To find the estimated y-intercept ($b_0$) of the regression line, we need to use the formula:

$b_0 = \bar{y} - b_1 \bar{x}$

Where:

• $\bar{y}$ is the mean of the $y$-values (number of bids),
• $\bar{x}$ is the mean of the $x$-values (list price),
• $b_1$ is the slope of the regression line.

Step 1: Calculate the means

We have the following data:

• Prices (x): 21, 27, 31, 43, 49
• Number of Bids (y): 4, 6, 7, 8, 9

First, calculate the mean of $x$ (price) and the mean of $y$ (number of bids).

$\bar{x} = \frac{21 + 27 + 31 + 43 + 49}{5} = \frac{171}{5} = 34.2$ $\bar{y} = \frac{4 + 6 + 7 + 8 + 9}{5} = \frac{34}{5} = 6.8$

Step 2: Calculate the slope ($b_1$)

To calculate the slope ($b_1$), we use the formula:

$b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$

Let's calculate the necessary terms:

$x_i$	$y_i$	$x_i - \bar{x}$	$y_i - \bar{y}$	$(x_i - \bar{x})(y_i - \bar{y})$	$(x_i - \bar{x})^2$
21	4	21 - 34.2 = -13.2	4 - 6.8 = -2.8	(-13.2)(-2.8) = 36.96	(-13.2)^2 = 174.24
27	6	27 - 34.2 = -7.2	6 - 6.8 = -0.8	(-7.2)(-0.8) = 5.76	(-7.2)^2 = 51.84
31	7	31 - 34.2 = -3.2	7 - 6.8 = 0.2	(-3.2)(0.2) = -0.64	(-3.2)^2 = 10.24
43	8	43 - 34.2 = 8.8	8 - 6.8 = 1.2	(8.8)(1.2) = 10.56	(8.8)^2 = 77.44
49	9	49 - 34.2 = 14.8	9 - 6.8 = 2.2	(14.8)(2.2) = 32.56	(14.8)^2 = 219.04

Now, we can calculate the sums:

$\sum (x_i - \bar{x})(y_i - \bar{y}) = 36.96 + 5.76 - 0.64 + 10.56 + 32.56 = 84.2$ $\sum (x_i - \bar{x})^2 = 174.24 + 51.84 + 10.24 + 77.44 + 219.04 = 532.8$

Now, we can calculate $b_1$:

$b_1 = \frac{84.2}{532.8} \approx 0.158$

Step 3: Calculate the y-intercept ($b_0$)

Now that we know $b_1$ and the means $\bar{x}$ and $\bar{y}$, we can substitute into the equation for $b_0$:

$b_0 = \bar{y} - b_1 \bar{x}$ $b_0 = 6.8 - (0.158 \times 34.2)$ $b_0 = 6.8 - 5.4 = 1.4$

Final Answer:

The estimated y-intercept is $\mathbf{1.400}$ (rounded to three decimal places).

Would you like to proceed with the next steps or need further details? Here are some questions that can expand on this:

1. How do you calculate the slope $b_1$ for a regression line in general?
2. What are the potential limitations of using the regression line for predictions?
3. How would you interpret the regression line equation once we have $b_0$ and $b_1$?
4. What are the assumptions made when using linear regression for predictions?
5. Can the y-intercept have a real-world meaning in this case?

Tip: Always check the correlation coefficient and the significance level before using a regression model for predictions, as it helps ensure reliability.