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

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

Where:

• $b_1$ is the slope of the regression line
• $\bar{x}$ is the mean of the age values
• $\bar{y}$ is the mean of the bone density values

Step 1: Calculate the necessary values.

1. Calculate the mean of the ages ($\bar{x}$):

$\bar{x} = \frac{37 + 39 + 43 + 63 + 67}{5} = \frac{249}{5} = 49.8$

2. Calculate the mean of the bone densities ($\bar{y}$):

$\bar{y} = \frac{356 + 349 + 336 + 328 + 318}{5} = \frac{1687}{5} = 337.4$

Step 2: Find the slope $b_1$ using the formula:

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

We will need to compute this value for each data point. Let’s calculate each term in the summation:

Age $x_i$	Bone Density $y_i$	$x_i - \bar{x}$	$y_i - \bar{y}$	$(x_i - \bar{x})(y_i - \bar{y})$	$(x_i - \bar{x})^2$
37	356	$37 - 49.8 = -12.8$	$356 - 337.4 = 18.6$	$(-12.8)(18.6) = -238.08$	$(-12.8)^2 = 163.84$
39	349	$39 - 49.8 = -10.8$	$349 - 337.4 = 11.6$	$(-10.8)(11.6) = -125.28$	$(-10.8)^2 = 116.64$
43	336	$43 - 49.8 = -6.8$	$336 - 337.4 = -1.4$	$(-6.8)(-1.4) = 9.52$	$(-6.8)^2 = 46.24$
63	328	$63 - 49.8 = 13.2$	$328 - 337.4 = -9.4$	$(13.2)(-9.4) = -123.28$	$(13.2)^2 = 174.24$
67	318	$67 - 49.8 = 17.2$	$318 - 337.4 = -19.4$	$(17.2)(-19.4) = -333.68$	$(17.2)^2 = 295.84$

Now, calculate the sums:

$\sum{(x_i - \bar{x})(y_i - \bar{y})} = -238.08 + (-125.28) + 9.52 + (-123.28) + (-333.68) = -810.8$

$\sum{(x_i - \bar{x})^2} = 163.84 + 116.64 + 46.24 + 174.24 + 295.84 = 796.8$

3. Calculate the slope $b_1$:

$b_1 = \frac{-810.8}{796.8} \approx -1.017$

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

$b_0 = \bar{y} - b_1 \bar{x} = 337.4 - (-1.017)(49.8)$ $b_0 = 337.4 + 50.6 = 388.0$

Conclusion:

The estimated y-intercept $b_0$ is approximately 388.0.

Let me know if you'd like further clarification or need help with the next steps! Here are 5 related questions to expand on this topic:

1. How do you calculate the slope of the regression line?
2. What is the significance of the correlation coefficient in linear regression?
3. How does the regression equation help in making predictions about bone density?
4. What are the limitations of using a regression line for prediction in this case?
5. How do you interpret the slope and y-intercept in the context of this problem?

Tip: Always check whether the correlation coefficient is statistically significant before making predictions with a regression model, as this ensures the model's reliability.