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# MTH302 GDB Solution & Discusion Last Date:28-01-2011

Construct a real world business problem and then apply simple linear regression analysis.

tomorrow is last date..!! anybody share the solution

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### Replies to This Discussion

tomorrow is last date..!! anybody share the solution

Demand for cars at its most basic is a function of the price and proxies for the economy, affordability, expectations and disposable income

Car demand = a + b(price) + cGDP + d average earnings + e interest rates + f unemployment + error

Obviously there could be many more things to include and they are likely to be correlated so judicious variable selection and model test is a must.

Errors have to be BLUE and you have to think carefully about including price without an instrumental variable.

I would not recommend a linear regression but an error correction model with a long run co-integrating relationship and short run stationary terms.
Question Example..

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

tariq bhai hm ny calculation he show karni hai k theory show karni hai??
karni parahay gi..

kia karni parhe ge calculation k theory show???:O

and agr theory show karni pare ge to phir os mein kia likhna pare ga???

Check this example or download the excel files for complete understanding and help

Regression Example:

To find the Simple/Linear Regression of

X Values Y Values
60 3.1
61 3.6
62 3.8
63 4
65 4.1

To find regression equation, we will first find slope, intercept and use it to form regression equation..

Step 1: Count the number of values.
N = 5

Step 2: Find XY, X2
See the below table

X Value Y Value X*Y X*X
60 3.1 60 * 3.1 = 186 60 * 60 = 3600
61 3.6 61 * 3.6 = 219.6 61 * 61 = 3721
62 3.8 62 * 3.8 = 235.6 62 * 62 = 3844
63 4 63 * 4 = 252 63 * 63 = 3969
65 4.1 65 * 4.1 = 266.5 65 * 65 = 4225

Step 3: Find ΣX, ΣY, ΣXY, ΣX2.
ΣX = 311
ΣY = 18.6
ΣXY = 1159.7
ΣX2 = 19359

Step 4: Substitute in the above slope formula given.
Slope(b) = (NΣXY - (ΣX)(ΣY)) / (NΣX2 - (ΣX)2)
= ((5)*(1159.7)-(311)*(18.6))/((5)*(19359)-(311)2)
= (5798.5 - 5784.6)/(96795 - 96721)
= 13.9/74
= 0.19

Step 5: Now, again substitute in the above intercept formula given.
Intercept(a) = (ΣY - b(ΣX)) / N
= (18.6 - 0.19(311))/5
= (18.6 - 59.09)/5
= -40.49/5
= -8.098

Step 6: Then substitute these values in regression equation formula
Regression Equation(y) = a + bx
= -8.098 + 0.19x.

Suppose if we want to know the approximate y value for the variable x = 64. Then we can substitute the value in the above equation.

Regression Equation(y) = a + bx
= -8.098 + 0.19(64).
= -8.098 + 12.16
= 4.06

This example will guide you to find the relationship between two variables by calculating the Regression from the above steps.
See the attached folder
Attachments:
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.
wow ... Tariq bhai, aap kitne Laik hain ...  ... Koooool ...
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

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