Naeem (g1139591) Qdm Individual Assignment
-
Upload
naeem-nasser -
Category
Documents
-
view
15 -
download
5
Transcript of Naeem (g1139591) Qdm Individual Assignment
INDIVIDUAL ASSIGNMENT
QUANTITATIVE DECISION MAKING
Submitted to: DR. RAFIKUL ISLAM
Date: 7.12.2012
Prepared by:NAEEM NASSER (G1139591)
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
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
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.
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.
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
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.
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
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.
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
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
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
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
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
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
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
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
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