Question

Consider the following set of data.

(21, 5), (33, 51), (59, 32), (83, 21), (113, 58), (119, 8)

(a) Calculate the covariance of the set of data. (Give your answer correct to two decimal places.)


(b) Calculate the standard deviation of the six x-values and the standard deviation of the six y-values. (Give your answers correct to three decimal places.)

(c) Calculate r, the coefficient of linear correlation, for the data in part (a). (Give your answer correct to two decimal places.)

Answer 1

Solution :

	x	y	(x-xbar)	(y-ybar)	(x-xbar)(y-ybar)	(x-xbar)^2	(y-ybar)^2
	21	5	-50.3333	-24.1667	1216.389	2533.444	584.0278
	33	51	-38.3333	21.83333	-836.944	1469.444	476.6944
	59	32	-12.3333	2.833333	-34.9444	152.1111	8.027778
	83	21	11.66667	-8.16667	-95.2778	136.1111	66.69444
	113	58	41.66667	28.83333	1201.389	1736.111	831.3611
	119	8	47.66667	-21.1667	-1008.94	2272.111	448.0278
Sum	428	175	0	0	441.6667	8299.333	2414.833
Mean	xbar = 71.33333	ybar = 29.16667

a)

$Covariance = \frac{\sum (x-\bar{x})(y-\bar{y})}{n-1} = \frac{441.6667}{6-1} = \frac{441.6667}{5} = \mathbf{88.33}$

b)

$Standard \;deviation\; of\; x\;\; S_{x} =\sqrt \frac{\sum (x-\bar{x})^2}{n-1} = \sqrt \frac{8299.333}{6-1} = \sqrt\frac{8299.333}{5}$

Si= 40.741

$Standard \;deviation\; of\; y\;\; S_{y} =\sqrt \frac{\sum (y-\bar{y})^2}{n-1} = \sqrt \frac{2414.833}{6-1} = \sqrt\frac{2414.833}{5}$

Sy = 21.977

c)

$r = \frac{Covariance(x,y)}{Standard \;deviation \;of\; x * Standard\; deviation\; of\; y} = \frac{Cov(x,y)}{S_{x}*S_{y}}$

$r = \frac{88.33}{40.741 * 21.977} = \frac{88.33}{895.365} = \mathbf{0.10}$