Let's solve this step by step.

Step 1: Find the Linear Regression Equation

The regression equation follows the form:
$y = mx + b$ where:

• $m$ is the slope
• $b$ is the y-intercept

We use the formulas:
$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$ $b = \frac{\sum y - m \sum x}{n}$

Step 2: Compute $m$ and $b$

We will calculate these values using the given data.

Step 3: Predict the Test Grade for $x = 64$

Once we obtain the equation, we substitute $x = 64$ to find $y$.

I'll do the calculations now. numpy as np

Given data

x = np.array([84, 71, 76, 80, 89, 72, 87, 86, 52]) y = np.array([72, 59, 71, 76, 82, 68, 88, 89, 50])

Number of data points

n = len(x)

Compute sums

sum_x = np.sum(x) sum_y = np.sum(y) sum_xy = np.sum(x * y) sum_x2 = np.sum(x ** 2)

Compute slope (m)

m = (n * sum_xy - sum_x * sum_y) / (n * sum_x2 - sum_x ** 2)

Compute y-intercept (b)

b = (sum_y - m * sum_x) / n

Round to the nearest hundredth

m = round(m, 2) b = round(b, 2)

Compute prediction for x = 64

x_pred = 64 y_pred = round(m * x_pred + b)

m, b, y_predThe linear regression equation is:
$y = 1.02x - 6.31$ For a homework grade of $x = 64$, the predicted test grade is 59.

Would you like a breakdown of the calculations?

Here are five related questions to expand your understanding:

1. How do you interpret the slope of the regression equation?
2. What is the meaning of the y-intercept in this context?
3. How can you calculate the correlation coefficient to measure the strength of the relationship?
4. How would adding more data points affect the accuracy of the regression model?
5. Can this model be used to predict values far outside the given range? Why or why not?

Tip: Always check the residuals (errors) of your regression model to assess its accuracy!