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Based on the data shown below, calculate the correlation coefficient (rounded to three decimal places).

\begin{tabular}{|r|r|}
\hline
[tex]$x$[/tex] & \multicolumn{1}{|c|}{[tex]$y$[/tex]} \\
\hline
4 & 29.72 \\
\hline
5 & 32 \\
\hline
6 & 37.98 \\
\hline
7 & 40.76 \\
\hline
8 & 43.34 \\
\hline
9 & 45.32 \\
\hline
10 & 49.6 \\
\hline
11 & 50.68 \\
\hline
12 & 55.06 \\
\hline
13 & 57.94 \\
\hline
14 & 57.72 \\
\hline
15 & 61.5 \\
\hline
16 & 63.68 \\
\hline
\end{tabular}


Sagot :

To calculate the correlation coefficient between the two sets of data [tex]\( x \)[/tex] and [tex]\( y \)[/tex], we use the Pearson correlation coefficient formula:

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

where [tex]\( n \)[/tex] is the number of data points, [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are the individual data points, and [tex]\( \sum \)[/tex] denotes summation.

Here is a step-by-step solution:

### Step 1: Calculate the sums and squares

Given data:
[tex]\[ x = [4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16] \][/tex]
[tex]\[ y = [29.72, 32, 37.98, 40.76, 43.34, 45.32, 49.6, 50.68, 55.06, 57.94, 57.72, 61.5, 63.68] \][/tex]

Number of data points [tex]\( n \)[/tex]:
[tex]\[ n = 13 \][/tex]

Calculate [tex]\( \sum x \)[/tex], [tex]\( \sum y \)[/tex], [tex]\( \sum xy \)[/tex], [tex]\( \sum x^2 \)[/tex], and [tex]\( \sum y^2 \)[/tex]:

[tex]\[ \sum x = 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 130 \][/tex]

[tex]\[ \sum y = 29.72 + 32 + 37.98 + 40.76 + 43.34 + 45.32 + 49.6 + 50.68 + 55.06 + 57.94 + 57.72 + 61.5 + 63.68 = 625.3 \][/tex]

[tex]\[ \sum xy = (4 \times 29.72) + (5 \times 32) + (6 \times 37.98) + (7 \times 40.76) + (8 \times 43.34) + (9 \times 45.32) + (10 \times 49.6) + (11 \times 50.68) + (12 \times 55.06) + (13 \times 57.94) + (14 \times 57.72) + (15 \times 61.5) + (16 \times 63.68) = 7211.84 \][/tex]

[tex]\[ \sum x^2 = 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 = 1496 \][/tex]

[tex]\[ \sum y^2 = 29.72^2 + 32^2 + 37.98^2 + 40.76^2 + 43.34^2 + 45.32^2 + 49.6^2 + 50.68^2 + 55.06^2 + 57.94^2 + 57.72^2 + 61.5^2 + 63.68^2 = 31402.2872 \][/tex]

### Step 2: Substitute these values into the Pearson correlation coefficient formula

[tex]\[ r = \frac{13(7211.84) - (130)(625.3)}{\sqrt{[13(1496) - (130)^2][13(31402.2872) - (625.3)^2]}} \][/tex]

### Step 3: Simplify the numerator and denominator

Numerator:
[tex]\[ 13 \times 7211.84 - 130 \times 625.3 = 93754.92 - 81329 = 12425.92 \][/tex]

Denominator:
[tex]\[ \sqrt{[13 \times 1496 - 130^2][13 \times 31402.2872 - 625.3^2]} \][/tex]
[tex]\[ = \sqrt{[19448 - 16900][408229.7336 - 390014.09]} \][/tex]
[tex]\[ = \sqrt{2548 \times 18197.8436} \][/tex]
[tex]\[ = \sqrt{46340378.7528} \][/tex]
[tex]\[ = 6806.62 \][/tex]

### Step 4: Calculate [tex]\( r \)[/tex]

[tex]\[ r = \frac{12425.92}{6806.62} \approx 0.9929317948924539 \][/tex]

### Step 5: Round to three decimal places

[tex]\[ r \approx 0.993 \][/tex]

The correlation coefficient, rounded to three decimal places, is [tex]\( 0.993 \)[/tex].
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