Question
In: Statistics and Probability
Individual Bettendorf Experience (X1) Education (X2) Sex (X3) 1 53600 5.5 4 F 2 52500 9...
| Individual | Bettendorf | Experience (X1) | Education (X2) | Sex (X3) |
| 1 | 53600 | 5.5 | 4 | F |
| 2 | 52500 | 9 | 4 | M |
| 3 | 58900 | 4 | 5 | F |
| 4 | 59000 | 8 | 4 | M |
| 5 | 57500 | 9.5 | 5 | M |
| 6 | 55500 | 3 | 4 | F |
| 7 | 56000 | 7 | 3 | F |
| 8 | 52700 | 1.5 | 4.5 | F |
| 9 | 65000 | 8.5 | 5 | M |
| 10 | 60000 | 7.5 | 6 | F |
| 11 | 56000 | 9.5 | 2 | M |
| 12 | 54900 | 6 | 2 | F |
| 13 | 55000 | 2.5 | 4 | M |
| 14 | 60500 | 1.5 | 4.5 | M |
1. What is the coefficient of determination between the three
predictors taken as a group and annual salary.
Select one:
a..323
b..772
c..522
d..771
2. Let X1 = experience, X2 = Education, and X3 = Sex, what is the multiple regression equation?
Select one:
a.Y= 2809 + 228.5(X1) + 560.6(X2) + 1287.4(X3)
b.Y=41462.6 + 337.3(X1) + 2169.3(X2) + 3097.0(X3)
c.Y=48951.9 + 195.3(X1) + 1480.6(X2) + 1595.1(X3)
d.Y= 42410.2 + 403.5(X1) + 1856.4(X2) + 2964.4(X3)
Solutions
Expert Solution
The R output of the given regresssion problem is given below.
Bettendorf<-c(53600,52500,58900,59000,57500,55500,56000,52700,65000,60000,56000,54900,55000,60500)
Experience<-c(5.5,9,4,8,9.5,3,7,1.5,8.5,7.5,9.5,6,2.5,1.5)
Education<-c(4,4,5,4,5,4,3,4.5,5,6,2,2,4,4.5)
Sex<-c(1,2,1,2,2,1,1,1,2,1,2,1,2,2)
model <- lm(Bettendorf~ Experience + Education +Sex)# Coding 1 for F and 2 for M
summary(model)
[1]
Call:
lm(formula = Bettendorf ~ Experience + Education + Sex)
Residuals:
Min 1Q Median 3Q Max
-5727.4 -2218.7 667.8 1632.5 5389.6
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 47356.8 4491.8 10.543 9.77e-07 ***
Experience 195.3 329.7 0.593 0.5667
Education 1480.6 809.0 1.830 0.0971 .
Sex 1595.1 1857.6 0.859 0.4106
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3249 on 10 degrees of freedom
Multiple R-squared: 0.3234, Adjusted R-squared: 0.1204
F-statistic: 1.593 on 3 and 10 DF, p-value: 0.2521(1) From this R code We get coefficient of determination
(Multiple 
) as 0.3234
0.323.
Therefore the corerct option is (a)0.323 .
(2) From the coefficients we get the estimates of the coefficients as follows,
| Variables | Estimates |
| Intercept | 47356.8 |
| Experience(X1) | 195.3 |
| Education(X2) | 1480.6 |
| Sex(X3) | 1595.1 |
Therefore the equation of the regression line is 
. Hence the correct option is (c).
(In the given answer they wrongly given the intercept as 48951.9.)
orchestra answered 1 year agoRelated Solutions
Individual Bettendorf Experience (X1) Education (X2) Sex (X3) 1 53600 5.5 4 F 2 52500 9...
Individual
Bettendorf
Experience (X1)
Education (X2)
Sex (X3)
1
53600
5.5
4
F
2
52500
9
4
M
3
58900
4
5
F
4
59000
8
4
M
5
57500
9.5
5
M
6
55500
3
4
F
7
56000
7
3
F
8
52700
1.5
4.5
F
9
65000
8.5
5
M
10
60000
7.5
6
F
11
56000
9.5
2
M
12
54900
6
2
F
13
55000
2.5
4
M
14
60500
1.5
4.5
M
1. At...
Individual Bettendorf Salary Experience (X1) Education (X2) Sex (X3) 1 53600 5.5 4.0 F 2 52500...
