Question

.

For a sample of 12 trees, the volume of lumber (in m³) and the diameter ( in cm ) at a fixed

height above the ground level was measured. The results were as follows.

Diameter    Volume    Diameter    Volume

35.1 0.81 33.8    0.80

48.4 1.39    45.3    1.69

47.9 1.31    25.2    0.30

35.3 0.67 28.5 0.19

47.3 1.46    30.1    0.63

26.4    0.47    30.0 0.64

a)Construct a scatterplot of volume ( y ) versus diameter ( x ). using Excel

b)Compute the least-square line for predicting volume from diameter.

c)Compute the fitted value and residual for each point.

d)If two trees differ in diameter by 8 cm, by how much would you predict their volume to

differ?

e)Predict the volume of a tree whose diameter is 44 cm.

f)For what diameter would you predict a volume of 1m³

Answer 1

a)

b)

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
35.1	0.81	1.016736	0.002844	0.053778
48.4	1.39	151.0851	0.277378	6.473611
47.9	1.31	139.0434	0.199511	5.266944
35.3	0.67	0.653403	0.037378	0.156278
47.3	1.46	125.2534	0.356011	6.677694
26.4	0.47	94.25174	0.154711	3.818611
33.8	0.8	5.328403	0.004011	0.146194
45.3	1.69	84.48674	0.683378	7.598444
25.2	0.3	118.9917	0.317344	6.145028
28.5	0.19	57.88674	0.453378	5.122944
30.1	0.63	36.10007	0.054444	1.401944
30	0.64	37.31174	0.049878	1.364194

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	433.3	10.36	851.4092	2.590267	44.22567
mean	36.10833	0.863333	SSxx	SSyy	SSxy

sample size ,   n =   12          
here, x̅ =   36.10833333       ȳ =   0.863333333  
                  
SSxx =    Σ(x-x̅)² =    851.4091667          
SSxy=   Σ(x-x̅)(y-ȳ) =   44.22566667          
                  
slope ,    ß1 = SSxy/SSxx =   0.051944081          
                  
intercept,   ß0 = y̅-ß1* x̄ =   -1.012280856          
                  
so, regression line is   Ŷ =   -1.0123   +   0.0519   *x
c)

S.no	X	Y				Ŷ	residual= (Y-Ŷ)
1	35.1	0.81				0.811	0.00
2	48.4	1.39				1.502	-0.11
3	47.9	1.31				1.476	-0.17
4	35.3	0.67				0.821	-0.15
5	47.3	1.46				1.445	0.02
6	26.4	0.47				0.359	0.11
7	33.8	0.8				0.743	0.06
8	45.3	1.69				1.341	0.35
9	25.2	0.3				0.297	0.00
10	28.5	0.19				0.468125	-0.28
11	30.1	0.63				0.551236	0.08
12	30	0.64				0.546042	0.09

d) Ŷ =   -1.0123   +   0.0519   *x

If two trees differ in diameter by 8 cm,

their volume to differ by 0.0519*8 = 0.41555

e)Ŷ =   -1.0123   +   0.0519   *x

Ŷ =   -1.0123   +   0.0519   *44=

1.273
f)

Ŷ =   -1.0123   +   0.0519   *x

1 =   -1.0123   +   0.0519   *x

X=38.74 cm