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statistics_for_prescription [2012/04/05 19:07] ddrummond created |
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====== Statistics for Prescription ====== | ====== Statistics for Prescription ====== | ||
+ | Statistics are numerical data about a subject or a name applied to | ||
+ | analysis of the reliability of statistical information by sampling. | ||
+ | \\ | ||
+ | The person who analyzes the forest land statistics commonly needs | ||
+ | only 6 items - Mean, Standard Deviation, Standard Error, Limit of Error, | ||
+ | Coefficient of Variation, and Number of Plots to Sample a Stand to a Desired Accuracy. | ||
+ | \\ | ||
+ | \\ | ||
+ | ====Mean- M or Average==== | ||
+ | A total of all sample counts divided by the number of samples.\\ | ||
+ | M = (sum of sample counts)/(number of samples) | ||
+ | \\ | ||
+ | \\ | ||
+ | ====Standard Deviation- SD==== | ||
+ | A statistician's measure of the spread between the individual | ||
+ | sample from the mean. The SD is just a tool in calculating error, CV and N.\\ | ||
+ | SD = √[(∑(Dev)<sup>2</sup>)/(n-1)] | ||
+ | \\ | ||
+ | \\ | ||
+ | ====Standard Error- SE==== | ||
+ | A statistician's stepping stone toward determination of LE.\\ | ||
+ | SE = (SD)/√N) | ||
+ | \\ | ||
+ | \\ | ||
+ | ====Limit of Error- LE==== | ||
+ | We cannot calculate the actual volume and cannot know the exact error. | ||
+ | We calculate a LE or a selected probability. We might say "using gambling odds of 67 out of 100, | ||
+ | (2 out of 3), we have a mean volume of 20 cords per acre, plus or minus 2 cords." The mean is 20, | ||
+ | the Standard Error is ±2, and the Limit of Error is .10 or 10%. Increasing the gambling odds to | ||
+ | 95 out of 100 doubles the LE. (20%). For odds of 99 out of 100 the LE triples (30%). \\ | ||
+ | LE = (SE)/(M) | ||
+ | \\ | ||
+ | \\ | ||
+ | ====Coefficient of Variation- CV==== | ||
+ | This describes the relative uniformity of he stand. The CV is | ||
+ | our most important statistical term. For planted stands the CV may vary .10 to .30 (10%-30%). | ||
+ | Natural stands may vary from about .30 to over 1.00. Different stands with the same CV require | ||
+ | the same number of samples to produce means of equal accuracy, regardless of the areas of these | ||
+ | stands. The CV comes from the SD and the calculated average or mean.\\ | ||
+ | CV = SD/M | ||
+ | \\ | ||
+ | \\ | ||
+ | ====Plots Needed to Sample a Stand to a Desired Accuracy==== | ||
+ | The number of samples (N) has a basic | ||
+ | relationship with CV, LE, and the probability.\\ | ||
+ | N = (t)<sup>2</sup>(CV)<sup>2</sup>/(LE)<sup>2</sup>\\ | ||
+ | "t" is the probability or gambling odds. Use "l" or "t" for one SD where values involved in the | ||
+ | cruise are low and when we can be satisfied with a 2 times out of 3 chance of arriving at a | ||
+ | figure that will vary within one SD plus or minus from the Mean.\\ | ||
+ | \\ | ||
+ | Use "2" for "t" for more accuracy. This variation of 2 SD's plus or minus makes gambling odds of | ||
+ | 95 in 100 that we'll not exceed 2 Standard Deviations. This will take 4 times as many samples as | ||
+ | for 1 SD. Here's and example:\\ | ||
+ | \\ | ||
+ | --To make a cruise of ±10% LE with a gambling chance of ± 2 SD (95 times in 100) CV = 30%.\\ | ||
+ | \\ | ||
+ | Use: \\ | ||
+ | N = (2<sup>2</sup>×.30<sup>2</sup>)/(.10<sup>2</sup>) = (4(.09))/(.01) = 36 plots\\ | ||
+ | For 1 SD, (t)<sup>2</sup> = 1 and N would be 9 plots.\\ | ||
+ | For 3 SD, (t)<sup>2</sup> = 9 and N would be 81 plots.\\ | ||
+ | If CV were 15%, N would be one-quarter as many.\\ | ||
+ | If SD were 1 and LE were .20, N would be 9 plots.\\ | ||
+ | \\ | ||
+ | Example: Tree counts on 9 sampling points-- | ||
+ | ^point no. ^tree count ^Deviation from mean ^(dev)^2 | | ||
+ | |1 |3 |2 |4 | | ||
+ | |2 |3 |2 |4 | | ||
+ | |3 |10 |5 |25 | | ||
+ | |4 |5 |0 |0 | | ||
+ | |5 |5 |0 |0 | | ||
+ | |6 |4 |1 |1 | | ||
+ | |7 |3 |2 |4 | | ||
+ | |8 |5 |0 |0 | | ||
+ | |9 |8 |3 |9 | | ||
+ | ^n = 9 ^9/46 ^ ^∑(D)^2 = 47| | ||
+ | M = 5.1 (call it 5)\\ | ||
+ | SD(Std. Dev) = √(∑(D)<sup>2</sup>)/(n-1) = √(47/8) = 2.42\\ | ||
+ | \\ | ||
+ | SE(std. Error) = SD/√n = 2.42/√9 = .81 \\ | ||
+ | \\ | ||
+ | LE (Limit of Error) = SE/M = .081/5 = .16 = 16%\\ | ||
+ | \\ | ||
+ | CV (Coef. of Var.) = SD/M = 2.42/5 = .48 = 48%\\ | ||
+ | \\ | ||
+ | N (No of plots): \\ | ||
+ | N (For 1 SD) = (1)<sup>2</sup>(CV)<sup>2</sup>/(% Acc)<sup>2</sup> = | ||
+ | (.48)<sup>2</sup>/(.10)<sup>2</sup> = 23 plots\\ | ||
+ | N (For 2 SD) = (2)<sup>2</sup>(CV)<sup>2</sup>/(% Acc)<sup>2</sup> = | ||
+ | (4×.23)<sup>2</sup>/(.01)<sup>2</sup> = 92 plots\\ | ||
+ | N (For 1 SD ±20%)= (CV)<sup>2</sup>/(% Acc)<sup>2</sup> = .23/.04 = 6 plots. | ||
+ | |||
+ | ===No of Samples to be Taken from Infinite Population = n=== | ||
+ | ^CV% ^Specified % Limit - SE ^^^| | ||
+ | ^ ^± 1-1 1/2% ^±5% ^±10% ^±20% | | ||
+ | ^ ^N ^^^| | ||
+ | |10 |45 |4 |1 |1 | | ||
+ | |20 |178 |16 |4 |1 | | ||
+ | |30 |400 |36 |9 |3 | | ||
+ | |40 |712 |64 |16 |4 | | ||
+ | |50 |1,112 |100 |25 |7 | | ||
+ | |60 |1,600 |144 |36 |9 | | ||
+ | |70 |2,178 |196 |49 |13 | | ||
+ | |80 |2,845 |256 |64 |16 | | ||
+ | |90 |3,600 |324 |81 |21 | | ||
+ | |100 |4,445 |400 |100 |25 | | ||
+ | |150 |10,000 |900 |225 |57 | |