Question

Compute the linear correlation coefficient between the two variables and determine whether a linear relation exists. Round to three decimal places X y 2 1.3 3 1.6 5 2.1 5 2.2 6 2.7 Click the icon to view the critical values table, A 0.983, a linear relation exists OB. r=0.883; a linear relation exists O C 0.883; no linear relation exists O D . r=0.983; no linear relation exists

Answer 1

Correct option:

A. r = 0.983. A linear relation exists.

Explanation:

From the given data, the following Table is calculated:

X	Y	XY	X2	Y2
2	1.3	2.6	4	1.69
3	1.6	4.8	9	2.56
5	2.1	10.5	25	4.41
5	2.2	11	25	4.84
6	2.7	16.2	36	7.29
Total = 21	9.9	45.1	99	20.79

Correlation Coefficient (r) is given by:

η ΣΧΥΣΥ ΣΥ Τ νηΣΧΣΧΡΣΥ ΣΥ n

= [(5 X 45.1) - (21 X 9.9) ] /$\sqrt{}$(5 X 99 - 21² ) $\sqrt{}$(5 X 20.79) - 9.9² = 0.9827

H0: $\rho$ = 0

HA: $\rho \neq$ 0

Test Statistics is given by:

rVn 2 t = 1-7

= 0.9827 X $\sqrt{}$5 - 2/$\sqrt{}$(1-0.9827²)

= 0.9827 X 1.7321/0.1852

= 9.1906

ndf = n - 2 = 5 - 2 = 3

$\alpha$ = 0.05

From Table, critical values of t = $\pm$ 3.1834

Since calculated value of t = 9.1906 is greater than critical value of t = 3.1834, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim the there is significant correlation.

So,

Correct option:

A. r = 0.983. A linear relation exists.