F-2, Block, Amity CampusSec-125, Nodia (UP)India 201303

ASSIGNMENTSPROGRAM:SEMESTER-ISubject Name : Master of Finance and Control

Study COUNTRY : Zambia

Permanent Enrollment Number (PEN) :

Roll Number : MFC001412014-2016002

Student Name : DERICK MWANSA

INSTRUCTIONSa) Students are required to submit all three assignment sets.

ASSIGNMENTDETAILSMARKS

Assignment AFive Subjective Questions10

Assignment BThree Subjective Questions + Case Study10

Assignment C40 Objective Questions10

b) Total weightage given to these assignments is 30%. OR 30 Marksc) All assignments are to be completed as typed in word/pdf.d) All questions are required to be attempted.e) All the three assignments are to be completed by due dates (specified from time to time) and need to be submitted for evaluation by Amity University.f) The evaluated assignment marks will be made available within six weeks. Thereafter, these will be destroyed at the end of each semester.g) The students have to attach a scan signature in the form.

Signature:______Date:__20/01/2015____( ) Tick mark in front of the assignments submittedAssignment AAssignment BAssignment C

QUATITATIVE APPLICATIONS IN MANAGEMENTASSIGNMENT AQuestion oneAnswer

No. of TabletsMid-Point (X)No. of People CuredFX

4- 861166

8- 121013130

12- 161416224

16- 201814252

20- 2422X22X

24- 28269234

28-323017510

32- 36346204

36- 40384152

Total90+X1772+22X

ASSIGNEMENT AQuestion twoAnswer

(a) The variances given show that boys have greater variability in individual heights. Boys have variance of 9 whereas Girls have variation of 4. (b) Common average heightSexAverage Frequency(f)

Boys68724896

Girls61382318

Totals1107214

Common Average Height = (f (f)= 7214 110 =65.58 inches

(c) Standard deviation Statistical DataDetailBoysGirls

Number (f)7238

Average Height( ) 6861

Variance 94

Standard deviation 32

Total ( f ) 48962318

332,928141,398

Variance = (Total Squared)/n - (x Squared)/nTnT-1

= 474,326 473,107.24109

= 11.18Therefore (for both heights) = 3.34

(d) Combined variability = 11.18

ASSIGNEMENT AQuestion threeAnswerYearTSales

XYXY

1965132321

1966247944

19673651959

196849236816

1969513266025

19706190114036

19717275192549

Total288334414140

Average4119

Trend Equation

ASSIGNEMENT AQuestion fourAnswer

XYXYy2

112121144

211224121

313399169

4156016225

5147025196

61710236289

71611249256

81915264361

91816281324

Total451357312852085

Average515

Standard deviation of X

Standard deviation for Y

Covariance of XY ()

Coefficient of correlation between X and Y ()

ASSIGNEMENT AQuestion 5AnswerYearIndex5 Yearly Cycles5 Yearly Trend Values

1941225

1942210

19432011074214.8

19442151094218.8

19452231119223.8

19462451143228.6

19472351161232.2

19482251187237.4

19492331207241.4

1950249

1951265

ASSIGNMENT BQuestion oneAnswerWe establish the null and alternative hypothesis. Ho: Die is not biasedH1: Die is biasedP (success) =1/6P (failure) = 5/6No. (X)Frequency(f) (Observed)ProbabilityP(X)ExpectedP(X)120O-E(O-(O-

1301/620101005

2251/6205251.25

3181/620-240.2

4101/620-101005

5221/620240.2

6151/620-5251.25

12012.9

Chi square test at 0.05 significance level.Degrees of freedom ( d f ) = (r 1) (c --1) = (6-1) (2-1) = (5) (1) = 5The Critical value is: Since the calculated chi square value of 12.9 is greater than the critical value of 11.1, we reject the null hypothesis and conclude that the die is biased.

ASSIGNEMENT BQuestion twoAnswerLet X1 be units of material A X2 be units of material B Minimize Subject to

Corner PointCoordinatesObjective function

Value

A (20,14)1520

B(2,14))1160

C (0,14))1200

Therefore, optimal solution is:No of units of product A = 2 UnitsNo of units of product B = 14 UnitsTotal cost, = 1160 which is the minimum

ASSIGNEMENT BQuestion threeAnswerBelow is the probability decision tree showing different alternatives and the associated benefits.

A critical analysis of the decision tree reviews that bidding would result in a loss in monetary terms of 8750. Therefore, the company should not bid.

ASSIGNEMENT BCase StudyAnswer Part (A)YearQuarterTime Population Level (Y)SM4SM2C%MASeasonal IndicesDeseasonalised

112931.11882261.88

1996222462630.94888259.25

332312652640.87500.86383267.41

44282266.5265.751.06111.07162263.15

15301265.52661.13161.11882269.03

199726252267.75266.6250.94510.94888265.58

37227268.5268.1250.84660.86383262.78

48291270.25269.3751.08031.07162271.55

19304273.25271.751.11871.11882271.72

1998210259274.5273.8750.94570.94888272.95

311239275274.750.86990.86383276.67

412296276.5275.751.07341.07162276.22

113306276.75276.6251.10621.11882273.50

1999214265277.75277.250.95580.94888279.28

3152400.86383277.83

4163001.07162279.95

Unadjusted Seasonal Indexes

Q11.118814964

Q20.948883544

Q30.863833919

Q41.071620734

SUM4.00315316

Adjusted Seasonal Indexes

Q11.117933708

Q20.948136137

Q30.863153504

Q41.070776651

SUM4.000000000

Case StudyPart (B)Answer

YEARTIME (X)SALES (Y)XYX2

12932931

199622464924

32316939

4282112816

5301150525

19976252151236

7227158949

8291232864

9304273681

1998102592590100

112392629121

122963552144

133063978169

1999142653710196

152403600225

163004800256

T0TALS1364332371351496

MEAN8.5270.75

Regression Line

Case StudyPart CAnswerWinter (X)Summer (Y)X-Y(X-Y)2

1293301-864

2246252-636

3231227416

4282291-981

530630424

6265259636

724023911

8300296416

Totals21632169-6254

Mean270.38271.13

H0: pollution levels are higher in winter than in summer and that they are increasing over the yearsH1: pollution levels are not higher in winter than in summer and that they are not increasing over the years

Level of significance: 0.05

Test statistic:

Since critical t of 1.895 is greater than the calculated t (-0.355). We reject the null hypothesis and conclude that there is not statistical significance in the levels of pollution in winter and summer.

ASSIGNEMENT CMULTIPLE CHOICEAnswer.

1C9A17D25B33A

2A10B18A26C34A

3C11B19D27C35B

4A12A20B28A36E

5B13A21B29C37C

6C14C22B30B38C

7C15A23B31E39B

8A16C24C32D40A