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|7240 | 15540 13500 20050 12920 * 2.20 3.80 3.12 4.06 2.70 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 02. Round your answer to the nearest integer (e.g. 9876). 131195554 = (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). ß = 3272 (c) Calculate the fitted value of y corresponding to x = 2.20. Find the corresponding residual. Round your answer to the nearest integer (e.g. 9876). û = 3739 Round your answer to the nearest integer (e.g. 9876). 3501 (d) Estimate the mean mass of the crop harvested from 5.0 hectares. Round your answer to the nearest integer (e.g. 9876). $ = 12901

Answer 1

Question:

The peanut crop was harvested form five field of various area. The following data are the mas of the crop from each field y (in kg) and field area x (in hectares).

y	x
7240	2.2
15540	3.8
13500	3.12
20050	4.06
12920	2.7

Answers:

First we we fit the regression model to the given data where y is dependent variable and x is independent variable.

By using excel we get,

	Coefficients	Standard Error	t Stat	P-value
Intercept	-4302.080642	3690.989	-1.16556	0.328036
Field area (x)	5715.39063	1135.952	5.031365	0.015131

Regression Statistics
Multiple R	0.945540963
R Square	0.894047713
Adjusted R Square	0.858730285
Standard Error	1742.537557
Observations	5

Residual output

Observation	observed y	Predicted y (ŷ)	Residuals=y -ŷ
1	7240	8271.778745	-1031.778745
2	15540	17416.40375	-1876.403753
3	13500	13529.93812	-29.9381247
4	20050	18902.40532	1147.594683
5	12920	11129.47406	1790.52594

The variable field area(x) is significant at 0.05 significance level. also the value of R- square and adjusted R-square is 0.89 and 0.85 respectively. So the model fit well for the data

The fitted model is,

ŷ= 5715.3906*x-4302.0806

a)  Fit the simple linear regression model using method of least square. find the estimate of σ².

σ² hat= SS residual/n-2

σ² hat =3036437.

b) What change in the mean mass is expected when the field area changes by 1 hectare?

  ^ˆβ₁₌  5715 kg.

c) Calculate the fitted value of y corresponding to x=2.20. find the corresponding residual.

  ŷ=8272

Residual corresponding to x=2.20 is,

e= y- ŷ= -1032

d) Estimate the mean mass of the crop harvested from 5 hectares.

   ŷ=24275

Explanation:

use To find out the estimation of o² we residual sum of squares and it's dif and a Shres h-k Where, n is 2 number of observation. k - number of regressos (including a constant) 3 (Y; - ) n-k goog311.413 (n=5 5-2 K=2 3036437.138 @z = 3036437 (b from model B = 5715. when field orea changes by t hectons, expected change in mean mass of yield is 5715 kilogrom. That is, © Calculate fitted value of y when x=2-20 The model is ŷ q 5715.3906** ST.15.3906** - 43020806

Y = 5715.39 (2.20) - 4303.0806 ŷ : 8271.778 and = 8273 the corresponding residual is e = y - y 7240-8271.778 -1031-7787 e - 1032 @ estimate the of mean mass of crop harvesteel from hectanesi ŷ = 5115-39 (5) -4302.0806 = 242740872 ŷ = 24275 The mean mass of cor yield the crop harvested from s hectares is 24275 kilograms