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