MTH302 GDB Solution
Thursday, January 27, 2011 Posted In Mth Edit ThisTotal Marks 5
Starting Date : Thursday, January 27, 2011
Closing Date : Friday, January 28, 2011
Starting Date : Thursday, January 27, 2011
Closing Date : Friday, January 28, 2011
Construct a real world business problem and then apply simple linear regression analysis.
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Question:-
Suppose that 4 randomly chosen plots were treated with various level of fertilizer in the following yield of corn:-
Fertilizer(kg/Acre) X 100 200 400 500
Production(Bushels/Acre) Y 70 70 80 100
Estimate the Linear Regression of production Y on fertilizer X.
Solution:-
X Y XY X2
100 70 7000 10000
200 70 14000 40000
400 80 32000 160000
500 100 5000 250000
å = 1200 å = 320 å =103000 å = 460000
Byx = nåXY – åXåY / nåX2 - (åX) 2
Byx = 4(103000)-((1200)(320)) / 4(460000)-(1200) 2
Byx = 0.07
ayx = y - Byx X
= ((åY/n) – ((0.07) (åX/n))
=(320/4) – (0.07)(1200/4)
= 59
y = 59 + 0.07 X is required regression equation.
Suppose that 4 randomly chosen plots were treated with various level of fertilizer in the following yield of corn:-
Fertilizer(kg/Acre) X 100 200 400 500
Production(Bushels/Acre) Y 70 70 80 100
Estimate the Linear Regression of production Y on fertilizer X.
Solution:-
X Y XY X2
100 70 7000 10000
200 70 14000 40000
400 80 32000 160000
500 100 5000 250000
å = 1200 å = 320 å =103000 å = 460000
Byx = nåXY – åXåY / nåX2 - (åX) 2
Byx = 4(103000)-((1200)(320)) / 4(460000)-(1200) 2
Byx = 0.07
ayx = y - Byx X
= ((åY/n) – ((0.07) (åX/n))
=(320/4) – (0.07)(1200/4)
= 59
y = 59 + 0.07 X is required regression equation.
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Following is the detail of investment made and profit gained in a business by an investor for the last 5 years. Calculate simple linear regression y = a + bx
Profit (y) Investment(x)
100 1800
200 1600
300 3000
400 2500
500 1200
Solution:
Let y = a + bx be the simple linear regression where ‘b’ is the slope of regression line and ‘a’ is y-intercept.
y x x2 xy
100 1800 3240000 180000
200 1600 2560000 320000
300 3000 9000000 900000
400 2500 6250000 1000000
500 1200 1440000 600000
sum of y = 1500 sum of x = 10100 sum of x2 = 22490000 sum of xy = 3000000
mean y = sum of y/n = 1500/5 = 300
And
Mean x = sum of x/n = 10100/5 = 2020
As we know that
b = (n . sum of xy - sum of x . sum of y) / n . sum of x2 - (sum of x)(sum of x)
b = [5(3000000)-(10100)(1500)] / [5(22490000)-(10100)(10100)]
b = -0.000146
now
a = mean of y - b . mean of x
a = 300 - (-0.000146)(2020)
a = 300 + 0.296349
a = 300.296
Now regression line y on x is
y = 300.296-0.000146X
Profit (y) Investment(x)
100 1800
200 1600
300 3000
400 2500
500 1200
Solution:
Let y = a + bx be the simple linear regression where ‘b’ is the slope of regression line and ‘a’ is y-intercept.
y x x2 xy
100 1800 3240000 180000
200 1600 2560000 320000
300 3000 9000000 900000
400 2500 6250000 1000000
500 1200 1440000 600000
sum of y = 1500 sum of x = 10100 sum of x2 = 22490000 sum of xy = 3000000
mean y = sum of y/n = 1500/5 = 300
And
Mean x = sum of x/n = 10100/5 = 2020
As we know that
b = (n . sum of xy - sum of x . sum of y) / n . sum of x2 - (sum of x)(sum of x)
b = [5(3000000)-(10100)(1500)] / [5(22490000)-(10100)(10100)]
b = -0.000146
now
a = mean of y - b . mean of x
a = 300 - (-0.000146)(2020)
a = 300 + 0.296349
a = 300.296
Now regression line y on x is
y = 300.296-0.000146X
