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<br />The Meeker Mines consist of three areas: <br />Northern No. 1 Mine (7.1 acres of reclaimed area) <br />Reinau Mine (5.0 acres) <br />Refuse Area (4.0 acres) <br />The Meeker permit contains a vegetative cover success standard of <br />70$. In order for vegetation to be considered successful (in terms <br />of cover), it must be demonstrated that cover on the reclaimed area <br />is not less than 90$ of the cover standard with 90$ statistical <br />confidence (Rule 4.15.8(3)(a)). Vegetative cover is the parameter <br />used by DMG to determine "successful establishment of vegetation", <br />as mentioned in Rule 3.03.1(2)(b), in regard to Phase II bond <br />release. Progress toward meeting other vegetation success <br />standards is considered although, it is not necessary to meet those <br />standards in order for Phase II bond to be released. <br />Matt Hayes collected cover data on the three areas listed above, <br />although only the portion of the Northern No. 1 Mine west of the <br />access road was sampled. DMG collected data on the Norther Mine <br />(both sides of the road) and the Reinau Mine. Due to time <br />Constraints, DMG did not collect data at the refuse area and did <br />not sample as many transects as Mr. Hayes. <br />Because of the small acreage and the apparent similarity of the <br />vegetation on the three reclaimed areas, DMG advised Mr. Hayes <br />(prior to sampling) to combine the data from the three areas and <br />treat them as one area. A summary of the data is presented below. <br /> Hayes DMG <br />Mean Cover 61.2$ 53.5$ <br />Std. Deviation 9.976 11.988 <br />No. Samples 17 8 <br />Only 1.2$ of the cover was attributed to annuals. As documented in <br />the Enron bond release application, the mean cover value (in Hayes' <br />data) is at least 90$ of the 70$ cover success standard with 90$ <br />statistical confidence. The DMG data is not. However, there are <br />several reasons for agreeing with Enron's conclusion that the cover <br />standard has been met. <br />1. A comparison of the two data sets with a t-test to determine <br />if the two means are the same (as in Danie1,1978) indicates that <br />the two means are the same, with a p value between .10 and .20. <br />Without a larger probability of error, I have no reason to question <br />the validity of Hayes' data. <br />