Question

Consider the following set of data.

(18, 14), (28, 45), (60, 33), (81, 24), (106, 56), (122, 6)

(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.)

Sx =

Sy =

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

Answer 1

(a) Calculate the covariance of the set of data.

Formula for covariance is given as below:

Cov. = ∑(X – Xbar)(Y – Ybar) / (n – 1)

Calculation table is given as below:

We are given n = 6

No.	x	y	(X - Xbar)	(Y - Ybar)	(X - Xbar)(Y - Ybar)
1	18	14	-51.16667	-15.66667	801.6113339
2	28	45	-41.16667	15.33333	-631.2221361
3	60	33	-9.16667	3.33333	-30.55553611
4	81	24	11.83333	-5.66667	-67.05557611
5	106	56	36.83333	26.33333	969.9442339
6	122	6	52.83333	-23.66667	-1250.388986
Total	415	178			-207.6666667
Mean	69.1666667	29.66666667

Cov. = ∑(X – Xbar)(Y – Ybar) / (n – 1)

Cov. = -207.6666667 / (6 – 1)

Cov. = -207.6666667 / 5

Cov. = -41.5333333

Cov. = -41.53

Part b

Formula for standard deviation is given as below:

S = sqrt[∑(X – Xbar)^2/(n – 1)]

Calculation tables are given as below:

No.	x	(X - Xbar)^2
1	18	2618.028119
2	28	1694.694719
3	60	84.02783889
4	81	140.0276989
5	106	1356.694199
6	122	2791.360759
Total	415	8684.833333

Sx = sqrt(8684.833333/5)

Sx = 41.67693207

Sx = 41.677

Calculation table for Sy

No.	y	(Y - Ybar)^2
1	14	245.4445489
2	45	235.1110089
3	33	11.11108889
4	24	32.11114889
5	56	693.4442689
6	6	560.1112689
Total	178	1777.333333

Sy = sqrt(1777.333333/5)

Sy = 18.85382366

Sy = 18.854

(c) Calculate r, the coefficient of linear correlation, for the data in part (a).

Formula for correlation coefficient is given as below:

r = Cov.(x, y) / (Sx*Sy)

r = -41.53 / (41.677*18.854)

r = -0.0528521

r = -0.05