: equivalence testing 0 Picture credit: Robinson and Froese 2004

32 V 1.0 Appendix

33 V 1.0 Carbon exists in: “Belowground” carbon Branches and foliage Bole and bark

34 V 1.0 How is carbon quantified?  Carbon is quantified in terms of mass – most commonly Mg/ac or Mg/ha  1 Mg = 1 metric tonne  1 Mg = 1,000 kg = 2,204.6 lbs  1 Mg CO 2 e (or CO 2 -e) = (44/12) Mg C ≈ 3.67 Mg C

35 V 1.0 How to estimate C in tree boles* * This exercise was inspired by page 40 of K. Iles, 2009, “The Compassman, The Nun, and the Steakhouse Statistician”  Using your thumb as an angle gauge, estimate the basal area (BA; ft 2 /ac).  Estimate average tree height (HT; ft)  Assuming conical tree form, (ft 3 ) (will not work well for hardwoods)

36 V 1.0 Appendix: How to estimate C in tree boles  Average Volume/Basal Area Ratio (VBAR; ft 3 /ft 2 ) =  Volume (ft 3 /ac) = VBAR * BA  Mass (lbs/ac) = Vol (ft 3 /ac) * WoodDens (lbs/ft 3 )

37 V 1.0 Appendix: How to estimate C in tree boles  Biomass (Mg/ac) = Mass (lbs/ac) / 2,204.6  C mass (Mg/ac) = Mass (Mg/ac) * 0.5  CO 2 -e mass (Mg/ac) = C * (44/12)  … and there you go!

38 V 1.0 An example from the California redwoods  BA = 240 ft 2 /ac  HT = 120 ft  VBAR = (1/3) * 120 = 40 ft 3 /ft 2  Vol/ac = 240 * 40 = 9,600 ft 3 /ac  Biomass = 9,600 * 21.22 = 203,712 lbs/ac = 92.4 Mg/ac  C mass = 92.4 * 0.5 = 46.2 Mg/ac  CO 2 -e mass = 46.2 * (44/12) = 169.4 Mg/ac


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