2 decimal places accuracy adopted for this linear regression.
Summary statistics for X: mean=1.00Tb, pop SD=1.29Tb, sample SD=1.38Tb.
Summary statistics for Y: mean=$154.11, pop SD=$141.85, sample SD=$151.64.
Correlation coefficient R=0.80 (suggests good correlation)
Linear regression equation: Y=64.46+87.98X.
X Tb |
$Y (actual) |
$Y (predicted) |
0.08 |
29.99 |
73.50 |
0.12 |
35 |
77.02 |
0.2 |
299.99 |
84.06 |
0.25 |
49.95 |
88.46 |
0.32 |
68.99 |
94.61 |
1 |
99.99 |
154.44 |
2 |
200 |
242.42 |
4 |
449 |
418.38 |
The table shows the actual measured values and values predicted by the linear equation. The cost of 0.2Tb at $299.99 clearly affected the correlation coefficient so that the predicted cost values tend to be higher than the actual values, except for 0.2Tb and 4Tb. The sensitivity of R has been influenced by the high cost for 0.2Tb, which has also shifted the mean cost. (A notably different result would be derived by treating the cost of 0.2Tb as an outlier. R then becomes 0.99 (very good correlation), the linear equation becomes Y=18.25+103.62X, producing (0.08Tb,$26.54), (0.12Tb,$30.68), (0.25Tb,$44.16), (0.32Tb,$51.41), (1Tb,$121.87), (2Tb,$225.49), ($Tb,$432.73).)