The following data on sale price, size, and land-to-building ratio for 10 large

Question

The following data on sale price, size, and land-to-building ratio for 10 large industrial properties appeared in a paper.

Property Sale Price
(millions
of dollars) Size
(thousands
of sq. ft.) Land-to-
Building
Ratio
1 10.5 2165 2.1
2   2.7   750 3.6
3 30.5 2421 3.6
4   1.8   224 4.6
5 19.9 3917 1.6
6   8.1 2866 2.2
7 10.1 1699 3.0
8   6.6 1045 4.7
9   5.7 1107 7.6
10   4.5   406 17.3  

(a) Calculate the value of the correlation coefficient between sale price and size. (Give the answer to three decimal places.)
r =  

(b) Calculate the value of the correlation coefficient between sale price and land-to-building ratio. (Give the answer to three decimal places.)
r =  
(c) If you wanted to predict sale price and you could use either size or land-to-building ratio as the basis for making predictions, which would you use?
SizeLand-to-building ratio    

(d) Based on your choice in Part (c), find the equation of the least-squares regression line you would use for predicting y = sale price. (Give answers to three decimal places.)

Explanation / Answer

PART A.

calculation procedure for correlation
sum of (x) = x = 100.4
sum of (y) = y = 16600
sum of (x^2)= x^2 = 1710.96
sum of (y^2)= y^2 = 40086898
sum of (x*y)= x*y = 232357.9
to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)
covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1
= 232357.9 - [ 10 * (100.4/10) * (16600/10) ]/10- 1
= 6569.39
and now to calculate r( x,y) = 6569.39/ (SQRT(1/10*232357.9-(1/10*100.4)^2) ) * ( SQRT(1/10*232357.9-(1/10*16600)^2)
=6569.39 / (8.3842*1119.4149)
=0.7
value of correlation is =0.7

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PART B.

calculation procedure for correlation
sum of (x) = x = 100.4
sum of (y) = y = 50.3
sum of (x^2)= x^2 = 1710.96
sum of (y^2)= y^2 = 447.03
sum of (x*y)= x*y = 382
to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)
covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1
= 382 - [ 10 * (100.4/10) * (50.3/10) ]/10- 1
= -12.3012
and now to calculate r( x,y) = -12.3012/ (SQRT(1/10*382-(1/10*100.4)^2) ) * ( SQRT(1/10*382-(1/10*50.3)^2)
=-12.3012 / (8.3842*4.4048)
=-0.3331
value of correlation is =-0.3331

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PART C.

We use size, as it gives positive corrleation and it would be better in understandig the structure of data

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PART D.

Line of Regression Y on X i.e Y = bo + b1 X

calculation procedure for regression

mean of X = X / n = 10.04

mean of Y = Y / n = 1660

(Xi - Mean)^2 = 702.944

(Yi - Mean)^2 = 12530898

(Xi-Mean)*(Yi-Mean) = 65693.9

b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2

= 65693.9 / 702.944

= 93.45538

bo = Y / n - b1 * X / n

bo = 1660 - 93.45538*10.04 = 721.70797

value of regression equation is, Y = bo + b1 X

Y'=721.708+93.455* X

( X) ( Y) X^2 Y^2 X*Y 10.5 2165 110.25 4687225 22732.5 2.7 750 7.29 562500 2025 30.5 2421 930.25 5861241 73840.5 1.8 224 3.24 50176 403.2 19.9 3917 396.01 15342889 77948.3 8.1 2866 65.61 8213956 23214.6 10.1 1699 102.01 2886601 17159.9 6.6 1045 43.56 1092025 6897 5.7 1107 32.49 1225449 6309.9 4.5 406 20.25 164836 1827

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