The given quantities below were calculated from a sample data set.


∑x=280     	∑y=291     	∑xy=2845
∑x^2=3906	            ∑y^2=3931          	n=29 

Determine the least squares regression line. Round values to four decimal places, if necessary.

y^=  

Determine the correlation coefficient. Round the solution to four decimal places, if necessary.

r=  

To find the least squares regression line \( y = \hat{y} = a + bx \) and the correlation coefficient \( r \), let's go through the calculations step-by-step.

### Given Data
- \( \sum x = 280 \)
- \( \sum y = 291 \)
- \( \sum xy = 2845 \)
- \( \sum x^2 = 3906 \)
- \( \sum y^2 = 3931 \)
- \( n = 29 \)

### Step 1: Calculate \( b \) (Slope) and \( a \) (Intercept) for the Regression Line
The formulas for the slope \( b \) and intercept \( a \) of the least squares regression line are:

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

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

#### 1. Calculate \( b \):
\[
b = \frac{29 \cdot 2845 - 280 \cdot 291}{29 \cdot 3906 - 280^2}
\]

Let's calculate each part:

- Numerator: \( 29 \cdot 2845 = 82405 \) and \( 280 \cdot 291 = 81480 \), so
  \[
  82405 - 81480 = 925
  \]

- Denominator: \( 29 \cdot 3906 = 113274 \) and \( 280^2 = 78400 \), so
  \[
  113274 - 78400 = 34874
  \]

Then,
\[
b = \frac{925}{34874} \approx 0.0265
\]

#### 2. Calculate \( a \):
\[
a = \frac{291 - (0.0265 \cdot 280)}{29}
\]

- \( 0.0265 \cdot 280 \approx 7.42 \), so
  \[
  291 - 7.42 = 283.58
  \]

Then,
\[
a = \frac{283.58}{29} \approx 9.7786
\]

### Therefore, the least squares regression line is:
\[
\hat{y} = 9.7786 + 0.0265x
\]

---

### Step 2: Calculate the Correlation Coefficient \( r \)

The formula for the correlation coefficient \( r \) is:

\[
r = \frac{n \sum xy - \sum x \sum y}{\sqrt{\left( n \sum x^2 - (\sum x)^2 \right) \left( n \sum y^2 - (\sum y)^2 \right)}}
\]

#### 1. Calculate the Numerator (which we already found above):
\[
n \sum xy - \sum x \sum y = 925
\]

#### 2. Calculate the Denominator:
- For \( x \):
  \[
  n \sum x^2 - (\sum x)^2 = 34874
  \]

- For \( y \):
  \[
  n \sum y^2 - (\sum y)^2 = 29 \cdot 3931 - 291^2
  \]

Calculating each part:

- \( 29 \cdot 3931 = 113999 \)
- \( 291^2 = 84681 \), so
  \[
  113999 - 84681 = 29318
  \]

Now we can compute \( r \):

\[
r = \frac{925}{\sqrt{34874 \cdot 29318}}
\]

Calculating the denominator:

\[
\sqrt{34874 \cdot 29318} \approx \sqrt{1022167412} \approx 31971.9
\]

So,
\[
r = \frac{925}{31971.9} \approx 0.0289
\]

---

### Final Answers

- **Least Squares Regression Line:** \( \hat{y} = 9.7786 + 0.0265x \)
- **Correlation Coefficient:** \( r \approx 0.0289 \)

Would you like further details on these calculations?

---

### Related Questions

1. What are the key assumptions behind using the least squares regression line?
2. How does the correlation coefficient help interpret the strength of a linear relationship?
3. Can a weak correlation still result in a significant regression line? Why or why not?
4. How would the regression line change if the values of \( x \) and \( y \) were reversed?
5. What are some limitations of using the least squares regression method?

---

**Tip:** When interpreting \( r \), keep in mind that values close to 0 indicate a weak linear relationship, while values closer to \(\pm1\) indicate stronger linear relationships.

The given quantities below were calculated from a sample data set.

∑x=280     	∑y=291     	∑xy=2845
∑x^2=3906	            ∑y^2=3931          	n=29

Determine the least squares regression line. Round values to four decimal places, if necessary.

y^=  

Determine the correlation coefficient. Round the solution to four decimal places, if necessary.
r=  

Learn how to calculate the least squares regression line and correlation coefficient for a sample data set. This guide covers essential statistical formulas and step-by-step calculations with values ∑x=280, ∑y=291, ∑xy=2845, ∑x²=3906, ∑y²=3931, and n=29. Suitable for students in Grades 11-12.

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