Question

a. using the pairs of values for all 10 points, find the equation of the regression line.
b. after removing the point with coordinates (2,4), use the pairs of values for the remaining 9 points and find the equation of the regression line.
c. compare the results from parts (a) and (b).
a. what is the equation of the regression line for all 10 points?
$hat{y} = \square + \square x$ (round to three decimal places as needed.)
b. what is the equation of the regression line for the set of 9 points?
$hat{y} = \square$ (round to three decimal places as needed.)
c. choose the correct description of the results below.
\bigcirc a. the removal of the point has a significant impact on the regression line.
\bigcirc b. the regression line changes, but the change is small.
\bigcirc c. the regression line is very similar in both cases.
\bigcirc d. there is no regression line for the second case because the data are in a pattern.

Explanation:

Step 1: Identify the data points

From the graph, the 10 points are: \((2,4)\) and nine points at \(x = 6,7,8\) and \(y = 7,8,9\) (assuming the blue dots are at \((6,7),(6,8),(6,9),(7,7),(7,8),(7,9),(8,7),(8,8),(8,9)\) and \((2,4)\)). Let's list all 10 points:

• \((2,4)\)
• \((6,7),(6,8),(6,9)\)
• \((7,7),(7,8),(7,9)\)
• \((8,7),(8,8),(8,9)\)

Step 2: Calculate the regression line for 10 points (part a)

The regression line is given by \(\hat{y}=b_0 + b_1x\), where:

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

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

First, calculate \(\sum x\), \(\sum y\), \(\sum xy\), \(\sum x^2\) for \(n = 10\):

• \(x\) values: \(2,6,6,6,7,7,7,8,8,8\)
• \(\sum x=2 + 6\times3+7\times3 + 8\times3=2 + 18+21 + 24 = 65\)
• \(\sum x^2=2^2+6^2\times3 + 7^2\times3+8^2\times3=4 + 108+147 + 192 = 451\)
• \(y\) values: \(4,7,8,9,7,8,9,7,8,9\)
• \(\sum y=4+(7 + 8+9)\times3=4 + 24\times3=4 + 72 = 76\)
• \(\sum xy=2\times4+(6\times7 + 6\times8+6\times9)+(7\times7 + 7\times8+7\times9)+(8\times7 + 8\times8+8\times9)\)
• \(2\times4 = 8\)
• \(6\times(7 + 8+9)=6\times24 = 144\)
• \(7\times(7 + 8+9)=7\times24 = 168\)
• \(8\times(7 + 8+9)=8\times24 = 192\)
• \(\sum xy=8 + 144+168 + 192 = 512\)

Now, calculate \(b_1\):

$$ b_1=\frac{10\times512-65\times76}{10\times451 - 65^2}=\frac{5120 - 4940}{4510 - 4225}=\frac{180}{285}\approx0.632 $$

$$ \bar{x}=\frac{\sum x}{n}=\frac{65}{10}=6.5 $$

$$ \bar{y}=\frac{\sum y}{n}=\frac{76}{10}=7.6 $$

$$ b_0=7.6-0.632\times6.5=7.6 - 4.108 = 3.492 $$

So, \(\hat{y}=3.492+0.632x\)

Step 3: Calculate the regression line for 9 points (part b)

Remove the point \((2,4)\), so \(n = 9\)

• New \(x\) values: \(6,6,6,7,7,7,8,8,8\)
• \(\sum x=6\times3+7\times3 + 8\times3=18+21 + 24 = 63\)
• \(\sum x^2=6^2\times3 + 7^2\times3+8^2\times3=108+147 + 192 = 447\)
• New \(y\) values: \(7,8,9,7,8,9,7,8,9\)
• \(\sum y=(7 + 8+9)\times3=24\times3 = 72\)
• \(\sum xy=(6\times7 + 6\times8+6\times9)+(7\times7 + 7\times8+7\times9)+(8\times7 + 8\times8+8\times9)\)
• \(6\times24 = 144\), \(7\times24 = 168\), \(8\times24 = 192\)
• \(\sum xy=144+168 + 192 = 504\)

Now, calculate \(b_1\) and \(b_0\) for \(n = 9\):

$$ b_1=\frac{9\times504-63\times72}{9\times447 - 63^2}=\frac{4536 - 4536}{4023 - 3969}=\frac{0}{54}=0 $$

$$ \bar{x}=\frac{63}{9}=7 $$

$$ \bar{y}=\frac{72}{9}=8 $$

$$ b_0=8-0\times7 = 8 $$

So, \(\hat{y}=8.000+0.000x=8.000\) (since \(b_1 = 0\))

Step 4: Compare the results (part c)

In part (a), the regression line is \(\hat{y}=3.492 + 0.632x\), and in part (b), it is \(\hat{y}=8.000\). The removal of the point \((2,4)\) (which is an outlier) has a significant impact on the regression line. So the correct option is A.

Answer:

s:
a. \(\hat{y}=\boxed{3.492}+\boxed{0.632}x\)
b. \(\hat{y}=\boxed{8.000}\) (or \(\hat{y}=\boxed{8.000}+0.000x\))
c. A. The removal of the point has a significant impact on the regression line.