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Reserve Problems Chapter 11 Section 2 Problem 2 The peanut crop was harvested from five fields of various area. The following

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Answer #1

x

y

X-X

y-ý

(y-7) (x - 1)

(x - 2)2

(y – 5)2

2.44

7450

-1.008

-6398

6449.184

1.016

40934404

4.09

15650

0.642

1802

1156.884

0.412

3247204

3.47

13480

0.022

-368

-8.096

0.000

135424

4.59

19840

1.142

5992

6842.864

1.304

35904064

2.65

12820

-0.798

-1028

820.344

0.637

1056784

Total

17.24

69240

0

0

15261.18

3.370

81277880

Mean

3.448

13848

The regression equation between two variables can be given by following equation

ỹ = 60 + b1 * *

The estimate of b0 and b1 can be obtained by least square method.

The least square estimate of b1 can be given by following expression

b1 : Σy - y) * (x – x) Σ(x – x)2

b1 15261.18 3.370

b1 = 4528.97

The estimate of b0 can be given by

b0 = 7 – b1*

b0 = 13848 - (4528.97) * 3.448

b0 = -1767.89

The final estimated simple linear regression model can be given as

ŷ = -1767.89+ 4528.97 * x

The ANOVA table of fitted regression is given by

Source

Df

SS

MS

F

p-value

Regression

1

b1*Sxy=4529*15261.18

         = 69117428

69117428

17.05

0.0258

Error

3

12160452

4053484

Total

n-1=5-1=4

SYY=81277880

Answer(a):

The estimate of σ2can be obtained by MSE.

Here we have, MSE =4053484

Hence

2 = 4053484

Answer(b):

The expected change in the mean mass when the field are changes by 1 hectare is given by the slope of simple linear regression model.

Hence

B1 = 4529

Answer(c):

The fitted value of y corresponding to x=4.59 can be obtained as

ŷ = -1767.89 +4528.97 * 4.59

y = 19020

The residual can be obtained as

e = y - ☺

e = 19840 – 19020

e = 820

Answer(d):

The estimate of mean mass of the crop harvested from 4 hectares can be obtained as

Y = -1767.89+ 4528.97 * 4

y = 16348

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