Question

The peanut crop was harvested from five fields of various area. The following data are the mass of the crop from each field y (in kilograms) and the field area X (in hectares). y | 7440 15560 13390 1993012860 4.17 2.67 x 2.38 3.78 3.61 Round your intermediate answers to four decimal places (e.g. 98.7654). (a) Fit the simple linear regression model using the method of least squares. Find the estimate of o. Round your answer to the nearest integer (e.g. 9876). (b) What change in the mean mass is expected when the field area changes by 1 hectare? Round your answer to the nearest integer (e.g. 9876). Ôi = (c) Calculate the fitted value of y corresponding to x = 3.78. Find the corresponding residual. Round your answer to the nearest integer (e.g. 9876). = Round your answer to the nearest integer (e.g. 9876). e = (d) Estimate the mean mass of the crop harvested from 5.0 hectares. Round your answer the nearest integer (e.g. 9876). ŷ

Answer 1

estimate for error variance is given by,

SSE c? = n = samplesize n-2


XY SSE = SSyy - Sxx

We'll Calculate the quantities as follows:

х у 2.38 7440 3.78 15560 3.61 1390 4.17 19930 2.67 12860 The independent variable is x, and the dependent variable is y. In order to compute the regression coefficients, the following table needs to be used: X y x*y x2 y2 2.38 7440 17707.2 5.6644 55353600 3.78 15560 58816.8 14.2884 242113600 3.61 1390 5017.9 13.0321 1932100 4.17 19930 83108.1 17.3889 397204900 2.67 12860 34336.2 7.1289 165379600 Sum = 16.61 57180 198986.2 57.5027 861983800 Based on the above table, the following is calculated: n 1 16.61 X = ΣΧ; = 3.322 п 5 i=1


n 57180 Y - Υ; 11436 n 5 1=1 2 SSXX -ΣΑ - (Σκ) = 57.5027 – 16.61? /5 = 2.32428 2 L 1 SSYy = ΣΥ? ΣΥ; = 861983800 – 57180? /5 = 208073320 n 1=1 1=1 η n n 1 SSXY = ΣΧΥ; ΣΧ; )(Σ ΣΥ; 198986.2 – 16.61 x 57180/5 9034.24 n 1=1 i=1 i=1 Therefore, based on the above calculations, the regression coefficients (the slope m, and the y- intercept n) are obtained as follows: 9034.24 m = SSXY SSxx - 3886.8983 2.32428 η =Y – Χ. m = 11436 – 3.322 x 3886.8983 : - –1476.2762 Therefore, we find that the regression equation is: y = -1476.2762 + 3886.89830

So , SSE = 208073320 - (9034.24)²/ 2.32428 = 172958147.8
Therefore,
2
= SSE/(n-2) = 172958147.8/(5-2) = 57652716.(nearest integer)

b)
$\widehat{\beta} = coefficient \: of \: x = 3886.8983$ = 3887

c)

y
for 3.78 is

y
= −1476.2762+3886.8983*3.78 = 13216.1993 = 13216
= y -
y
= 15560 - 13216.1993 = 2343.8007 = 2344(nearest integer)

d)

y
for x=5 is

y
=  −1476.2762+3886.8983*5 = 17958.215 = 17958(nearest integer)