From variance to 95% confidence interval: Is this done right?

#1
Hello

I have a forest inventory project where I am trying to estimate the total forest wood volume (in cubic feet) with a 95% confidence interval. I am following a textbook method, but I am not 100% confident that the textbook has it right.

The sampling of individual trees is done with 1/10-acre circular plots, and a plot might have from 0 to 20 trees of 8 inches diameter or larger.

First, here are the actual plot volumes calculated on the first ten plots I’ve done so far, along with the volume per acre (VAC):

Plot Volumes VAC
198.4 1984
168.2 1682
49.8 498
210.2 2102
212.1 2121
37.8 378
155.0 1550
268.9 2689
99.6 996
105.4 1054

Note that the VAC amounts are ten times the volume per plot because there are ten plots in an acre.

Next I calculate the mean VAC which happens to be 1,505 cu. ft. The total forest is estimated to be 99 acres, so the total forest wood volume is estimated to be 1,505 * 99 = 148,585 cu. ft.

The sample variance of the list of VACs is computed using Excel’s VAR.S function and comes to 571,733. The variance of the mean of the VAC is defined by my textbook as the 571,733 divided by the number of plots (so far = 10), so, 57,173. This figure really is the VAC variance per plot. I don’t understand the “variance of the mean” terminology, but that’s another discussion.

Now the standard error of the mean is the square root of the 57,173, which comes to 239. But we want the standard error for the entire tract which requires the 239 to be multiplied by 99 acres, which equals 23,600.

The z-value for the 95% confidence level is 1.96, so the lower limit of the interval for our total volume estimate is 148,585 – 1.96 * 23,600, or 102,328, and the upper limit is 148,585 + 1.96 * 23,600, or 194,841. (I know this interval is wide, but the plot volumes are quite variable (0-20) and with only ten plots in so far.)

Are the procedure and result correct?

Thanks for your help.
 
#2
Yes ... but strictly you should use the t distribution rather than the normal, For 10 samples the factor is 2.26 rather than 1.96 but the difference gets less as the sample size increases. If you want the official factor use =TINV(0.05,samples - 1) in Excel.
 
#4
For sample sizes up to 100 or more, you are better off using 2 than 1.96 for 95% confidence intervals, if you haven't come across the t distribution yet.
 
#5
For sample sizes up to 100 or more, you are better off using 2 than 1.96 for 95% confidence intervals, if you haven't come across the t distribution yet.
Actually I don't think my sample sizes will ever exceed 30 samples. That's when the margin of error levels out after following a reverse-J progression. Also, the estimated total volume, after a chaotic beginning, settles down to a smooth horizontal line where additional samples rarely change it. If I had a client I'd have to ask how much they would be willing pay for all the sampling needed to get the margin of error down another point or two.