Question 1190695
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Here's the original table<table border = "1" cellpadding = "5"><tr><td>x</td><td>y</td></tr><tr><td>-1</td><td>3</td></tr><tr><td>0</td><td>3</td></tr><tr><td>1</td><td>1</td></tr><tr><td>2</td><td>0</td></tr><tr><td>4</td><td>-3</td></tr></table>


Form the column of xy values by multiplying each x and y number for each separate row. 
Example: Row 1 has x*y = -1*3 = -3<table border = "1" cellpadding = "5"><tr><td>x</td><td>y</td><td>xy</td></tr><tr><td>-1</td><td>3</td><td>-3</td></tr><tr><td>0</td><td>3</td><td>0</td></tr><tr><td>1</td><td>1</td><td>1</td></tr><tr><td>2</td><td>0</td><td>0</td></tr><tr><td>4</td><td>-3</td><td>-12</td></tr></table>


Now compute the column of x^2 values<table border = "1" cellpadding = "5"><tr><td>x</td><td>y</td><td>xy</td><td>x^2</td></tr><tr><td>-1</td><td>3</td><td>-3</td><td>1</td></tr><tr><td>0</td><td>3</td><td>0</td><td>0</td></tr><tr><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>2</td><td>0</td><td>0</td><td>4</td></tr><tr><td>4</td><td>-3</td><td>-12</td><td>16</td></tr></table>
We'll get the sum of each column
A = sum of the x values = -1+0+1+2+4 = 6
B = sum of the y values = 3+3+1+0+(-3) = 4
C = sum of the xy values = -3+0+1+0+(-12) = -14
D = sum of the x^2 values = 1+0+1+4+16 = 22


Then we can compute the slope m based on those values computed
{{{m = (C*n - A*B)/(D*n - A^2)}}}
where n is the number of points
In this case, n = 5


So the slope is,
{{{m = (C*n - A*B)/(D*n - A^2)}}}


{{{m = (-14*5 - 6*4)/(22*5 - 6^2)}}}


{{{m = -1.27027}}} approximately


And the y intercept b is
{{{b = (B-m*A)/n}}}


{{{b = (4-(-1.27027)*6)/5}}}


{{{b = 2.324324}}} also approximate.


The regression line is approximately {{{y = -1.27027x + 2.324324}}}
For larger data sets, I recommend using a calculator to make quick work of finding the regression line. Also, they are handy to check your answer.


I'll let you plug in x = 3 to find the corresponding y estimate value.
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