Correlation and Regression Practice Problems

Correlation and Regression are statistical techniques used to study the relationship between two or more variables.

Problem 1: Given two variables, X and Y, calculate the Pearson correlation coefficient.

• X: [1, 2, 3, 4, 5]
• Y: [2, 4, 6, 8, 10]

Solution:

r = \frac{\sum (X - \bar{X})(Y - \bar{Y})}{\sqrt{\sum (X - \bar{X})^2 \sum (Y - \bar{Y})^2}}

Calculating the mean of X and Y:

• Mean of X (\bar{X}) = 3
• Mean of Y (\bar{Y}) = 6

Calculating the Pearson correlation coefficient:

r = \frac{(1-3)(2-6) + (2-3)(4-6) + (3-3)(6-6) + (4-3)(8-6) + (5-3)(10-6)}{\sqrt{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2} \sqrt{(2-6)^2 + (4-6)^2 + (6-6)^2 + (8-6)^2 + (10-6)^2}}= \frac{2*4 + 1*2 + 0 + 1*2 + 2*4}{\sqrt{4 + 1 + 0 + 1 + 4} \sqrt{16 + 4 + 0 + 4 + 16}} = \frac{20}{\sqrt{10} \cdot \sqrt{40}} = \frac{20}{\sqrt{400}} = \frac{20}{20} = 1

So, the Pearson correlation coefficient ( r = 1 ), indicating a perfect positive linear relationship between X and Y.

Problem 2: Calculate the Spearman's rank correlation coefficient for the following data:

• X: [10, 20, 30, 40, 50]
• Y: [30, 40, 10, 20, 50]

Solution:

1. Rank the data points in X and Y:

• Ranks of X: [1, 2, 3, 4, 5]
• Ranks of Y: [3, 4, 1, 2, 5]

2. Calculate the difference between ranks (d_i):

Differences (d): [2, 2, -2, -2, 0]

3. Calculate the square of differences (d_i^2)):

Squared differences (d²): [4, 4, 4, 4, 0]

4. Sum the squared differences:

\sum d_i^2 = 4 + 4 + 4 + 4 + 0 = 16

5. Use the Spearman's rank correlation formula:

r_s = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}

Here, n = 5:

r_s = 1 - \frac{6 \cdot 16}{5(5^2 - 1)} = 1 - \frac{96}{120} = 1 - 0.8 = 0.2

So, the Spearman's rank correlation coefficient n =5, indicating a weak positive rank correlation between X and Y.

Problem 3: Given data points, predict the value of Y for X = 6 using the equation y = a + bX.

• Data points: (1, 2), (2, 4), (3, 6), (4, 8), (5, 10)

Solution:

Calculate the slope (b) and intercept (a):

b = \frac{\sum (X - \bar{X})(Y - \bar{Y})}{\sum (X - \bar{X})^2}

\bar{X} = 3, \quad \bar{Y} = 6

b = \frac{(1-3)(2-6) + (2-3)(4-6) + (3-3)(6-6) + (4-3)(8-6) + (5-3)(10-6)}{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2} = \frac{20}{10} = 2

a = \bar{Y} - b\bar{X} = 6 - 2 \cdot 3 = 0

Equation: Y = 0 + 2X

For X = 6:

Y = 2 × 6 = 12

So, the predicted value of Y for X = 6 is 12.

Problem 4: Given the following data, predict the value of Y for X₁ = 3 and X₂ = 4:

Data:

• (X₁, X₂, Y)
• (1, 2, 3)
• (2, 3, 4)
• (3, 4, 5)
• (4, 5, 6)
• (5, 6, 7)

Solution:

The multiple linear regression equation is .

Using a statistical software or calculation tool (e.g., Excel, R), we can determine:

• b₁ = 0.5
• b₂ = 0.5
• a = 1.5

So, the equation becomes:

Y = 1.5 + 0.5X₁ + 0.5X₂

For X₁ = 3 and X₂ = 4:

Y = 1.5 + 0.5 × 3 + 0.5 × 4 = 1.5 + 1.5 + 2 = 5

So, the predicted value of Y for X₁ = 3 and X₂ = 4 is 5.

Problem 5: Given the following data, determine the probability of Y being 1 for X = 4:

Data:

• (X, Y)
• (1, 0)
• (2, 0)
• (3, 1)
• (4, 1)
• (5, 1)

Solution:

The logistic regression model is:

\log\left(\frac{p}{1-p}\right) = a + bX

Using a statistical software or calculation tool, we can determine:

• b = 1.1
• a = -3

So, the equation becomes:

\log\left(\frac{p}{1-p}\right) = -3 + 1.1X

For X = 4:

\log\left(\frac{p}{1-p}\right) = -3 + 1.1 \times 4 = 1.4

Solving for p:

\frac{p}{1-p} = e^{1.4} \approx 4.055

p \approx \frac{4.055}{1 + 4.055} \approx 0.802

So, the probability of Y being 1 for X = 4 is approximately 0.802, or 80.2%.