Project Management Analysis and Statistical Investigation of Insurance Product Implementation and Student Accommodation Factors

Overview

This coursework assignment comprises two distinct tasks: a critical path method analysis for implementing a new insurance product with project scheduling and timeline optimization, and a chi-square test of independence examining the relationship between accommodation type and gender among university students. The analysis demonstrates project management techniques and statistical hypothesis testing methodologies.

What this coursework covers

Preview

Faculty of Business and Law

Assignment Mode E and R Regulations

Task 1: (25 Marks)

Project: Implementation of a New Insurance Product

Activity

Duration (weeks)

Code

Predecessor(s)

Market Research

2

A

-

Product Design

3

B

A

Actuarial Analysis

4

C

B

Pricing Strategy

2

D

C

Marketing Plan

3

E

B

Underwriting Guidelines

3

F

B

System Development

6

G

C

Policy Administration System Integration

4

H

G

Testing and Quality Assurance

4

I

G

Employee Training

2

J

H, I

Product Launch

2

K

E, F, J

Monitoring and Feedback

1

L

K

Earliest and Latest Starting and Finishing Time

Activity

ES

EF

LS

LF

A

0

2

0

2

B

2

5

2

5

C

5

9

5

9

D

9

11

9

11

E

5

8

8

11

F

5

8

8

11

G

9

15

9

15

H

15

19

15

19

I

15

19

15

19

J

19

21

19

21

K

21

23

21

23

L

23

24

23

24

Duration, Total Float, and Critical Path:

Activity

Duration

Total Float

Critical Path

A

2

0

Yes

B

3

0

Yes

C

4

0

Yes

D

2

4

No

E

3

3

No

F

3

3

No

G

6

0

Yes

H

4

0

Yes

I

4

0

Yes

J

2

2

No

K

2

0

Yes

L

1

0

Yes

The critical path is A-B-C-G-H-I-K-L with a total duration of 24 weeks. Any delay in the critical path activities will result in a delay in the project completion date. Activities D, E, F, and J have total float and can be delayed without affecting the project completion date.

Task 2: (35 Marks)

  1. Tests

Time To University?

A-Level: Total Points?

Mean

40.80795

Mean

167.2715

Standard Error

2.589616

Standard Error

9.300654

Median

32

Median

200

Mode

32

Mode

0

Standard Deviation

31.82174

Standard Deviation

114.2883

Sample Variance

1012.623

Sample Variance

13061.83

Kurtosis

-0.10313

Kurtosis

-1.18802

Skewness

0.932481

Skewness

-0.36524

Range

132

Range

380

Minimum

5

Minimum

0

Maximum

137

Maximum

380

Sum

6162

Sum

25258

Count

151

Count

151

To investigate whether the factor 'Accommodation Type' influences the type of gender, conduct a chi-square test of independence.

First, we need to create a contingency table with 'Accommodation Type' as rows and 'Gender' as columns. We will count the number of males and females in each accommodation type.

Male

Female

House

9

8

Flat

7

4

Halls

2

4

Other

1

1

Next, we will perform the chi-square test of independence using a significance level of 0.05.

The null hypothesis is that there is no association between 'Accommodation Type' and 'Gender', while the alternative hypothesis is that there is an association between the two variables.

The test statistic is calculated as follows:

χ² = Σ (Observed - Expected)² / Expected

where Observed is the observed frequency count and Expected is the expected frequency count under the null hypothesis.

Using a calculator, we can obtain the expected frequency counts as follows:

Male

Female

House

8.06

8.94

Flat

5.03

5.97

Halls

2.41

3.59

Other

0.5

1.

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