Step 1) Enter the (x,y) pairs of values into a table. In the attached image, this is in columns B and C
Step 2) Make a new column of the squared x values. This is column D in that same attached image
Step 3) Make another column of the x*y values. This is the result of multiplying each (x,y) pair together. This is column E.
Step 4) Add up everything in their respective columns. The sums are highlighted in yellow at the bottom. Those sums are important since we'll use them later.
Let
P = sum of the x values = 22
Q = sum of the y values = 56
R = sum of the x^2 values = 75
S = sum of the x*y values = 111
We'll use P,Q,R,S in the formulas below
m = (n*S-P*Q)/(n*R-P^2)
b = (Q - m*P)/n
where m is the slope, b is the y intercept and n is the number of points (in this case, n = 8). The regression line is in the form y = mx+b
So let's compute the slope m first
m = (n*S-P*Q)/(n*R-P^2)
m = (8*111-22*56)/(8*75-22^2)
m = -2.96551724137931
which rounds to m = -3
Then use this m value to find b
b = (Q - m*P)/n
b = (56 - (-2.96551724137931)*22)/8
b = 15.1551724137931
which rounds to b = 15
So y = mx+b turns into y = -3x+15
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Final Answer: Choice A) y = -3x+15This regression line is approximate to the nearest whole number
Side Notes:
1) If your teacher doesn't want you using that formula I mentioned above, then you can try looking in your book for similar versions of the formula
2) If your teacher requires you only use two points, then I suggest you find the equation of the line through (1,12) and (4.5,0) to get roughly the same answer if you rounded to the nearest whole number
3) Using a calculator, such as a TI83, will help speed up the process immensely. The LINREG function will automatically find the linear regression equation after you type in the x,y values into the table. Let me know if you need me to go over this in greater detail.
