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Statisticshelpdesk
2014
Statistics Assignment Help Statistics Homework Help
Alex Gerg
Statisticshelpdesk
Copyright © 2012 Statisticshelpdesk.com, All rights reserved
Statistics Assignment Help | Statistics Homework Help
About Statistics: Statisticshelpdesk provides solution to all kind
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Statistics assignment help will be viewed in numerous ways
that. Normally it's thought-about to subsume graphs, charts,
percentages, and averages. It consists of rules and strategies of collecting and presenting
numerical data. It conjointly consists of constructing inferences from a given knowledge.
The statistical data will be used to elucidate unexplained things, to form and justify a claim,
to form comparisons, to seek out unknown quantities, to predict data regarding future and
to ascertain relationship between quantities. Thus, it's a subject matter that consists of
quite numbers.
Methods of Simple Regression Analysis:
Q 1. From the following data from the regression equations,
Y𝑒 = a + bX and 𝑋𝑒 = a + bY
Use the normal equation method:
X:
Y:
1
15
3
18
5
21
7
23
9
22
Also, estimate the value of Y when X=4, and the value of X when Y=24.
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Solution:
Formulation of the regression equations
X Y 𝑋2 𝑌2 XY
1
3
5
7
9
15
18
21
23
22
1
9
25
49
81
225
324
441
529
484
15
54
105
161
198
Total ∑X=25 ∑Y=99 ∑𝑋2=165 ∑𝑌2=2003
∑XY=533
N=5
(i) Regression equation of X on Y. This is given by
𝑋𝑒 = a + bY
To find the value of the constants a and b in the above formula, the following two normal
equation are to the simultaneously solved.
∑X=Na+b∑Y
∑XY=a∑Y+b∑𝑌2
Substituting the respective values in the above formula we get,
25=5a+99b
533=99a+2003b
Multiplying the equation (i) by 99 and eqn. (ii) by 5 and presenting them in the form of a
subtraction we get,
2475=495a+9801b
(-) 2665=495a+10015b
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_______________________________________
Thus -190= -214b
Or 214=190
b=190/214= .888 approx.
Putting the above values of b in the equ.(i) we get,
25=5a+99(.888)
Or 5a=25-87.912=-62.912
a= -62.912/5=-12.5824=-12.5824
Thus,
Substituting the above values of the constants a and b, we get the regression
equation of X on Y as,
𝐗𝐞 = -12.5824 + 0.888Y
Thus, when Y=24, 𝑋𝑒 = -12.5824 + 0.888(24)
=-12.5824+21.312
=8.7296
(ii) Regression equation of Y on X. this is given by
𝐘𝑒 = a + bX
as under:
To find the values of the constants a and b in the above formula, the following two normal
equations are to be simultaneously solved as under:
∑Y=Na+b∑X
∑XY=a∑X+b∑X2
Substituting the respective values in the above formula we get,
99=5a+25b
533=25a+165b
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Multiplying the equation (1) by 5 and getting the same subtracted from the equation
(ii) we get,
533=25a+165b
(-) 495=25a+125b
_______________________________________
Thus 38= 40b
Or b=38/40=15.05
Putting the above values of b in the equ.(i) we get,
99=5a+25(0.95)
Or 5a=99-23.75=75.25
a= 75.25/5=15.05
thus, a=15.05 and b=0.95
Substituting the above values of the constants a and b, we get the
regression equation of Y on x as,
𝑌𝑒 = 15.05 + 0.95X
Thus, when X=4, 𝑌𝑒 = 15.05 + 0.95(24)
=15.05+3.80
=18.85.
Note:
It may be noted that the above normal equation method of formulating the two regression
equation is verity lengthy and tedious. In order to do away with such difficulties, any of the
following two method of deviation may be used advantageously.
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