Let’s recalculate the correlation coefficient \( r \) to verify:

Given data: \( x = [1, 2, 3, 4] \), \( y = [2, 3, 7, 8] \), \( n = 4 \).

Calculations:

• \(\sum x = 1 + 2 + 3 + 4 = 10\)
• \(\sum y = 2 + 3 + 7 + 8 = 20\)
• \(\sum x^2 = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30\)
• \(\sum y^2 = 2^2 + 3^2 + 7^2 + 8^2 = 4 + 9 + 49 + 64 = 126\)
• \(\sum xy = (1 \cdot 2) + (2 \cdot 3) + (3 \cdot 7) + (4 \cdot 8) = 2 + 6 + 21 + 32 = 61\)
• Numerator: \( n \sum xy – \sum x \sum y = 4 \cdot 61 – 10 \cdot 20 = 244 – 200 = 44 \)
• Denominator: \(\sqrt{[n \sum x^2 – (\sum x)^2][n \sum y^2 – (\sum y)^2]} = \sqrt{[4 \cdot 30 – 10^2][4 \cdot 126 – 20^2]} = \sqrt{[120 – 100][504 – 400]} = \sqrt{20 \cdot 104} = \sqrt{2080} \approx 45.61\)
• \( r = \frac{44}{45.61} \approx 0.965 \)

The correlation coefficient \( r \approx 0.965 \) indicates:

\(\boxed{\text{A very strong positive linear relationship between } x \text{ and } y}\)

Using least squares regression with updated \( r \):

Slope \( a = r \cdot \frac{s_y}{s_x} \), where \( s_x = \sqrt{\frac{\sum x^2 – (\sum x)^2/n}{n-1}} = \sqrt{\frac{30 – 10^2/4}{3}} = \sqrt{\frac{5}{3}} \approx 1.291 \),

\( s_y = \sqrt{\frac{\sum y^2 – (\sum y)^2/n}{n-1}} = \sqrt{\frac{126 – 20^2/4}{3}} = \sqrt{\frac{26}{3}} \approx 2.943 \),

\( a = 0.965 \cdot \frac{2.943}{1.291} \approx 2.2 \),

Intercept \( b = \bar{y} – a \bar{x} = 5 – 2.2 \cdot 2.5 = 5 – 5.5 = -0.5 \).

\(\boxed{y = 2.2x – 0.5}\)

The coefficient \( a = 2.2 \) represents:

\(\boxed{\text{For each 1 unit increase in } x, y \text{ increases by 2.2 units}}\)

The constant \( b = -0.5 \) represents:

\(\boxed{\text{The predicted value of } y \text{ is -0.5 when } x = 0}\)