The table below gives the age and bone density for five randomly selected women. Using this data, consider the equation of the regression line, yˆ=b0+b1x
, for predicting a woman's bone density based on her age. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant.

Age	42
43
51
56
65
Bone Density	353
344
333
324
322
Table

Step 2 of 6 :  Find the estimated y-intercept. Round your answer to three decimal places.

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

\[
\hat{y} = b_0 + b_1x
\]

where:
- \(b_0\) is the y-intercept,
- \(b_1\) is the slope of the regression line,
- \(x\) is the independent variable (age),
- \(\hat{y}\) is the dependent variable (bone density).

To compute \(b_0\), we need the following formulas:

\[
b_1 = \frac{n\sum{xy} - \sum{x}\sum{y}}{n\sum{x^2} - (\sum{x})^2}
\]
\[
b_0 = \frac{\sum{y} - b_1\sum{x}}{n}
\]

Where:
- \(n\) is the number of data points (5 in this case),
- \(\sum{x}\) is the sum of the x-values (ages),
- \(\sum{y}\) is the sum of the y-values (bone densities),
- \(\sum{xy}\) is the sum of the products of corresponding x and y values,
- \(\sum{x^2}\) is the sum of the squares of the x-values.

### Data
\[
\text{Age} (x): 42, 43, 51, 56, 65
\]
\[
\text{Bone Density} (y): 353, 344, 333, 324, 322
\]

### Step 1: Calculate the necessary sums:

\[
\sum{x} = 42 + 43 + 51 + 56 + 65 = 257
\]
\[
\sum{y} = 353 + 344 + 333 + 324 + 322 = 1676
\]
\[
\sum{x^2} = 42^2 + 43^2 + 51^2 + 56^2 + 65^2 = 1764 + 1849 + 2601 + 3136 + 4225 = 11575
\]
\[
\sum{xy} = (42 \times 353) + (43 \times 344) + (51 \times 333) + (56 \times 324) + (65 \times 322) = 14826 + 14792 + 16983 + 18144 + 20930 = 85775
\]

### Step 2: Calculate \(b_1\) (slope):

\[
b_1 = \frac{5 \times 85775 - 257 \times 1676}{5 \times 11575 - (257)^2}
\]
\[
b_1 = \frac{428875 - 430132}{57875 - 66049}
\]
\[
b_1 = \frac{-1257}{-816} = 1.54
\]

### Step 3: Calculate \(b_0\) (y-intercept):

\[
b_0 = \frac{\sum{y} - b_1\sum{x}}{n}
\]
\[
b_0 = \frac{1676 - 1.54 \times 257}{5}
\]
\[
b_0 = \frac{1676 - 396.38}{5}
\]
\[
b_0 = \frac{1279.62}{5} = 255.924
\]

### Final Answer:
The estimated y-intercept (\(b_0\)) is **255.924** (rounded to three decimal places).

---

Would you like to move on to the next step, or need more details about this calculation? Here are some related questions:

1. How do you interpret the slope (\(b_1\)) in this regression equation?
2. What is the next step after finding the y-intercept and slope?
3. How do you check if the correlation coefficient is statistically significant?
4. How do you calculate the correlation coefficient for this data?
5. What role does the residual plot play in evaluating the regression line?

**Tip:** Always check for outliers or influential data points that might skew the regression results.

The table below gives the age and bone density for five randomly selected women. Using this data, consider the equation of the regression line, yˆ=b0+b1x, for predicting a woman's bone density based on her age. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant. Age 42 43 51 56 65 Bone Density 353 344 333 324 322. Step 2 of 6: Find the estimated y-intercept. Round your answer to three decimal places.

Learn how to calculate the estimated y-intercept (b0) of a regression line using real data. This step-by-step solution involves the least squares method and linear regression concepts, focusing on predicting bone density based on age. Suitable for high school students studying statistics.

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