Question

The data below are a variation of the data presented in Exercise 12-4 Age (X) 50 34 12 36 18 6.20 1.40 6.05 3.30 8.05 (a) Calculate the correlation coefficient by hand. Use formula (12.1) on page 488. (A negative value should be indicated by a minus sign. Round your final answer to 3 decimal places). calc

Answer 1

- 0.352

Explanation:

Data provided in the question:

Age (X) 50 34 12 36 18

(Y) 6.20 1.40 6.05 3.30 8.05

Now,

correlation coefficient, r =
`(\sum((X-My)(Y-Mx)))/(√(((SSx)(SSy))))`

Here,

∑(X - Mx)² = SSx

∑(Y - My)² = SSy

Mx: Mean of X Values =
`(50+34+12+36+18)/(5)` = 30

My: Mean of Y Values =
`(6.20+1.40+6.05+3.30+8.05)/(5)` = 5

X - Mx & Y - My: Deviation scores

(X - Mx)² & (Y - My)²: Deviation Squared

(X - Mx)(Y - My): Product of Deviation Scores

Thus,

( X - Mx ) ( Y - My ) (X - Mx)² (Y - My)² (X - Mx)(Y - My)

20.0 1.20 400.0 1.44 24.0

4.0 -3.60 16.0 12.96 -14.4

- 18.0 1.05 324.0 1.102 -18.9

6.0 -1.70 36.0 2.89 -10.2

- 12.0 3.05 144.0 9.303 -36.6

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∑(X - Mx)² = 920.0

∑(Y - My)² = 27.695

∑(X - Mx)(Y - My) = -56.1

thus,

r =
`(-56.1)/(√(((920)(27.695))))`

or

r = - 0.352