Part 1
1)	Plot numbers	Vol M^3	y^2
	50.00	12.40
	47.00	19.70
	95.00	20.60
	99.00	17.80
	56.00	21.90
	116.00	17.40
	52.00	16.30
	102.00	22.10
	54.00	12.60
	68.00	12.10
2)	Est pop mean vol (Ybar)	17.29	m^3/0.04 ha
	Degrees of freedom = 9
	Est Pop. Variance (s^2)=	13.52	m^3/0.04 ha
	Standard error of mean (sybar) =	#ERROR!	m^3/0.04 ha
	95% CI for mean vol per plot
	t.inv (0.975,9)	2.26
	t*sybar	#ERROR!	m^3/0.04 ha
	Lower CI = Ybar-t*sybar	#ERROR!	m^3/0.04 ha
	Upper CI = Ybar+t*sybar	#ERROR!	m^3/0.04 ha
	95% CI	14.75 m^3/0.04ha < 17.2 m^3/0.04ha < 19.83 m^3/0.04ha
3)	YBar/ha = Ybar per plot x 25 plots/ha
	Uy	432.25	m^3/0.04 ha
	Lower CI	#ERROR!	m^3/0.04 ha
	Upper CI	#ERROR!	m^3/0.04 ha
	95% CI	368.87 m^3/0.04ha 432.25 m^3/0.04ha 495.63 m^3/0.04ha
4)	Ytotal = Ybar per plot x N	17.29 m^3/0.04 ha * 140	2420.6	m^3
	Sytotal = Sybar per plot x N	1.12 m^3/0.04ha*140	156.8	m^3
	95% CI for total area (5.6ha)
	t.inv (0.975,9)	2.26
	t*Sytotal	354.37
	Lower CI = ytotal-t*sytotal	2066.23
	Upper CI = ytotal+t*sytotal	2774.97
	95% CI	2066.23m^3/0.04ha 2420.26m^3/0.04ha 2774.97 m^3/0.04ha
5)	H0: Uy = 2000m^3
	H1: Uy < 2000 m^3
	t calculated = (Ybar - C)/Sybar	8.42
	t critical =	1.833112933
	t calculated > t critical, therefore we accept the null hypothesis because there is enough total volume to harvest that area.
6)	Estimating sample size for SRS
	a) +/- 35m^3/0.04ha = 1.6m^3/0.04ha
	1st example of estimating sample size for SRS
	1/N=1/140	0.007142857143
	PE	E/Ybar	9.25	%
	CV	Sy/Ybar*100	21.28	%
	Sy	17.29*0.05	3.68
	Let Ybar= 17.29; Sy^2= 13.52; E = 35%; =0.05; N = 140
	Let df = 100000		t	1.96
	n=	17.75	rounded to 18
	Let df = 17		t	2.11
	n=	20.16	rounded to 20
	let df= 19			2.09
	n=	19.88	rounded to 20
	Sample size of 20 required to achieve +/- 35m^3 ha
	2nd example of estimating sample size for SRS
	1/N=1/140	0.007142857143
	PE	E/Ybar	7.00	%
	CV	Sy/Ybar*100	21.28	%
	Sy	17.29*0.05	3.68
	Let Ybar= 17.29; Sy^2= 13.52; E = 7.00%; =0.05; N = 140
	Let df = 100000		t	1.96
	n=	28.33	rounded to 61
	Let df = 60		t	2.00
	n=	29.26	rounded to 63
	Let df= 62		t	2.00
	n=	29.23	rounded to 62
	Sample size of 63 is required to achieve an error of +/- 7.00%

part 2

Random start point	126
Plot #	83	62
	12	131
	94	45
	107	30
	21	127

2)
a)	0.1 x 20 ha	2 ha
b)	n=(I * A)/a	50 plots
c)	1)	B(m)=A(m^2)/n*L(m)
		200000
		4
		d(m) = sqrt(A(m^2)/n
		50/4 is 12.5 lines or 13 plots
		1000/13 is 76.92m is length
	2	sqrt (200000m^2/50 is 63.25m^2
		spacing between plots is 63.25m^2

copy of part 2 instructions

2. A forester is planning a survey with the following specifications: Intensity = 0.10 (i.e., 10 %) Total area = 20 ha (1000 m X 200 m; 1 ha = 10,000 m2 ) Size of plot = 0.04 ha a. Calculate the area of all plots of the forester's survey, combined (Ap).

b. Calculate the number of samples required for this desired sampling intensity. c.

(1) Given that the length between plots is fixed at 50 m, what is the length between lines?

(2) What would the spacing between plots be if you used square spacing?

d. Graph the sampling layout using the calculations above for square spacing. Make sure that the number of samples required equals the number to be sampled on your layout. Indicate on the layout how you would attempt to insure that the systematic sample approximates a random sample. Since the lines for the systematic survey will repeat, you may wish to only show the first two starting lines and the last two lines, and indicate that this will repeat over the 20 ha area.

e. Given a fixed cost of $1200 (i.e., truck rental), and a cost per plot of $60, what will be the cost of this survey

basically, just wondering if you can double check my math for part 1 and 2, part 1 if it is correct and part 2 what it will cost for e