In: Math
7. In an area of the Great Plains, records were kept
on the relationship between the rainfall (in inches)
and the yield of wheat (bushels per acre).
Rainfall (in inches) x |
Yield (Bushels per acre) y |
10.5 |
50.5 |
8.8 |
46.2 |
13.4 |
58.8 |
12.5 |
59.0 |
7.0 |
31.9 |
16.0 |
78.8 |
7a. Using the linear
regression feature on your calculator,
find a linear equation that models the
miles per gallon as the respone
variable (y) and the engine size in
liters as the explanatory variable (x). (Use 2 decimal places in
the regression equation.)
7b. Using the line you obtained in 7a. above, compute the sum of
the squared
residuals of the least squares line for the given data. (Use 2
decimal places and show your
calculations, by hand!)
a)
x | y | (x-x̅)² | (y-ȳ)² | (x-x̅)(y-ȳ) |
10.5 | 50.5 | 0.75 | 13.69 | 3.21 |
8.8 | 46.2 | 6.59 | 64.00 | 20.53 |
13.4 | 58.8 | 4.13 | 21.16 | 9.35 |
12.5 | 59 | 1.28 | 23.04 | 5.44 |
7 | 31.9 | 19.07 | 497.29 | 97.38 |
16 | 78.8 | 21.47 | 605.16 | 113.98 |
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 68.2 | 325.2 | 53.29333333 | 1224.3 | 249.89 |
mean | 11.37 | 54.20 | SSxx | SSyy | SSxy |
sample size , n = 6
here, x̅ = Σx / n= 11.37 ,
ȳ = Σy/n = 54.20
SSxx = Σ(x-x̅)² = 53.2933
SSxy= Σ(x-x̅)(y-ȳ) = 249.9
estimated slope , ß1 = SSxy/SSxx = 249.9
/ 53.293 = 4.68895
intercept, ß0 = y̅-ß1* x̄ =
0.90222
so, regression line is Ŷ = 0.90
+ 4.69 *x
7b)
x | y | Ŷ | residual,ei=y-yhat | (Y-Ŷ)² | |||
10.5 | 50.5 | 50.1362 | 0.3638 | 0.13 | |||
8.8 | 46.2 | 42.1650 | 4.0350 | 16.28 | |||
13.4 | 58.8 | 63.7342 | -4.9342 | 24.35 | |||
12.5 | 59 | 59.5141 | -0.5141 | 0.26 | |||
7 | 31.9 | 33.7249 | -1.8249 | 3.33 | |||
16 | 78.8 | 75.9255 | 2.8745 | 8.26 |
SSE = Σ(Y-Ŷ)² = 52.62