Correlation & Regression: Scatterplots, Lines, and Predictions

TAKEAWAYS:

• Correlation measures how two variables move together (positive, negative, or no relationship).
• Regression finds the best-fit line to predict values (linear regression is most common).
• Pearson’s r quantifies correlation strength (–1 to +1), while r² shows how much variation is explained.
• Least squares method minimizes errors to find the regression line’s slope and intercept.
• Residuals help check how well the model fits the data.
• Applications include economics, biology, and engineering for forecasting and decision-making.

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What is Correlation?

Correlation studies how two variables change together. For example:

• Height vs. Weight: As height increases, weight usually increases → positive correlation.
• Study Hours vs. Exam Scores: More study time → higher scores → positive correlation.
• Ice Cream Sales vs. Temperature: Higher temperature → more ice cream sold → positive correlation.
• Rainfall vs. Sunburn Cases: More rain → fewer sunburns → negative correlation.

Types of Correlation

• Positive Correlation (r > 0): Both variables increase together.
• Negative Correlation (r < 0): One increases as the other decreases.
• No Correlation (r ≈ 0): No clear relationship.

Measuring Correlation: Pearson’s Correlation Coefficient (r)

Pearson’s r measures the strength and direction of a linear relationship: $$r=\frac{n(\sum XY)-(\sum X)(\sum Y)}{\sqrt{[n\sum X^2-(\sum X{)}^2][n\sum Y^2-(\sum Y{)}^2]}}$$ Where:

• $n$ = number of data points,
• $\sum XY$ = sum of the product of paired scores,
• $\sum X$ and $\sum Y$ = sums of X and Y,
• $\sum X^2$ and $\sum Y^2$ = sums of squared X and Y.

Example 1: Calculate r for Height (X) and Weight (Y)

• $n=5$
• $\sum X=850$, $\sum Y=300$
• $\sum XY=41500$
• $\sum X^2=145500$, $\sum Y^2=18250$

$$r=\frac{5(41500)-(850)(300)}{\sqrt{[5(145500)-850^2][5(18250)-300^2]}}$$ $$r=\frac{207500-255000}{\sqrt{(727500-722500)(91250-90000)}}$$ $$r=\frac{-47500}{\sqrt{(5000)(1250)}}=\frac{-47500}{\sqrt{6,250,000}}=\frac{-47500}{2500}=-1.9$$ Correction: The calculation above has an error. Let’s re-calculate carefully: $$\text{Numerator}=5(41500)-(850)(300)=207500-255000=-47500$$ $$\text{Denominator part 1}=5(145500)-850^2=727500-722500=5000$$ $$\text{Denominator part 2}=5(18250)-300^2=91250-90000=1250$$ $$\text{Denominator}=\sqrt{5000\times 1250}=\sqrt{6,250,000}=2500$$ $$r=\frac{-47500}{2500}=-19\text{(This is impossible!)}$$ Mistake Identified: The values in the table are unrealistic (e.g., height 150 cm for a 12th grader is too short). Let’s use realistic data:

Revised Example:

Calculations:

• $n=5$
• $\sum X=850$, $\sum Y=325$
• $\sum XY=55500$
• $\sum X^2=144750$, $\sum Y^2=21375$

$$r=\frac{5(55500)-(850)(325)}{\sqrt{[5(144750)-850^2][5(21375)-325^2]}}$$ $$r=\frac{277500-276250}{\sqrt{(723750-722500)(106875-105625)}}$$ $$r=\frac{250}{\sqrt{(1250)(1250)}}=\frac{250}{1250}=0.2$$ Interpretation: A weak positive correlation (r = 0.2) exists between height and weight in this data.

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What is Regression?

Regression finds the best-fit line (or curve) to predict one variable (dependent variable, Y) from another (independent variable, X). The linear regression equation is: $$Y=a+bX$$ Where:

• $a$ = y-intercept (value of Y when X = 0),
• $b$ = slope (change in Y for a unit change in X).

Calculating the Regression Line

The slope ($b$) and intercept ($a$) are calculated using: $$b=\frac{n(\sum XY)-(\sum X)(\sum Y)}{n(\sum X^2)-(\sum X{)}^2}$$ $$a=\frac{\sum Y-b(\sum X)}{n}$$

Example 2: Find the Regression Line for Height (X) and Weight (Y) Using the same data as above:

• $n=5$, $\sum X=850$, $\sum Y=325$,
• $\sum XY=55500$, $\sum X^2=144750$.

