If you are in a hurry, then you can use technology to quickly compute the r and r^2 values.
Two examples would be the LinReg command on a TI83 (or similar) and using the the CORREL command in a spreadsheet.
There are many other options to choose from. Feel free to search out your favorite.
You should find these approximations
r = 0.9878
r^2 = 0.9757
Since r^2 is very close to 1, this makes the linear regression a good fit. Approximately 97.57% of the variation in x explains the variation in y.
If you have more time to go over the math, then there are various ways to calculate the correlation coefficient.
I'll go over two slightly different methods.
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Method 1
x = capacity of hard drive in terabytes (TB)
y = price in dollars
Given info:
n = 9 = sample size = number of x,y pairs
xbar = 7.611
ybar = 786.49
SD(x) = 9.854
SD(y) = 1417.82
The term "xbar" refers to the horizontal bar over the x, i.e.
A similar story is with ybar as well.
Given Data
| x | y |
| 0.5 | 60.99 |
| 1 | 77.99 |
| 2 | 112.97 |
| 3 | 110.99 |
| 4 | 151.99 |
| 6 | 425.34 |
| 8 | 597.11 |
| 12 | 1081.99 |
| 32 | 4459 |
Eg: 0.5*60.99 = 30.495 in the first row
| x | y | xy |
| 0.5 | 60.99 | 30.495 |
| 1 | 77.99 | 77.99 |
| 2 | 112.97 | 225.94 |
| 3 | 110.99 | 332.97 |
| 4 | 151.99 | 607.96 |
| 6 | 425.34 | 2552.04 |
| 8 | 597.11 | 4776.88 |
| 12 | 1081.99 | 12983.88 |
| 32 | 4459 | 142688 |
It's not only fast and efficient, but also something that is expected in real world applications.
I'm using LibreOffice but you could use Excel or Google Sheets or whichever app you prefer most.
Add up the values in the xy column to get 164,276.155
Then we subtract off the value of n*xbar*ybar = 9*7.611*786.49 = 53,873.77851
So we have:
Sum(xy) - n*xbar*ybar = 164,276.155 - 53,873.77851 = 110,402.37649
We'll divide that result over the product of the given standard deviation values, multiplied with (n-1)
So,
(n-1)*SD(x)*SD(y) = (9-1)*(9.854)*(1417.82) = 111,769.58624
Therefore,
r = (110,402.37649)/(111,769.58624)
r = 0.98776760480204
r^2 = (0.98776760480204)^2
r^2 = 0.97568484109636
r^2 = 0.9757
which is approximate.
Since r^2 is very close to 1, this makes the linear regression a good fit. Approximately 97.57% of the variation in x explains the variation in y.
Note: The formula I used just now is
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Method 2
x = capacity of hard drive in terabytes (TB)
y = price in dollars
Given info:
n = 9 = sample size = number of x,y pairs
xbar = 7.611
ybar = 786.49
SD(x) = 9.854
SD(y) = 1417.82
Given Data
| x | y |
| 0.5 | 60.99 |
| 1 | 77.99 |
| 2 | 112.97 |
| 3 | 110.99 |
| 4 | 151.99 |
| 6 | 425.34 |
| 8 | 597.11 |
| 12 | 1081.99 |
| 32 | 4459 |
Zx = (x - xbar)/(SD(x))
We're computing the z score for each x term
For instance, in the first row we have
Zx = (x-xbar)/(SD(x))
Zx = (0.5-7.611)/(9.854)
Zx = -0.72163588390501
Zx = -0.721636
Do the same thing for each item in the x column. The values of xbar and SD(x) will remain constant.
This is what the updated table looks like
| x | y | Zx |
| 0.5 | 60.99 | -0.721636 |
| 1 | 77.99 | -0.670895 |
| 2 | 112.97 | -0.569413 |
| 3 | 110.99 | -0.467932 |
| 4 | 151.99 | -0.36645 |
| 6 | 425.34 | -0.163487 |
| 8 | 597.11 | 0.039476 |
| 12 | 1081.99 | 0.445403 |
| 32 | 4459 | 2.475036 |
For example, we'll have the following calculation for the 1st row.
Zy = (y - ybar)/(SD(y))
Zy = (60.99 - 786.49)/(1417.82)
Zy = -0.511701
We have this so far
| x | y | Zx | Zy |
| 0.5 | 60.99 | -0.721636 | -0.511701 |
| 1 | 77.99 | -0.670895 | -0.499711 |
| 2 | 112.97 | -0.569413 | -0.475039 |
| 3 | 110.99 | -0.467932 | -0.476436 |
| 4 | 151.99 | -0.36645 | -0.447518 |
| 6 | 425.34 | -0.163487 | -0.254722 |
| 8 | 597.11 | 0.039476 | -0.133571 |
| 12 | 1081.99 | 0.445403 | 0.208419 |
| 32 | 4459 | 2.475036 | 2.590251 |
Eg: Zx*Zy = (-0.721636)*(-0.511701) = 0.369262 in row one
This is what the fully completed table looks like
| x | y | Zx | Zy | ZxZy |
| 0.5 | 60.99 | -0.721636 | -0.511701 | 0.369262 |
| 1 | 77.99 | -0.670895 | -0.499711 | 0.335254 |
| 2 | 112.97 | -0.569413 | -0.475039 | 0.270493 |
| 3 | 110.99 | -0.467932 | -0.476436 | 0.22294 |
| 4 | 151.99 | -0.36645 | -0.447518 | 0.163993 |
| 6 | 425.34 | -0.163487 | -0.254722 | 0.041644 |
| 8 | 597.11 | 0.039476 | -0.133571 | -0.005273 |
| 12 | 1081.99 | 0.445403 | 0.208419 | 0.09283 |
| 32 | 4459 | 2.475036 | 2.590251 | 6.410964 |
Add up the values in the final column and you should get roughly 7.902107
So,
r = Sum(ZxZy)/(n-1)
r = 7.902107/(9-1)
r = 7.902107/8
r = 0.987763375
r^2 = (0.987763375)^2
r^2 = 0.9756764849914
r^2 = 0.9757
Since r^2 is very close to 1, this makes the linear regression a good fit. Approximately 97.57% of the variation in x explains the variation in y.
Answer: r^2 = 0.9757 approximately