Naeem (g1139591) Qdm Individual Assignment
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INDIVIDUAL ASSIGNMENT
QUANTITATIVE DECISION MAKING
Submitted to: DR. RAFIKUL ISLAM
Date: 7.12.2012
Prepared by:NAEEM NASSER (G1139591)
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Problem # 45 (chapter- 10)
∑ x=(29+40+12+10+6+41+25+21+5+4+19+2+7+10+8+3+6+52+4+12+33+6+2+17+21+8+38+2+13+8+14+11+2)
= 491 n = 35
x = ∑ x
n=
49135 = 14.028
Step 1 Identify the null and alternative hypothesis H0: µ = 13 H1: µ≠ 13 (2 tailed test)
Step 2 Rejection region α = 0.05 from z table, z = 0.05 = 1.96
Step 3 Test statistics z = x−μ
S
√n= 14.028−13
11.6 /√ 35
¿ 1.028
1.962 ¿0.523
Step 4 Make the decisionAccept the H0
0.523
13 1.96
Here the computed value is less than table value, so we do not reject null hypothesis
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The mean travel time to work for all north Dakota residents did not change from the 1990 mean 13 minutes.
Problem # 46 (chapter 10)
∑ x=(15+16+15.3+14.6+17+18.1+12.6+9.4+11.9+6.3+14.4+16.6+19.5+12.5+16.8+ 14.6+20.7+18.3+18.6+12.5+10.9+19.8+14.5+13.1+16.3+18.1+11.6+16.6+12.1+14.7+18.2+12.8+11.5+10.7+12.7+18.3+15.6+16.4+17.3+16.3+15+11+12.5+12.4+11.5)
= 660.6 n = 45
x = ∑ x
n=
660.645 = 14.68
Step 1 Identify the null and alternative hypothesis H0: µ = 18 H1: µ¿18 (1 tailed test)
Step 2 Rejection region α = 0.01 from z table, z = 0.01 = -2.33
Step 3 Test statistics z = x−μ
S
√n= 14.68−18
4.2/√ 45
¿ −3.32
0.62 ¿−5.35
Step 4 Make the decisionReject the H0
-5.35
-2.33 18
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Here the computed value is more than table value, there fore reject null hypothesis. RDA of iron for adult female under the age 51 is not 18 mg.
Problem # 50 (chapter 11)
Step 1 Identify the null and alternative hypothesis H0: µ≥ µm H1: µw<µm
Step 2 Rejection region α = 0.001 from z table, z = 0.001 = 3.8
n1=87 n2=76x1=3343 x2=5568S1=1226 S2=1716
Step 3 Test statistics
z = x1−x2
√ S12
n1
+S2
2
n2
=−2225
√56022.2 = -9.40
Step 4 Make the decisionReject the H0
-9.40
-3.8
Here computed z value is more than the table value, so reject null hypothesis.
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There fore women does not annually pay the same or more than men into private pension funds.
Problem # 51 (chapter 11)
Step 1 Identify the null and alternative hypothesis H0:µ1 ¿ µ2 H1: µ1≠ µ2
Step 2 Rejection region α = 0.01 from z table, z = 0.01 = 2.58
n1=50 n2=50x1=1.96 x2=3.02S1=1 S2=0.917
Step 3 Test statistics
z = x1−x2
√ S12
n1
+S2
2
n2
=−1.06
√0.0368 = -5.54Step 4 Make the decision
Reject the H0
-2.58
-5.54
Step 6 At 1% significance level, there is no difference between people in these two industries computer/ electronics and food/ beverage.
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Problem # 65 (chapter 12)
Step 1 Identify the null and alternative hypothesis H0:µ1 ¿ µ2= µ3=µ4
H1: µ1≠ µ2≠ µ3≠ µ4
BRAND A BRAND B BRAND C BRAND D4230392829
2836313227
2436282833
2032382825
ΣX=168X 1=33.6Σ x2=5810
s1=41.3
ΣX=154X 2=30.8Σ x2=4794
s2=12.7
ΣX=149X 3=29.8Σ x2=4529
s3=22.2
ΣX=143X 4=28.6Σ x2=4277
s4=46.8
Mean of x = 30.7
Step 2 Rejection region 5% from f table =3.24
Step 3 Test statistics
SSTR= n1(X 1−X ¿2+n2(X 2−X ¿2+n3(X 3−X ¿2+n4(X 4−X ¿2
= 5(33.6-30.7¿2+5(30.8-30.7¿2+5(29.8-30.7¿2+5(28.6-30.7¿2
= 42.05+0.05+4.05+22.05 = 68.2
MSTR = SSTRK−1
= 68.24−1
= 22.73
SSE = (n1-1)s1+(n2-1)s2+(n3-1)s3+(n4-1)s4
=4(41.3)+4(12.7)+4(22.2)+4(46.8) = 492
MSE = SSEn−k
= 49216
= 30.75
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F =MSTRMSE
=22.7330.75
= 0.739
F Table value, Numerator = k-1 = 3 Denominator = n – k = 16There fore, table value = 3.24
Step 4 Make the decision Here F table value is 3.24 and the computed value is 0.739. there fore we accept the null hypothesis.At 5% significance level, there appear to be a difference in mean lifetime among the four brands of batteries.
