Let's analyze the given data and answer each part of the question step by step.

Data:

$x = \{-18, -15, -13, -12, -8\}$ $y = \{102, 122, 120, 132, 142\}$

Part (a): Find the equation of the line of best fit (linear regression equation).

The line of best fit is typically found using the equation:

$y = mx + b$

Where:

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

To find the equation, we need to calculate:

1. The slope $m$,
2. The y-intercept $b$.

We will use the following formulas:

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

Where $n$ is the number of data points.

Let me calculate these values first.The equation of the line of best fit (linear regression equation) is:

$y = 3.861x + 174.569$

Part (b): Predict $y$ when $x = -10$.

Using the equation $y = 3.861x + 174.569$, let's substitute $x = -10$ and calculate the value of $y$.When $x = -10$, the predicted value of $y$ is approximately $135.96$.

Part (c): Calculate the residual for $x = -13$.

The residual is the difference between the actual $y$ value and the predicted $y$ value. For $x = -13$, the actual $y$ is 120.

Let's calculate the predicted $y$ value using the regression equation, then find the residual:

$\text{Residual} = \text{Actual } y - \text{Predicted } y$The residual for $x = -13$ is approximately $-4.37$. This means the predicted value is about 4.37 units higher than the actual value.

Part (d): Interpret the slope in the context of the problem.

The slope $m = 3.861$ indicates that for each 1-unit increase in $x$, the value of $y$ is predicted to increase by approximately 3.861 units.

Part (e): Interpret the $y$-intercept in the context of the problem.

The $y$-intercept $b = 174.569$ represents the predicted value of $y$ when $x = 0$. If $x$ were 0 (though it's outside the given range), $y$ would be about 174.569.

Part (f): Determine if the line of best fit is a good model for the data.

To assess how well the line fits the data, we can calculate the correlation coefficient $r$. A value of $r$ close to 1 or -1 indicates a good fit. Let's calculate $r$.It seems like I can’t do more advanced data analysis right now. Please try again later.

Let me know if you'd like any clarification on the earlier parts or if you have additional questions!

Here are some related questions for further practice:

1. How would the residuals look for other $x$ values?
2. Can you calculate the residual sum of squares (RSS) to evaluate the fit?
3. How does changing the slope affect predictions for extreme $x$ values?
4. What happens to the $y$-intercept if the data range changes?
5. Can you estimate the correlation visually by plotting the data?

Tip: Always plot your data when doing regression analysis to visually inspect how well the line fits!