Calculations of Covariance, Correlation, and Measures of Central Tendency

Question 1(a).
x y x−x y− y (x−x)( y− y)
5 20 -0.5 1.5 -0.75
3 23 -2.5 4.5 -11.25
7 15 1.5 -3.5 -5.25
9 11 3.5 -7.5 -26.25
2 27 -3.5 8.5 -29.75
4 21 -1.5 2.5 -3.75
6 17 0.5 -1.5 -0.75
8 14 2.5 -4.5 -11.25
According to (Cunden, 2014) Covariance between x and y is given by;
cov ( x , y ) =
∑
i=1
n
(x ¿¿ i−x)( yi − y)
n−1 ¿
Where;
x Is the mean of the independent variable.
y isthe dependent variable
x is the independent variable
y is the dependent variable
n is number of data points in the sample
x=∑
i=1
n
xi
= 5+3+7+ 9+2+4 +6+8
8 =5.5
y=∑
i=1
n
yi
= 20+23+15+11+ 27+21+17+14
8 =18.5
∑
i=1
n
(x ¿¿ i−x)( yi − y)¿=-0.89

Covariance between x and y = ∑
i=1
n
(x ¿¿ i−x )( yi− y )
n−1 ¿=
= (-0.75) + (-11.25) + (-5.25) + (-26.25) + (-29.75) + (-3.75) + (-0.75) + (-
11.25)
= −89
7 =
-12.7143
Since the value is negative (-12.7143), it reveals that there is negative
relationship between variable x and y as shown
Question 1(b).
Covariance measures the variability of two variables. A negative covariance is
due that greater values of one variable results to a smaller value of the other
therefore, the two variables move in opposite direction. This, therefore, imply
that when one variable is decreased by one unit, the other variable increase
proportionately.
Question 1(c).
The coefficient of the correlation is calculated as follows
r ( x , y ) =cov (x , y )
sx s y
Where;
r (x, y) is correlation of the variables x and y
COV (x, y) is covariance of the variables x and y
Sx is the sample standard deviation of the random variable x
Sy is the sample standard deviation of the random variable y
x y x−x y− y (x−x)( y− y) (x ¿¿ i−x)2 ¿ ( y ¿¿ i− y )2 ¿
5 20 -0.5 1.5 -0.75 0.25 2.25
3 23 -2.5 4.5 -11.25 6.25 20.25
7 15 1.5 -3.5 -5.25 2.25 12.25
9 11 3.5 -7.5 -26.25 12.25 56.25

2 27 -3.5 8.5 -29.75 12.25 72.25
4 21 -1.5 2.5 -3.75 2.25 6.25
6 17 0.5 -1.5 -0.75 0.25 2.25
8 14 2.5 -4.5 -11.25 6.25 20.25
∑ (x ¿¿ i−x)2 ¿= 0.25 +6.25 +2.25+ 12.25 +12.25 +2.25+ 0.25 6.25
= 42
∑ ( y ¿¿ i− y )2=¿ ¿2.25+ 20.25 +12.25 +56.25+ 72.25+ 6.25+ 2.25
= 192
sx=
2
√ ∑
i=1
n
(x ¿¿ i−x)2
n−1 ¿ = √42/7= 2.449
And
sy=
2
√ ∑
i=1
n
( y ¿¿ i− y)2
n−1 ¿ =√192/7 = 5.237
From question 1(a), cov (x , y) = - 0.89
r ( x , y )= −0.89
2.449∗5.237 = - 0.06939
The result for the correlation coefficient is - 0.06939. The negative sign implies
that there is no linear relationship between variable x and y.
Question 1(d).
Negative correlation happens due to the imbalance between the two variables.
In the case of supply and demand, an increase of one variable result to the
corresponding decrease of the other variable by a proportionate unit.
Question 2(a).
The hypothesis is formulated as follows;
H0: P = 0.1