Question

based on the data shown below, calculate the correlation coefficient (to three decimal places)

x	y
6	17.8
7	17.1
8	15.8
9	16.3

Explanation:

Step1: List the data points

We have the following pairs of \((x, y)\): \((5, 20)\), \((6, 17.8)\), \((7, 17.1)\), \((8, 15.8)\), \((9, 16.3)\)

Step2: Calculate the necessary sums

First, we calculate \(n = 5\) (number of data points).

Calculate \(\sum x\):
\(\sum x=5 + 6+7 + 8+9=35\)

Calculate \(\sum y\):
\(\sum y = 20+17.8 + 17.1+15.8+16.3=87\)

Calculate \(\sum xy\):
\(5\times20 + 6\times17.8+7\times17.1+8\times15.8+9\times16.3\)
\(= 100+106.8 + 119.7+126.4+146.7\)
\(=100 + 106.8=206.8\); \(206.8+119.7 = 326.5\); \(326.5+126.4=452.9\); \(452.9 + 146.7=599.6\)

Calculate \(\sum x^{2}\):
\(5^{2}+6^{2}+7^{2}+8^{2}+9^{2}=25 + 36+49+64+81 = 255\)

Calculate \(\sum y^{2}\):
\(20^{2}+17.8^{2}+17.1^{2}+15.8^{2}+16.3^{2}\)
\(=400+316.84+292.41+249.64+265.69\)
\(400+316.84 = 716.84\); \(716.84+292.41=1009.25\); \(1009.25+249.64 = 1258.89\); \(1258.89+265.69=1524.58\)

Step3: Use the correlation coefficient formula

The formula for the correlation coefficient \(r\) is:

$$r=\frac{n\sum xy-\sum x\sum y}{\sqrt{[n\sum x^{2}-(\sum x)^{2}][n\sum y^{2}-(\sum y)^{2}]}}$$

Substitute the values we calculated:

First, calculate the numerator:
\(n\sum xy-\sum x\sum y=5\times599.6-35\times87\)
\(=2998 - 3045=- 47\)

Calculate the denominator part 1: \(n\sum x^{2}-(\sum x)^{2}=5\times255-35^{2}\)
\(=1275 - 1225 = 50\)

Calculate the denominator part 2: \(n\sum y^{2}-(\sum y)^{2}=5\times1524.58-87^{2}\)
\(=7622.9-7569 = 53.9\)

Then the denominator is \(\sqrt{50\times53.9}=\sqrt{2695}\approx51.913\)

Now, calculate \(r\):
\(r=\frac{- 47}{51.913}\approx - 0.905\)

Answer:

\(-0.905\)