Individual
Bettendorf Salary
Experience (X1)
Education (X2)
Sex (X3)
1
53600
5.5
4.0
F
2
52500
9.0
4.0
M
3
58900
4.0
5.0
F
4
59000
8.0
4.0
M
5
57500
9.5
5.0
M
6
55500
3.0
4.0
F
7
56000
7.0
3.0
F
8
52700
1.5
4.5
F
9
65000
8.5
5.0
M
10
60000
7.5
6.0
F
11
56000
9.5
2.0
M
12
54900
6.0
2.0
F
13
55000
2.5
4.0
M
14
60500
1.5
4.5
M
What...
Let X1, X2, X3 be continuous random variables with joint pdf f(X1, X2, X3)= 2 if...
Let X1, X2, X3 be continuous random variables with joint pdf
f(X1, X2, X3)= 2 if 1<X1<2 -1<X2<0
-X2-1<X3<0
0 otherwise
Find Cov(X2, X3)
(1) z=ln(x^2+y^2), y=e^x. find ∂z/∂x and dz/dx. (2) f(x1, x2, x3) = x1^2*x2+3sqrt(x3), x1 = sqrt(x3),...
(1) z=ln(x^2+y^2), y=e^x. find ∂z/∂x and dz/dx.
(2) f(x1, x2, x3) = x1^2*x2+3sqrt(x3), x1 = sqrt(x3), x2 = lnx3.
find ∂f/∂x3, and df/dx3.
DATA 2 ID X1 X2 X3 Y A 0 2 4 9 B 1 0 8...
DATA 2
ID
X1
X2
X3
Y
A
0
2
4
9
B
1
0
8
10
C
0
1
0
5
D
1
1
0
1
E
0
0
8
10
CORRELATION MATRIX
Y
X1
X2
X3
Y
1
?
-0.304
+0.889
X1
?
1
-0.327
0
X2
-0.304
-0.327
1
-0.598
X3
+0.889
0
-0.598
1
1. What is the sum of squares regression for the full model?
(Correct answer is 58, please show me how to get...
DATA 2 ID X1 X2 X3 Y A 0 2 4 9 B 1 0 8...
DATA 2
ID
X1
X2
X3
Y
A
0
2
4
9
B
1
0
8
10
C
0
1
0
5
D
1
1
0
1
E
0
0
8
10
CORRELATION MATRIX
Y
X1
X2
X3
Y
1
?
-0.304
+0.889
X1
?
1
-0.327
0
X2
-0.304
-0.327
1
-0.598
X3
+0.889
0
-0.598
1
Comparing the zero order model and full
model
1. Did the addition of X2 and X3
significantly increase R2? (correct answer is...
Let X1, X2, X3 be independent having N(0,1). Let Y1=(X1-X2)/√2, Y2=(X1+X2-2*X3)/√6, Y3=(X1+X2+X3)/√3. Find the joint pdf...
Let X1, X2, X3 be independent having N(0,1). Let Y1=(X1-X2)/√2,
Y2=(X1+X2-2*X3)/√6, Y3=(X1+X2+X3)/√3. Find the joint pdf of Y1, Y2,
Y3, and the marginal pdfs.
4.Maximize: Z = 2X1+ X2-3X3 Subject to: 2X1+ X2= 14 X1+ X2+ X3≥6 X1, X2, X3≥0...
4.Maximize: Z = 2X1+ X2-3X3
Subject to: 2X1+ X2= 14
X1+ X2+ X3≥6
X1, X2, X3≥0
Solve the problem by using the M-technique.
3. Given is the function f : Df → R with F(x1, x2, x3) = x...
3. Given is the function f : Df → R with F(x1, x2, x3) = x 2 1 +
2x 2 2 + x 3 3 + x1 x3 − x2 + x2 √ x3 . (a) Determine the gradient
of function F at the point x 0 = (x 0 1 , x0 2 , x0 3 ) = (8, 2,
4). (b) Determine the directional derivative of function F at the
point x 0 in the direction given...
Consider the problem maximize Z = 5 x1 + 3 x2 + 2 x3 + 4...
Consider the problem maximize Z = 5 x1 +
3 x2 + 2 x3 + 4 x4
subject
to
5 x1 + x2 + x3 + 8 x4 = 10
2 x1 + 4 x2 + 3 x3 + 2 x4 =
10
X j > 0, j=1,2,3,4
(a) Make the necessary row reductions to have the tableau ready
for iteration 0. On this tableau identify the corresponding initial
(artificial) basic feasible solution. Also, identify the initial
entering and...
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