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:

$\hat{y} = 9.7786 + 0.0265x$

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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, $r = \frac{925}{31971.9} \approx 0.0289$

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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?

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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?

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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.