In: Statistics and Probability
1. As a result of running a simple regression on a data set, the following estimated regression equation was obtained:
= 9.7 + 13.4x
Furthermore, it is known that SST = 622, and SSE = 150.
2. You are given the following information about y and x:
y |
x |
|
Dependent Variable |
Independent Variable |
|
11 |
6 |
|
15 |
5 |
|
10 |
2 |
|
14 |
2 |
Linear regression using least squares method yielded the
following equation:
= 12.06 + 0.12x
What is the predicted value of y when
x = 1? Round your answer to two decimal
places.
Calculate the correlation coefficient R; round your answer to three decimal places.
3. You are given the following information about variables y and x:
y |
x |
|
Dependent Variable |
Independent Variable |
|
-10.0 |
-9.1 |
|
8.2 |
-7.8 |
|
-5.5 |
6.4 |
|
10.3 |
-9.0 |
In addition, it is known that the slope of the
regression line b1= -6.6
The y-intercept b0for
the estimated regression equation equals ____ (round your answer to
two decimal places).
2)
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 15 | 50 | 12.75 | 17.0 | 1.50 |
mean | 3.75 | 12.50 | SSxx | SSyy | SSxy |
sample size , n = 4
here, x̅ = Σx / n= 3.75 ,
ȳ = Σy/n = 12.50
SSxx = Σ(x-x̅)² = 12.7500
SSxy= Σ(x-x̅)(y-ȳ) = 1.5
estimated slope , ß1 = SSxy/SSxx = 1.5
/ 12.750 = 0.1176
intercept, ß0 = y̅-ß1* x̄ =
12.0588
so, regression line is Ŷ =
12.0588 + 0.117647 *x
SSE= (SSxx * SSyy - SS²xy)/SSxx =
16.8
std error ,Se = √(SSE/(n-2)) =
2.90
correlation coefficient , r = Sxy/√(Sx.Sy)
= 0.102
Predicted Y at X= 1 is
Ŷ = 12.05882 +
0.117647 * 1 =
12.18
3)
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | -19.5 | 3 | 170.5475 | 301.3 | -86.24 |
mean | -4.88 | 0.75 | SSxx | SSyy | SSxy |
sample size , n = 4
here, x̅ = Σx / n= -4.88 ,
ȳ = Σy/n = 0.75
SSxx = Σ(x-x̅)² = 170.5475
SSxy= Σ(x-x̅)(y-ȳ) = -86.2
estimated slope , ß1 = SSxy/SSxx = -86.2
/ 170.548 = -0.5056
intercept, ß0 = y̅-ß1* x̄ =
-1.72
THANKS
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