Round the values of
a and
b to the nearest hundredth, then use the equation to solve for
y when
x

To solve this problem, we need to find the linear regression line $y = a + bx$ that best fits the given data. This involves determining the values of the slope $b$ and the intercept $a$. The general formulas for calculating $a$ and $b$ are:

$b = \frac{n \sum (xy) - \sum x \sum y}{n \sum x^2 - (\sum x)^2}$

$a = \frac{\sum y - b \sum x}{n}$

Where:

• $n$ is the number of data points
• $x$ and $y$ are the individual values from the data set
• $\sum x$, $\sum y$, $\sum x^2$, and $\sum xy$ are the summations of the respective values

Let's go step by step:

Step 1: Calculate the necessary summations

Given the data:

x & y \\ \hline 3 & 3.44 \\ 4 & 5.84 \\ 5 & 8.23 \\ 6 & 6.38 \\ 7 & 9.06 \\ 8 & 9.06 \\ 9 & 9.87 \\ 10 & 11.36 \\ \end{array}$$ We will compute: - $$\sum x$$ - $$\sum y$$ - $$\sum x^2$$ - $$\sum xy$$ ### Step 2: Plug these summations into the formulas for $$a$$ and $$b$$ Once we have the values of $$a$$ and $$b$$, we will use the equation $$y = a + bx$$ to estimate the value of $$y$$ when $$x = 12$$. I'll go ahead and compute these for you. numpy as np # Data points x_values = np.array([3, 4, 5, 6, 7, 8, 9, 10]) y_values = np.array([3.44, 5.84, 8.23, 6.38, 9.06, 9.06, 9.87, 11.36]) # Calculate necessary summations n = len(x_values) sum_x = np.sum(x_values) sum_y = np.sum(y_values) sum_x_squared = np.sum(x_values ** 2) sum_xy = np.sum(x_values * y_values) # Calculate slope (b) and intercept (a) b = (n * sum_xy - sum_x * sum_y) / (n * sum_x_squared - sum_x ** 2) a = (sum_y - b * sum_x) / n # Use the equation to estimate y when x = 12 x_estimate = 12 y_estimate = a + b * x_estimate # Round values of a, b, and the estimated y to the nearest hundredth a_rounded = round(a, 2) b_rounded = round(b, 2) y_rounded = round(y_estimate, 2) a_rounded, b_rounded, y_roundedThe linear regression line is given by: $$y = 1.66 + 0.96x$$ Now, using this equation to estimate the value of $$y$$ when $$x = 12$$, we get: $$y = 1.66 + 0.96(12) = 13.19$$ Thus, the estimated value of $$y$$ when $$x = 12$$ is **13.19**. Would you like more details on the process, or do you have any other questions? Here are a few related questions to explore further: 1. How can you calculate the correlation coefficient $$r$$ for this data? 2. How would you interpret the slope $$b$$ and the intercept $$a$$ in the context of this data? 3. What does the residual plot tell you about the quality of the linear regression? 4. How would you estimate $$y$$ for another value of $$x$$, say $$x = 7.5$$? 5. What are the potential limitations of using linear regression with this dataset? **Tip**: When interpreting the slope $$b$$, remember that it indicates the rate of change in $$y$$ for each unit increase in $$x$$.