To find the equation of the regression line, we need to calculate the slope ($b$) and the y-intercept ($a$) of the line using the least-squares method. The formulas for the slope and intercept are given by:

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

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

Where:

• $n$ is the number of data points
• $\sum{xy}$ is the sum of the product of each pair of $x$ and $y$
• $\sum{x}$ is the sum of $x$ values
• $\sum{y}$ is the sum of $y$ values
• $\sum{x^2}$ is the sum of the squares of $x$ values

Let's start by calculating these sums:

&\sum{x} = 9 + 12 + 8 + 10 + 11 + 6 + 14 + 15 + 13 + 7 + 5 = 110 \\ &\sum{y} = 8.47 + 9.51 + 7.51 + 9.12 + 9.47 + 4.69 + 8.69 + 7.82 + 9.26 + 6.26 + 2.82 = 83.62 \\ &\sum{xy} = (9 \times 8.47) + (12 \times 9.51) + (8 \times 7.51) + (10 \times 9.12) + (11 \times 9.47) + (6 \times 4.69) + (14 \times 8.69) + (15 \times 7.82) + (13 \times 9.26) + (7 \times 6.26) + (5 \times 2.82) \\ &= 76.23 + 114.12 + 60.08 + 91.20 + 104.17 + 28.14 + 121.66 + 117.30 + 120.38 + 43.82 + 14.10 = 891.20 \\ &\sum{x^2} = 9^2 + 12^2 + 8^2 + 10^2 + 11^2 + 6^2 + 14^2 + 15^2 + 13^2 + 7^2 + 5^2 = 81 + 144 + 64 + 100 + 121 + 36 + 196 + 225 + 169 + 49 + 25 = 1210 \end{aligned}$$ Now we can calculate the slope $$b$$: $$\begin{aligned} b &= \frac{11 \cdot 891.20 - 110 \cdot 83.62}{11 \cdot 1210 - 110^2} \\ &= \frac{9803.20 - 9198.20}{13310 - 12100} \\ &= \frac{605}{1210} \\ &= 0.50 \end{aligned}$$ Next, we calculate the intercept $$a$$: $$\begin{aligned} a &= \frac{\sum{y} - b\sum{x}}{n} \\ &= \frac{83.62 - 0.50 \cdot 110}{11} \\ &= \frac{83.62 - 55}{11} \\ &= \frac{28.62}{11} \\ &= 2.60 \end{aligned}$$ So, the equation of the regression line is: $$\hat{y} = 2.60 + 0.50x$$ To identify a characteristic of the data that is ignored by the regression line, we examine the scatterplot. Typically, the regression line minimizes the sum of squared residuals but doesn't account for the following potential data characteristics: - **Outliers**: Points that are significantly different from others can heavily influence the regression line. - **Non-linearity**: The regression line assumes a linear relationship between $$x$$ and $$y$$, which may not be true for all data. - **Heteroscedasticity**: The regression line assumes constant variance of residuals; deviations can indicate this is not the case. Would you like to see the detailed calculations or have any questions? Here are some related questions for further exploration: 1. How would the regression line change if an outlier was removed from the data set? 2. What is the correlation coefficient for this data set, and how does it relate to the regression line? 3. How can we assess the goodness-of-fit of the regression line to the data? 4. How would you modify the model if the relationship between $$x$$ and $$y$$ was non-linear? 5. What are some methods to identify and handle outliers in a data set? 6. How does heteroscedasticity affect the interpretation of the regression results? 7. What are the assumptions of linear regression, and how can we test if they are met? 8. How would you use this regression model to make predictions for new $$x$$ values? **Tip:** When examining a scatterplot for regression analysis, always check for patterns that suggest non-linear relationships or the presence of outliers.