Question

Reserve Problems Chapter 11 Section 2 Problem 2 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 | 7420 | 15500 13620 19740 12800 x 2.20 3.86 3.58 4.29 3.02 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). Bi = (c) Calculate the fitted value of y corresponding to x = 3.02. 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 4.5 hectares. Round your answer to the nearest integer (e.g. 9876).

Answer 1

ANSWER :

The given information is

The given data is on the harvesting done of peanut crop. x is the field area in hectares and y is the mass of the crop in kilograms.

The data is:

x	y
2.20	7420
3.86	15500
3.58	13620
4.29	19740
3.02	12800

We are to first determine the linear regression on this equation. The regression model is:

y = Bo + B12

The formulae for the constants are:

βι ηΣαμ - ΣτΣ! ηΣ2 – (Στ)2 and Σ» - βιΣ Υ Bo n

Computing the required sums for the above formulae, we get:

	x	y	x^2	xy
	2.20	7420	4.84	16324
	3.86	15500	14.90	59830
	3.58	13620	12.82	48759.6
	4.29	19740	18.40	84684.6
	3.02	12800	9.12	38656
SUM	16.95	69080	60.16	248254.2

Substituting the values in the formulae, we get:

B1

5213.1876

Bo

—3856.7061

Therefore, the equation for the regression line is:

6

5213.1876.0 – 3856.7061

y = 5213.c – 3857

a)

We use the regression model to estimate $\sigma ^2$

S Sres
follows a chi-square distribution with n - 2 degrees of freedom, since 2 degrees of freedom are used in
Bo and 81

Therefore, a good estimate of $\sigma ^2$ is:

S Sres S 11 - 2

The formula for is:

S Sres

SSres => -Συ; – 3:02 e

Calculating the error:

	x	y	x^2	xy
	2.20	7420	4.84	16324
	3.86	15500	14.90	59830
	3.58	13620	12.82	48759.6
	4.29	19740	18.40	84684.6
	3.02	12800	9.12	38656
SUM	16.95	69080	60.16	248254.2

y	( - )
7611.6	36710.56
16265.18	585500.432
14805.54	1405505.09
18506.77	1520856.23
11886.26	834920.788
sum=69075.35	4383493.1

Therefore, estimate of $\sigma ^2$ is

s= 1461164.37

o2 = 1461164

b)

The change in mean mass that would be expected when the field area changes by 1 hectare is captured by the coefficient

B1

Therefore, this value is

B1 = 5213

c)

In order to determine the fitted value of y for x = 3.02,

we substitute x = 3.02 in the linear regression model.

Doing so, we get:

$\widehat{y}=11886.26$

The residual can be computed by subtracting the actual value of y_i from the predicted value y_hat.

Doing so, we get:

$\epsilon =y_{i}-\widehat{y}$

€= 913.74

d) In order to predict the mean mass of the crop harvested from 4.5 hectares, substitute x = 4.5

Substituting x = 4.5 in the equation, we get:

y = 19601.5