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Problem # 66 (chapter 12)
Step 1 Identify the null and alternative hypothesis H0:µ1 ¿ µ2= µ3
H1: µ1≠ µ2≠ µ3
P1 P2 P3222026212422
212525202226
292431322627
ΣX=135X 1=22.5Σ x2=3061
s1=4.7
ΣX=139X 2=23.1Σ x2=3235
s2=6.16
ΣX=169X 3=28.1Σ x2=4807
s3=9.36
Mean of x = 24.56
Step 2 Rejection region 1% from f table = 6.36
Step 3 Test statistics
SSTR= n1(X 1−X ¿2+n2(X 2−X ¿2+n3(X 3−X ¿2
= 6(22.5-24.56¿2+6(23.1-24.56¿2+6(28.1-24.8¿2
= 25.44+12.78+75.18 = 113.4
MSTR = SSTRK−1
= 113.43−1
= 56.7
SSE = (n1-1)s1+(n2-1)s2+(n3-1)s3
=5(47)+4(12.7)+5(6.16)+5(9.36) = 101.1
MSE = SSEn−k
= 101.1
15 = 6.74
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F =MSTRMSE
=56.76.74
= 8.41
F Table value, Numerator = k-1 = 2 Denominator = n – k = 15There fore, table value = 6.36
Step 4 Make the decision
Reject H0
At 1% significance level, the data does not provide evidence of a difference in mean monthly sales among three policies.
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Problem # 71 (chapter 13)
RAW STEEL PRODUCTIONx
NEW OREDERSy
99.997.998.987.992.997.9100.6104.9105.3108.6
2.742.872.932.872.983.093.363.613.753.95
ΣX=994.8X=99.48
ΣX=139X=3.21
y = a + bx
a = y - bx
b = rsx
s y
sx = 6.029 ; sy = 0.4241
r = Σ (x−X ) ¿¿ = 18.5932
23.01 = 0.808
b = rs y
sx = (0.808)
0.42416.029
=¿ 0.0568
a = y - bx = 3.21- 5.650 = -2.440
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y = - 2.440 + 0.0568x
Model Summary
Model R R Square Adjusted R Square Std. Error of the Estimate
1 .794a .630 .584 .27350
a. Predictors: (Constant), x
ANOVAb
Model Sum of Squares df Mean Square F Sig.
1 Regression 1.021 1 1.021 13.646 .006a
Residual .598 8 .075
Total 1.619 9
a. Predictors: (Constant), x
b. Dependent Variable: y
Coefficientsa
Model
Unstandardized Coefficients
Standardized
Coefficients
t Sig.B Std. Error Beta
1 (Constant) -2.313 1.499 -1.543 .161
x .056 .015 .794 3.694 .006
a. Dependent Variable: y
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85 90 95 100 105 1100
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Y-Value 1Linear (Y-Value 1)
RAW STEEL PRODUCTION
itle
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Problem # 72 (chapter 13)
X Y262733292934304022
259274294296325380457523215
ΣX=270X=30
ΣX=3023X=335.88
y = a + bx
a = y - bx
b = rsx
s y
sx = 5.19 ; sy = 99.65
r = Σ (x−X ) ¿¿ = 3433
4137.468 = 0.829
b = rs y
sx = (0.829)
99.655.19
=15.930
a = y - bx = 335.88- 15.930 = -142.047
y = - 142.047 + 15.930x
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Model Summary
Model R R Square Adjusted R Square Std. Error of the Estimate
1 .829a .687 .642 59.62075
a. Predictors: (Constant), x
ANOVAb
Model Sum of Squares df Mean Square F Sig.
1 Regression 54562.449 1 54562.449 15.350 .006a
Residual 24882.440 7 3554.634
Total 79444.889 8
a. Predictors: (Constant), x
b. Dependent Variable: y
Coefficientsa
Model
Unstandardized Coefficients
Standardized
Coefficients
t Sig.B Std. Error Beta
1 (Constant) -140.917 123.312 -1.143 .291
x 15.894 4.057 .829 3.918 .006
a. Dependent Variable: y
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20 25 30 35 40 450
100
200
300
400
500
600
Y-Value 1Linear (Y-Value 1)
SQUARE FEET ( hundreds)
PR
ICE
(th
ousa
nd
s o
f dol
lars
)
y = a + bx
= -142.047 + 15.930x = -142.047 + 15.930(2600)
= 272.13
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Problem no 78 (chapter 14)
Model Summary
Model R R Square Adjusted R Square Std. Error of the Estimate
1 .068a .005 -.244 6.54405
a. Predictors: (Constant), dividend, debtratio
ANOVAb
Model Sum of Squares df Mean Square F Sig.
1 Regression 1.580 2 .790 .018 .982a
Residual 342.596 8 42.825
Total 344.176 10
a. Predictors: (Constant), dividend, debtratio
b. Dependent Variable: Insider
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Coefficientsa
Model
Unstandardized Coefficients
Standardized
Coefficients
t Sig.B Std. Error Beta
1 (Constant) 17.677 17.560 1.007 .344
debtratio -.059 .315 -.097 -.189 .855
dividend -.118 1.050 -.058 -.113 .913
a. Dependent Variable: Insider
Problem no 79 (chapter 14)
Model Summary
Model R R Square
Adjusted R
Square
Std. Error of the
Estimate
1 .982a .965 .955 .23310
a. Predictors: (Constant), x3, x1, x2
ANOVAb
Model Sum of Squares df Mean Square F Sig.
1 Regression 16.378 3 5.459 100.472 .000a
Residual .598 11 .054
Total 16.976 14
a. Predictors: (Constant), x3, x1, x2
b. Dependent Variable: y
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Coefficientsa
Model
Unstandardized Coefficients
Standardized
Coefficients
t Sig.B Std. Error Beta
1 (Constant) 3.981 1.573 2.531 .028
x1 .073 .021 .876 3.505 .005
x2 -.032 .021 -.424 -1.547 .150
x3 -.004 .004 -.315 -1.014 .332
a. Dependent Variable: y