Ans:
a)
Froma above data,it best fit the linear regression function.
b)
Area(x) | cost(y) | xy | x^2 | y^2 | |
1 | 90 | 246 | 22140 | 8100 | 60516 |
2 | 85 | 230 | 19550 | 7225 | 52900 |
3 | 105 | 269 | 28245 | 11025 | 72361 |
4 | 110 | 284 | 31240 | 12100 | 80656 |
5 | 120 | 298 | 35760 | 14400 | 88804 |
6 | 130 | 322 | 41860 | 16900 | 103684 |
7 | 140 | 355 | 49700 | 19600 | 126025 |
8 | 95 | 242 | 22990 | 9025 | 58564 |
Total= | 875 | 2246 | 251485 | 98375 | 643510 |
slope,b=(8*251485-875*2246)/(8*98375-875^2)=2.1815
y-intercept,a=(2246-2.18152*875)/8=42.146
Regression eqn:
y'=2.1815x+42.146
c)when x=125 m^2
y'=2.1815*125+42.146=314.83
d)
correlation cofficient,r=(8*251485-875*2246)/SQRT((8*98375-875^2)*(8*643510-2246^2))=0.9911
cofficient of determination,r^2=0.9911^2=0.9822 or 98.22%,which indicates that 98.22% of the linear variation in monthly cost is explained by the area.
Consider the following data. Monthly natural gas cost, TL 246 230 269 284 298 322 355...
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