UNIT-1
CORRELATION REGRESSION AND CURVE-FITTING
Q1) Ten students got the following percentage of marks in Economics and Statistics
Calculate the 
of correlation.
Roll No. |
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Marks in Economics |
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Marks in |
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A1)
Let the marks of two subjects be denoted by 
and 
respectively.
Then the mean for 
marks 
and the mean ofy marks 
and
are deviations ofx’s and 
’s from their respective means, then the data may be arranged in the following form:
x | y | X=x-65 | Y=y-66 | X2 | Y2 | XY |
78 36 98 25 75 82 90 62 65 39 | 84 51 91 60 68 62 86 58 53 47 | 13 -29 33 -40 10 17 25 -3 0 -26 | 18 -15 25 -6 2 -4 20 -8 -13 -19 | 169 841 1089 1600 100 289 625 9 0 676 | 324 225 625 36 4 16 400 64 169 361 | 234 435 825 240 20 -68 500 24 0 494
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Q 2) Compute coefficient of correlation by Karl Pearson Method for the following data.
X: | 1800 | 1900 | 2000 | 2100 | 2200 | 2300 | 2400 | 2500 | 2600 |
F: | 5 | 5 | 6 | 9 | 7 | 8 | 6 | 8 | 9 |
A2)
Let the A.M.s 
and 
be 2200 and 6 for X and Y series respectively.
X | Y | dx | (i=100) dx | dy | d | d | dxdy |
1800 1900 2000 2100 2200 2300 2400 2500 2600 | 5 5 6 9 7 8 6 8 9 | -400 -300 -200 -100 0 100 200 300 400 | -4 -3 -2 -1 0 1 2 3 4 | -1 -1 0 3 1 2 0 2 3 | 16 9 4 1 0 1 4 9 16 | 1 1 0 9 1 4 0 4 9 | 4 3 0 -3 0 2 0 6 12 |
N=9 |
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(Note: We can also proceed dividing X by 100)
Q 3) Let 
random variables. Define
X = 
Y = 
where 
is real number in (-1, 1).
A3)
first, note that since 
are normal and independent, they are jointly normal, with the joint PDF.
= 
(a). we need to show that aX + bY is normal for all a,b 
R, we have
aX + bY = 
= 
Which is linear combination of 
and thus it is normal.
(b). we can use the method of transformations to find the joint PDF of X and Y.
The inverse transformation is given by
Where J = det
= det 
Thus, we conclude that
= 
= 
(c). To find 
(X, Y), first note that
Var(X) = Var(Z1) = 1,
Var(Y) = 
Therefore, 
(X, Y) = Cov (X, Y)
= Cov( 
=
Cov( 
.
= 
.1 + 
. 0
= 
.
Q 4) Let X and Y be jointly normal random variables with parameters 
Find the conditional distribution of Y given X =x.
A4) one way to solve this problem is by using PDF formula. In particular, since X 
N (
x, 
), we can use
.
Thus, given X = x we have,
and 
.
.+
.
Since 
are independent, knowing 
does not provide any information on 
. We have shown that given X =x, Y is a linear function of 
, thus it is normal. In particular
E[Y|X =x ]= 
+ 
E
= 
,
Var (Y|X = x) = 
= (1 - 
We conclude that given X = x, Y is normally distributed with mean 
and variance (1 - 
.
Q 5: Let X and Y be jointly normal random variables with parameters 
A5) a. Since X and Y are jointly normal, the random variable V=2X+Y is normal. We have
Thus, V 
Therefore,
b. Note that Cov (X, Y) = 
Cov (X+Y, 2X-Y) = 2Cov (X, X)-Cov (X, Y) +2Cov (Y, X)-Cov (Y, Y)
= 2-1+2-4 = -1.
d. Using Properties, we conclude that given X = 2, Y is normally distributed with
Thus,
Q 6:Two Way Frequency Tables:
Student Grades in Science Projects.
| Male | Female |
A | 9 | 12 |
B | 18 | 14 |
C | 8 | 11 |
D | 2 | 3 |
F | 1 | 2 |
A6)
| Male | Female | Total |
A | 9 | 12 | 21 |
B | 18 | 14 | 32 |
C | 8 | 11 | 19 |
D | 2 | 3 | 5 |
F | 1 | 2 | 3 |
Total | 38 | 42 | 80 |
Q: How Many students earned a grade of A?
