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). Y7120 | 1553013600 19860 13090 x 2.65 4.06 3.39 4.01 2.60 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 o2 Round your answer to the nearest integer (e.g. 9876). 2 = (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). B1- (c) Calculate the fitted value of y corresponding to x = 3.39. Find the corresponding residual. Round your answer to the nearest integer (e.g. 9876). Round your answer to the nearest integer (e.g. 9876). (d) Estimate the mean mass of the crop harvested from 4.0 hectares. Round your answer to the nearest integer (e.g. 9876). y =

Answer 1

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
2.65	7120
4.06	15530
3.39	13600
4.01	19860
2.6	13090

We are to first determine the linear regression on this equation. The regression model is: $y = \beta_0 + \beta_1 x$

The formulae for the constants are:

$\\\beta_1 = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}\\\\ and\\\\ \beta_0 = \frac{\sum y - \beta_1 \sum x}{n}$

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

	x	y	x^2	xy
	2.65	7120	7.0225	18868
	4.06	15530	16.4836	63051.8
	3.39	13600	11.4921	46104
	4.01	19860	16.0801	79638.6
	2.6	13090	6.76	34034
SUM	16.71	69200	57.8383	241696.4

Substituting the values in the formulae, we get:

$\\\beta_1 = \frac{5\times 241696.4 - 16.71\times 69200}{5\times 57.8383 - 16.71^2}\\\\ \therefore \beta_1 = \frac{1208482.0 - 1156332.0}{289.1915 - 279.2241}\\\\ \therefore \beta_1 = \frac{52150}{9.9674}\\\\ \therefore \beta_1 = 5232.0565\\\\ and\\\\ \beta_0 = \frac{69200 - 5232.0565\times 16.71}{5}\\\\ \therefore \beta_0 = \frac{69200 - 87427.6641}{5}\\\\ \therefore \beta_0 = \frac{-18227.6641}{5}\\\\ \therefore \beta_0 = -3645.5328$

Therefore, the equation for the regression line is:

9 = 5232.0565 - 3645.5328 :. 9 = 5232.1 5232.c – 3646

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

$SS_{res}$ follows a chi-square distribution with n - 2 degrees of freedom, since 2 degrees of freedom are used in

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

$s = \frac{SS_{res}}{n - 2}$

The formula for $SS_{res}$ is:

SSres = Σ? =Σ(9 – 9:) e

Calculating the error:

	x	y	x^2	xy	$\hat{y}$	$e_i^2$
	2.65	7120	7.0225	18868	10218.8	9602561.44
	4.06	15530	16.4836	63051.8	17595.92	4268025.446
	3.39	13600	11.4921	46104	14090.48	240570.6304
	4.01	19860	16.0801	79638.6	17334.32	6379059.462
	2.6	13090	6.76	34034	9957.2	9814435.84
SUM	16.71	69200	57.8383	241696.4	69196.72	30304652.82

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

$\\s = \frac{30304652.82}{5 - 2}\\\\ \therefore s = \frac{30304652.82}{3}\\\\ \therefore s = 10101550.94\\\\ \therefore \sigma ^2 \approx 10101551$

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 $\beta_1 = 5232$

c) In order to determine the fitted value of y for x = 3.39, we substitute x = 3.39 in the linear regression model. Doing so, we get:

$\\\therefore \hat{y} = 5232\times 3.39 - 3646\\\\ \therefore \hat{y} = 17736.48 - 3646\\\\ \therefore \hat{y} = 14090.48$

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

Doing so, we get:

$\\e = y_i - \hat{y}\\\\ \therefore e = 13600 - 14090.48\\\\ \therefore e = -490.48$

d) In order to predict the mean mass of the crop harvested from 4.0 hectares, substitute x = 4.

Substituting x = 4 in the equation, we get:

$\\\therefore \hat{y} = 5232\times 4 - 3646\\\\ \therefore \hat{y} = 20928 - 3646\\\\ \therefore \hat{y} = 17282$

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