In: Statistics and Probability
1. We have the data as follows. There are three independent variables and three dependent variables
(You may use the following table to solve this problem)
x |
y |
|||||||||
3 |
11 |
|||||||||
5 |
6 |
|||||||||
7 |
4 |
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Total |
15 |
21 |
a) Calculate b1 and b0, and write the equation of the least squares line.
b) Determine the values of SSE and SST.
c) Calculate the standard error.
d) Find the rejection point for the t statistic at α = .05 and test H0: β1 = 0 vs. Ha: β1 ≠ 0.
e) What is the correlation coefficient?
2. Consider the following partial computer output from a simple linear regression analysis.
Regression Statistics
R Square ( )
Standard Error 0.4449
ANOVA df SS MS F P
Regression 1 ( ) ( ) 0.000
Residual 13 ( ) ( )
Total 14 12.3240
Coefficient Standard Error t Stat P-Value
Intercept 4.8615 ( ) ( ) ( )
Ind. Variable -0.3475 0.0487 ( )
a) Write the equation of the least squares line.
b) Calculate the t statistic used to test H0: β1 = 0 versus Ha: β1 ≠ 0 at α = 0.01.
What do you conclude about the relationship between y and x?
c) What is the predicted value of y
when x = 9.00?
d) Calculate the MSE and SSE.
e) What are the unexplained variance and the explained variance?
f) What is the coefficient of determination and the correlation coefficient?
3. Consider the following partial computer output from a multiple regression analysis.
ANOVA df SS MS F P
Regression 4 12.5268 4.0089 ( )
Residual 25 1.4318 0.0551
Total 29 13.9586
a) Calculate the F(model) statistic by using the explained and unexplained variations and other relevant quantities. Find F(model) on the output to check your answer.
b) Use the F(model) statistic and the appropriate critical value to test the significance of the linear regression model under consideration by setting α equal to 0.05.
1)
a)
ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 15.00 | 21.00 | 8.00 | 26.00 | -14.00 |
mean | 5.00 | 7.00 | SSxx | SSyy | SSxy |
sample size , n = 3
here, x̅ = Σx / n= 5.00 ,
ȳ = Σy/n = 7.000
SSxx = Σ(x-x̅)² = 8.0000
SSxy= Σ(x-x̅)(y-ȳ) = -14.0
estimated slope , ß1 = SSxy/SSxx =
-14.0 / 8.000 =
-1.7500
intercept, ß0 = y̅-ß1* x̄ =
15.75000
so, regression line is Ŷ =
15.750 + -1.750 *x
..
b)
SSE= (SSxx * SSyy - SS²xy)/SSxx =
1.50
SST = 26.000
................
c)
std error ,Se = √(SSE/(n-2)) =
1.2247
...
d)
Ho: ß1= 0
H1: ß1╪ 0
n= 3
alpha = 0.05
estimated std error of slope =Se(ß1) = Se/√Sxx =
1.225 /√ 8.00 =
0.4330
t stat = estimated slope/std error =ß1 /Se(ß1) =
-1.7500 / 0.4330 =
-4.0415
t-critical value= 12.7062 [excel function:
=T.INV.2T(α,df) ]
rejection poitn:
|test stat| > |critical |
.......
e)
correlation coefficient , r = SSxy/√(SSx.SSy)
= -0.9707
.........................................
.....................
2)
a)
Y =4.8615 -0.3175*x
..........
b)
t-test statistic = t = estimated slope / std error = -0.3475 / 0.0487 = -7.14
Df = n - 2 = 13
p-value = 0.0000 [excel function:
=t.dist.2t(t-stat,df) ]
decision: p value < α , so, reject the null
hypothesis
linear relation exists
.
c)
Y =4.8615 -(0.3175*9)
=2.004
...
d)
SSE =2.5732
MSE =2.572/13
=0.1978
............
THANKS