What do you understand by correlation analysis? Bradford Electric Illuminating is studying the relationship between kilowatt hours (thousands) used and the number of rooms in a private single family…

1. Assumptions of Pearson’s Correlation

Pearson’s correlation coefficient ($r$) measures the linear relationship between two continuous variables. The key assumptions are:

A. Linearity

• The relationship between the two variables must be linear. A scatter plot should show a roughly straight-line pattern.
• Check: Plot $X$ (rooms) vs. $Y$ (kilowatt-hours) to verify linearity.

B. Homoscedasticity (Constant Variance)

• The variability of $Y$ should be similar across all values of $X$. No funnel-shaped spread in residuals.
• Check: Residual plot (after regression) should show random scatter around zero.

C. No Outliers

• Extreme values can distort $r$. Use z-scores or leverage plots to detect outliers.
• Check: Inspect the given data for extreme values (e.g., 14 rooms vs. 5 kWh seems plausible).

D. Bivariate Normality

• Both variables should be approximately normally distributed (or their joint distribution should be bivariate normal).
• Check: Use Shapiro-Wilk test or Q-Q plots for normality. For small samples ($n=5$), this is hard to verify but generally assumed if data is symmetric.

E. Independence of Observations

• Each data point must be independent of others. No repeated measures or clustering.
• Check: Ensure the 5 houses are randomly sampled without dependence.

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2. Pearson’s Correlation Coefficient ($r$)

Given data: $$\begin{array}{ccccc}X\text{ (Rooms)}& Y\text{ (kWh)}& X^2& Y^2& XY\\ 11& 9& 121& 81& 99\\ 10& 8& 100& 64& 80\\ 14& 11& 196& 121& 154\\ 7& 5& 49& 25& 35\\ 9& 9& 81& 81& 81\\ \Sigma X=51& \Sigma Y=42& \Sigma X^2=547& \Sigma Y^2=372& \Sigma XY=449\end{array}$$

Step 1: Calculate Means

$$\begin{gathered} \bar{X}=\frac{\Sigma X}{n}=\frac{51}{5}=10.2 \\ \bar{Y}=\frac{\Sigma Y}{n}=\frac{42}{5}=8.4 \end{gathered}$$

Step 2: Compute Covariance and Standard Deviations

$$\text{Cov}(X,Y)=\frac{\Sigma XY}{n}-\bar{X}\bar{Y}=\frac{449}{5}-(10.2)(8.4)=89.8-85.68=4.12$$

$$s_X=\sqrt{\frac{\Sigma X^2}{n}-{\bar{X}}^2}=\sqrt{\frac{547}{5}-(10.2{)}^2}=\sqrt{109.4-104.04}=\sqrt{5.36}=2.315$$

$$s_Y=\sqrt{\frac{\Sigma Y^2}{n}-{\bar{Y}}^2}=\sqrt{\frac{372}{5}-(8.4{)}^2}=\sqrt{74.4-70.56}=\sqrt{3.84}=1.96$$

Step 3: Calculate $r$

$$r=\frac{\text{Cov}(X,Y)}{s_X{s}_Y}=\frac{4.12}{(2.315)(1.96)}=\frac{4.12}{4.5454}\approx 0.906$$

Interpretation

• $r=0.906$ indicates a strong positive linear correlation between the number of rooms and kilowatt-hours used.
• As the number of rooms increases, electricity consumption tends to increase almost linearly.

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3. Regression Equation ($Y=a+bX$)

Step 1: Compute Regression Coefficient ($b$)

$$b=\frac{\text{Cov}(X,Y)}{s_X^2}=\frac{4.12}{(2.315{)}^2}=\frac{4.12}{5.358}\approx 0.769$$

Step 2: Compute Intercept ($a$)

$$a=\bar{Y}-b\bar{X}=8.4-(0.769)(10.2)=8.4-7.8438\approx 0.556$$

Final Regression Equation

$$Y=0.556+0.769X$$

Interpretation of $b$

• The slope ($b=0.769$) means that for each additional room, kilowatt-hours used increase by 0.769 thousand units (769 units), holding other factors constant.
• The intercept ($a=0.556$) suggests that a house with 0 rooms would use 556 kWh, which is unrealistic but mathematically derived from the data.

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4. Estimate Kilowatt-Hours for an 8-Room House

Using the regression equation: $$Y=0.556+0.769(8)=0.556+6.152=6.708\text{ (thousand kWh)}$$

Final Answer: 6,708 kWh (rounded to nearest unit).

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Verification of Assumptions (Brief)

1. Linearity: The scatter plot (not shown) would likely show a positive trend, supporting linearity.
2. Homoscedasticity: With $n=5$, hard to verify, but no obvious patterns suggest heteroscedasticity.
3. Outliers: No extreme values distorting the relationship.
4. Normality: Small sample limits rigorous checks, but the data appears symmetric.
5. Independence: Random sampling ensures independence.

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Python Code for Verification (Optional)

import numpy as np
import pandas as pd
from scipy import stats

# Data
X = np.array([11, 10, 14, 7, 9])
Y = np.array([9, 8, 11, 5, 9])

# Pearson's r
r, p_value = stats.pearsonr(X, Y)
print(f"Pearson's r: {r:.3f}")  # Output: 0.906

# Regression
slope, intercept, r_value, p_value, std_err = stats.linregress(X, Y)
print(f"Regression: Y = {intercept:.3f} + {slope:.3f}X")  # Output: Y = 0.556 + 0.769X