Calculate slope (b): $$b=\frac{5(55500)-(850)(325)}{5(144750)-850^2}$$ $$b=\frac{277500-276250}{723750-722500}=\frac{1250}{1250}=1$$

Calculate intercept (a): $$a=\frac{325-(1)(850)}{5}=\frac{325-850}{5}=\frac{-525}{5}=-105$$

Regression Equation: $$Y=-105+1X\text{or}Y=X-105$$

Prediction: If height (X) is 170 cm, predicted weight (Y) is: $$Y=170-105=65\text{ kg}$$

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Residuals and Goodness of Fit

Residuals are the differences between observed Y values and predicted Y values: $$\text{Residual}=Y_{\text{observed}}-Y_{\text{predicted}}$$ A good regression line has small residuals.

Example 3: Calculate Residuals Using the regression line $Y=X-105$:

Observation: In this case, the line fits perfectly (residuals = 0), which is rare in real data.

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Coefficient of Determination (r²)

r² (R-squared) measures how much of the variation in Y is explained by X: $$r^2={\left(\frac{n(\sum XY)-(\sum X)(\sum Y)}{\sqrt{[n\sum X^2-(\sum X{)}^2][n\sum Y^2-(\sum Y{)}^2]}}\right)}^2$$ From Example 1 (corrected), $r=0.2$, so: r^2 = (0.2)^2 = 0.04 \quad \text{(4% of variation in Y is explained by X)}

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Applications of Correlation and Regression

1. Economics: Predicting sales based on advertising spend.
2. Biology: Studying the relationship between drug dosage and effectiveness.
3. Engineering: Optimizing machine performance based on input variables.
4. Agriculture: Estimating crop yield from rainfall data.
5. Medicine: Analyzing the effect of exercise on heart rate.

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Common Mistakes to Avoid

1. Assuming Causation: Correlation does not imply causation. Just because two variables are correlated does not mean one causes the other.
   • Example: Ice cream sales and drowning incidents both increase in summer, but ice cream does not cause drowning.
2. Ignoring Outliers: Outliers can heavily influence correlation and regression results.
3. Misinterpreting r: A correlation of –0.9 is stronger than +0.3, even if both are "strong" in absolute terms.
4. Extrapolating Beyond Data Range: Predicting values far outside the observed X range can be unreliable.

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NEB-Style Questions and Solutions

Question 1: Short Answer

Define correlation and regression. Give one example of each. Solution:

• Correlation: A statistical measure that expresses the extent to which two variables are related (e.g., height and weight).
• Regression: A method to find the best-fit line to predict one variable from another (e.g., predicting exam scores from study hours).

Question 2: Calculation

Given the following data, calculate Pearson’s correlation coefficient (r):

Solution:

• $n=4$
• $\sum X=20$, $\sum Y=32$
• $\sum XY=180$
• $\sum X^2=120$, $\sum Y^2=276$

$$r=\frac{4(180)-(20)(32)}{\sqrt{[4(120)-20^2][4(276)-32^2]}}$$ $$r=\frac{720-640}{\sqrt{(480-400)(1104-1024)}}$$ $$r=\frac{80}{\sqrt{(80)(80)}}=\frac{80}{80}=1$$ Answer: $r=1$ (perfect positive correlation).

Question 3: Regression Line

Find the regression line for the data in Question 2. Solution: Using the same data: $$b=\frac{4(180)-(20)(32)}{4(120)-20^2}=\frac{720-640}{480-400}=\frac{80}{80}=1$$ $$a=\frac{32-(1)(20)}{4}=\frac{12}{4}=3$$ Regression Equation: $Y=3+1X$ or $Y=X+3$.

Question 4: Interpretation

What does an r value of –0.8 indicate? Solution: An r value of –0.8 indicates a strong negative correlation between the two variables. As one variable increases, the other decreases significantly.

Question 5: Real-World Application

How can regression be used in business? Solution: Regression can be used in business to:

1. Predict future sales based on past trends.
2. Determine the relationship between advertising expenditure and sales revenue.
3. Optimize pricing strategies based on demand elasticity.

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Exam Tip

1. Understand the Difference: Know when to use correlation (measuring relationship) vs. regression (prediction).
2. Formulas: Memorize the formulas for r, b, and a. Practice calculations step-by-step.
3. Interpretation: Always interpret the meaning of r and r² in context (e.g., "r = 0.9 means a strong positive correlation").
4. Graphs: Draw scatterplots and regression lines to visualize relationships.
5. Avoid Common Errors: Never assume causation from correlation. Check for outliers.
6. Units: Ensure all units are consistent when calculating slopes and intercepts.
7. Residuals: Understand that residuals help assess the fit of the regression line.

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Practice Problems for NEB Exam:

1. Calculate r for the following data:

   X	Y
   1	2
   2	3
   3	5
   4	4

2. Find the regression line for the data in Q1 and predict Y when X = 5.

3. Explain why a correlation of –0.5 is weaker than a correlation of +0.9.

4. What is the coefficient of determination, and how is it useful?

5. A company finds that r = 0.7 between employee training hours (X) and productivity (Y). What does this mean?

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Good luck with your NEB exam preparation! 🚀