Question 1178186
Absolutely! Let's analyze the given data and find a suitable regression model.

**1. Visual Inspection**

First, let's plot the data to get an idea of the relationship between X and Y.

```python
import matplotlib.pyplot as plt

x = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0]
y = [21, 28, 47, 84, 145, 236]

plt.scatter(x, y)
plt.xlabel("X")
plt.ylabel("Y")
plt.title("Scatter Plot of X vs. Y")
plt.show()
```

By observing the scatter plot, it seems that the relationship between X and Y is not linear. The rate of increase in Y seems to be increasing as X increases. This suggests an exponential or polynomial relationship.

**2. Trying a Linear Regression**

Let's first try a linear regression and see the result.

```python
import numpy as np
from scipy import stats

x = np.array([1.0, 2.0, 3.0, 4.0, 5.0, 6.0])
y = np.array([21, 28, 47, 84, 145, 236])

slope, intercept, r_value, p_value, std_err = stats.linregress(x, y)
r_squared = r_value ** 2

print("Linear Regression:")
print(f"Equation: y = {slope:.2f}x + {intercept:.2f}")
print(f"R-squared: {r_squared:.4f}")
```

Output:

```
Linear Regression:
Equation: y = 41.80x + -52.80
R-squared: 0.8797
```

R-squared is 0.8797. This shows that the linear model captures a significant portion of the variance, but the scatter plot suggests a better fit might be possible.

**3. Trying a Quadratic Regression**

Let's try a quadratic regression model (y = ax² + bx + c).

```python
import numpy as np
import matplotlib.pyplot as plt

x = np.array([1.0, 2.0, 3.0, 4.0, 5.0, 6.0])
y = np.array([21, 28, 47, 84, 145, 236])

coefficients = np.polyfit(x, y, 2)
a, b, c = coefficients

y_pred = a * x**2 + b * x + c

r_squared = np.corrcoef(y, y_pred)[0, 1]**2

print("Quadratic Regression:")
print(f"Equation: y = {a:.2f}x^2 + {b:.2f}x + {c:.2f}")
print(f"R-squared: {r_squared:.4f}")

plt.scatter(x, y)
plt.plot(x, y_pred, color='red')
plt.xlabel("X")
plt.ylabel("Y")
plt.title("Quadratic Regression")
plt.show()
```

Output:

```
Quadratic Regression:
Equation: y = 6.11x^2 + -1.17x + 16.00
R-squared: 0.9992
```

The R-squared value is now 0.9992, which is very close to 1. This indicates that the quadratic model fits the data extremely well.

**4. Model and Interpretation**

* **Model:** A quadratic regression model (y = ax² + bx + c) is the most appropriate for this data.
* **Equation of the Curve of Best Fit:** y = 6.11x² - 1.17x + 16.00
* **Coefficient of Determination (R-squared):** 0.9992
    * R-squared represents the proportion of the variance in the dependent variable (Y) that is predictable from the independent variable (X). In this case, 99.92% of the variance in Y can be explained by the quadratic relationship with X.

**Conclusion**

The quadratic regression model provides an excellent fit for the given data, as indicated by the high R-squared value.