The critical value for this two tailed test at 5% level of significance =
1.96.
Now Z =
P1 P2 1 2
S P1 P2
But since the null hypothesis is 1 = 2, the second part of the numerator
disappear i.e.
1  2 = 0 which will always be the case at this level.
The
F=
S12
S 22
which is the test statistic.
Which follows F distribution with V 1 and V2 degrees of freedom. The
larger sample variance is placed in the numerator and the smaller one in
the denominator
If the computed value of F exceeds the table value of F,
The following is used when sample mean > population mean
Acceptance region
Critical region (rejection region)
5% = 0.05
0
Z = 1.65 (critical value)
NB: If the sample mean standardized value < 1.65, we accept the null
hypothesis but reject the alternative.
Where:
pq pq
n1 n2
S P1 P2
p=
(e)
p1n1 p2 n2
n1 n2
q=1P
Chisquare test
X2 =
O E
2
E
Where O = observed frequency
E=
= expected frequency
Column total Row total
SampleSize
(f)
F test (variance test)
F=
S12
S22
here the bigger value between the standard
Therefore we can state with 99% confidence that the population mean is
between 41.93 and 58.07 grams
Note: To use the t distribution tables it is important to find the degrees of
freedom (v = n 1). In the example above v = 12 1 = 11
From the tables we fin
i.
Sample mean 2.575 includes 99% of the population
1.
LARGE SAMPLES
These are samples that contain a sample size greater than 30(i.e. n>30)
(a)
Estimation of population mean
Here we assume that if we take a large sample from a population then
the mean of
The computer generated solution for this problem is as follows;
Objective value = 71666.7
Variable
Value
Obj. Coef
1666.7
1500
0
25
20
24
Constraint
RHS
1(<)
2(<)
3(<)
4(<)
5(>)
500
1000
800
600
1500
Slack/Surpl
us+
0
250316.670
166.7+
X1: Xtragrow
X2: Y
i.
A risky contract promising shs 7 million with probability 0.6 and
shs 4 million with probability 0.4 and
ii.
A diversified portfolio consisting of two contracts with
independent outcomes each promising Shs 3.5 million with
probability 0.6 and shs 2 mil
LESSON EIGHT
Operation Research

Linear programming
Transport and Assignments
Network Analysis
8.1
Linear programming
Linear programming is a technique of decision making used by managers
to allocate limited resources eg machinery, raw materials and labo
Step 2. Reduce each row by smallest figures in that row.
Table 10
W
X
Y
Z
A
3
0
20
6
B
0
13
0
7
C
23
11
12
0
D
12
0
17
1
Step 3.Cover zeros by minimum possible number of lines.
Table 11
W
X
Y
Z
A
3
0
20
6
B
0
13
0
7
C
23
11
12
0
D
12
0
17
1
Step 4. If a n
2.
A company has four salesmen who have to visit four clients. The
profits records from previous visits are shown in the table and it is
required to maximise profits by the best assignments.
Customer 1
A
B
C
D
6
22
12
16
12
18
16
8
20
15
18
12
12
20
15
20
QUESTION SEVEN
Torch bulbs are packed in boxes of 5 and 100 boxes are selected
randomly to test for the number of defectives
Number of
Number
Total
Defectives
of boxes
defectives
0
40
0
1
37
37
2
17
34
3
5
15
4
1
4
5
0
0
100
90
The number of any individua
Since the computed value of Z is less than the critical value of Z = 1.96
at 5% level of significance therefore we accept the hypothesis and
conclude that there is no significant difference in the habit of taking tea
in the two villages A and B
t distribu
Then the net cost for 12 day duration = 1,300 + (140 50) =
1,390.
The network becomes
1
3
A
D
3
3
9
3 (crash)
12
5
C
E
4
0
0
(d)
B
0
8
Next we reduce D & E
Project duration = 11 days
Project cost = 1,510
Critical activities = All
2
7
7
12
(e)
Final reduct
Example
Given two samples A and B of 100 and 400 items respectively, they have
the means
= 7 ad
= 10 and standard deviations of 2 and 3
X1
X2
respectively. Construct confidence interval at 70% confidence level?
Solution
Sample
A
= 7
B
= 10
X1
X2
n1 = 100
Requirements
Availabl
e
A
B
C
D
Dummy
25
25
42
8
10
17
8
10
I
40
5+
II
20
20
III
50
8+
42
The maximum transferable number is the lowest number in the minus
cell, i.e. 17. after the transfer is made we get;
Availabl
e
A
B
C
D
Dummy
25
25
42
8
10
0
8
1
Option 2
Expected value of Giuoco
Node (A): 0.7(90,000 6) + 0.3(30,000 6)
= 378,000 + 54,000 = Shs. 432,000
Note that the figures a multiplied by 6 to account for the 6 years.
Option 3
Expected value of market research
Node (B): 0.95(100,000 6) + 0.05(25,
LESSON 7 REINFORCING QUESTIONS
QUESTION ONE
An Oil Company has recently acquired rights in a certain area to conduct
surveys and test drillings to lead to lifting oil where it is found in
commercially exploitable quantities. The area is already considered
(25 8) X1 +(8 15) X2 1800 (This can be simplified further as)
200 X1 +120 X2 1800
(dividing through by 40)
5X1 +3 X2 45
X1 2
X1, X2 0
Thus the LP model is:
Minimize
40X1 + 36X2
Subject to:
X1 8
X2 10
5X1 +3 X2 45
X1 2
X1, X2 0
Plotting this on a graph we
a) Start from the start event giving it 0 values,
b) For the rest of the events EST is obtained by summing the EST of
the tail event and the activity duration
c) Where two or more routes converge into an activity, calculate
individual EST per route and th
Table 3
Requirement
Available
A
B
C
D
3
3
4
5
X
2 Units
+
2
Y
6 Units
1
1+
Z
7 Units
2
4
5
Table 3 is a reproduction of Table 2 with a number of + and  inserted.
These were inserted for the following reasons.
Cell X : A + indicates a transfer in as ind
Solving linear programming problems
The question requires us to optimize (in our case, maximize) the
objective (the contribution function), or in simple terms we are required
to solve the linear programming model.
Solving linear programming model entails
there are 15 machines available. The management wish to minimise
delivery costs by dispatching the computers from the appropriate branch
for each customer.
Details of the availabilities, 'requirements, and transport costs per
computer are given in the fol
Note:
for examination purposes, float always refers to total float
The total float can be calculated separately for each activity but it is
often useful to find the total float over chains of noncritical activities
between critical events
Example.
The fo
Now including the NonNegativity constraints since no negative product
can be produced;
X1 0; x2 0
We must now consider how to choose the production which will maximize
contribution. This we do by plotting a line representing the objective
function (4x1 +
a) The appropriate probability tree is shown in the figure below. The
alternatives available to the management of KAM are identified.
The joint probabilities are the result of the path sequence that is
followed. For example, the sequence enter market inst
LESSON SEVEN
Decision Theory
7.1
Decision theory
Decision trees and sequential decisions
Game theory
Decision Theory
Types of decisions
There are many types of decision making
1. Decision making under uncertainty
These refer to situations where more than