Ans: 21 Students
Q: How many males were surveyed?
Ans: 38 Male Students
Q: How many males earned a grade of A?
Ans: 9 Male Students
Q: How many students earned a grade of B or C:
Ans: 51 Students
Q 7:
x | 1 2 3 4 5 6 7 |
y | 0.5 2.5 2.0 4.0 3.5 6.0 5.5 |
A7)
: y = 0.07143+0.8393x.
Q 8: Fit a least square line for the following data. Also find the trend values and show that ∑(Y–
)=0 ∑(Y–
)=0.
X | 1 | 2 | 3 | 4 | 5 |
Y | 2 | 5 | 3 | 8 | 7 |
A8)
X | Y | XY | X2 |
| Y– |
1 | 2 | 2 | 1 | 2.4 | -0.4 |
2 | 5 | 10 | 4 | 3.7 | +1.3 |
3 | 3 | 9 | 9 | 5.0 | -2 |
4 | 8 | 32 | 16 | 6.3 | 1.7 |
5 | 7 | 35 | 25 | 7.6 | -0.6 |
∑X=15 | ∑Y=25 | ∑XY=88 | ∑X 2=55 | Trend Values | ∑(Y-
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The equation of least square line Y=a +bX
Normal equation for ‘a’ ∑Y=na +b 25=5a+15b —- (1)
Normal equation for ‘b’ ∑XY = a∑X+b∑X2 88=15a+55b —-(2)
Eliminate a a from equation (1) and (2), multiply equation (2) by 3 and subtract from equation (2).
Eliminate a from equation (1) and (2), multiply equation (2) by (3) and subtract from equation (2). Thus, we get the values of a and b
Here a=1.1 and b=1.3, the equation of least square line becomes
Y=1.1+1.3X
Q 9: Using least square method to fit a straight line of the following data
X | 8 | 2 | 11 | 6 | 5 | 4 | 12 | 9 | 6 | 1 |
Y | 3 | 10 | 3 | 6 | 8 | 12 | 1 | 4 | 9 | 14 |
A9)
First, we calculate 
for the given data
Now we calculate 
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1 | 8 | 3 | 1.6 | -4 | -6.4 | 2.56 |
2 | 2 | 10 | -4.4 | 3 | -13.2 | 19.36 |
3 | 11 | 3 | 4.6 | -4 | -18.4 | 21.16 |
4 | 6 | 6 | -0.4 | -1 | 0.4 | 0.16 |
5 | 5 | 8 | -1.4 | 1 | -1.4 | 1.96 |
6 | 4 | 12 | -2.4 | 5 | -12 | 5.76 |
7 | 12 | 1 | 5.6 | -6 | -33.6 | 31.36 |
8 | 9 | 4 | 2.6 | -3 | -7.8 | 6.76 |
9 | 6 | 9 | -0.4 | 2 | -0.8 | 0.16 |
10 | 1 | 14 | -5.4 | 7 | -37.8 | 29.16 |
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Calculate the slope
m = 
= -131/118.4 
calculate the y-intercept
use the formula to calculate the y-intercept
b = 
= 7-(-1.1*6.4)
The required line equation is
Y= -1.1x+14.0
Q 10: Determine the constants a and b by the method of least square such that 
X | 2 | 4 | 6 | 8 | 10 |
Y | 4.077 | 11.084 | 30.128 | 81.897 | 222.62
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A10)
The given relation is 
Taking logarithms on both sides we get,
log y = log a+ bx…. (1)
let,
log y = Y
x = X
log a = A
b = B
now we have,
…. (2)
…. (3)
Now we need to find 
X=x | Y =ln(y) |
| xy |
2 | 1.405 | 4 | 2.810 |
4 | 2.405 | 16 | 9.620 |
6 | 3.405 | 36 | 20.430 |
8 | 4.405 | 44 | 35.240 |
10 | 5.405 | 100 | 54.050 |
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The normal equations to fit the straight line is
Y = logey
Y= ln(y)
17.025 = 5A +30B…. (4) 
122.150 = 30A+220B…. (5)
By solving 4 and 5 we get
30A +180B = 102.15… (4)
30A+220B = 122.150… (5)
We get a = 0.405, b = 0.5
A =log a 
a = 1.499
since we have X=x and Y=y
log y=Y,
And we know y= aebx
Y = (1.499) e0.5x is the required exponential curve.
