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:
= [(5 X 45.1) - (21 X 9.9) ] /(5 X 99 -
212 )
(5 X 20.79)
- 9.92 = 0.9827
H0: = 0
HA: 0
Test Statistics is given by:
= 0.9827 X 5 -
2/
(1-0.98272)
= 0.9827 X 1.7321/0.1852
= 9.1906
ndf = n - 2 = 5 - 2 = 3
= 0.05
From Table, critical values of t = 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.
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