Q 11) Fit the curve of the form y= aebx for the following data
X | 0 | 2 | 4 |
Y | 8.12 | 10 | 31.82 |
A11)
The given relation is 
Taking logarithms on both sides we get,
log y = log a+ bx logee…. (1)
the required normal equations are,
…. (2)
…. (3), We have n=3
X | y | Y= logey | xy | X2 |
0 | 8.12 | 2.0943 | 0 | 0 |
2 | 10 | 2.3026 | 4.6052 | 4 |
4 | 31.82 | 3.4601 | 13.8404 | 16 |
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The normal equations become
3A +6b = 7.8750
6A + 20 b = 18.4456
By solving the above two equations we get
A = 1.361 and b = 0.3415
Since A =logea 
a = e1.361 = 6.9317
The curve of the fit is 
Thus, the required equation is,
Q 12:
X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Y | 2 | 6 | 7 | 8 | 10 | 11 | 11 | 10 | 9 |
A12)
X |
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1 | -4 | 2 | 16 | -64 | 256 | -8 | 32 |
2 | -3 | 6 | 9 | -27 | 81 | -8 | 54 |
3 | -2 | 7 | 4 | -8 | 16 | -14 | 28 |
4 | -1 | 8 | 1 | -1 | 1 | -8 | 8 |
5 | 0 | 10 | 0 | 0 | 0 | 0 | 0 |
6 | 1 | 11 | 1 | 1 | 1 | 11 | 11 |
7 | 2 | 11 | 4 | 8 | 16 | 22 | 44 |
8 | 3 | 10 | 9 | 27 | 81 | 30 | 90 |
9 | 4 | 9 | 16 | 64 | 256 | 36 | 144 |
N=0 |
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∑Y i =Na + b ∑X i +c∑
∑X i Y i =a ∑X i +b∑
+c∑
∑
Y i =a∑
+b∑
+c∑
The required parabola is of the form y= ax2+bx+c
∴74=9a+b (0) +60c∴9a+60c=74…(i)
51=a (0) +60b+0c ∴60b=51 ∴b=5160 =0.85411=60a+0b+708 c∴60a+708c=411…(ii)
Solving (i) and (ii) simultaneously, we get
a=10.004, c=-0.267
The Equation of parabola is therefore,
y=10.004+0.85X−0.267X 2
=10.004+0.85(x−5) −0.267(x−5) 2
=10.004+0.85x−4.25−0.267(x 2 −10x+25)
=10.004+0.85x−4.25−0.267x 2 +2.67x−6.675
∴ y = −0.921+3.52x−0.267x 2
Q13) Find the least square approximation of degree two to the data
X | 0 | 1 | 2 | 3 | 4 |
Y | -4 | -1 | 4 | 11 | 20 |
A13)
X | y | xy |
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0 | -4 | 0 | 0 | 0 | 0 | 0 |
1 | -1 | -1 | 1 | -1 | 1 | 1 |
2 | 4 | 8 | 4 | 16 | 8 | 16 |
3 | 11 | 33 | 9 | 99 | 27 | 81 |
4 | 20 | 80 | 16 | 320 | 64 | 256 |
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the normal equations are:
Here,
n = 5, 
by substituting all the above values in normal equations, we get,
30 = 5a+10b+30c
120=10a+30b+100c
434=30a+100b+354c
By solving the above equations, we get
a = -4, b=2, c=1.
Therefore, the required polynomial is
Y= -4x+2x+x2 and errors =0.
