Business Mathematics and Statistic (523)
First Assignment
1
Q.1 (a)
The following table presents the number of new business
opened in Pakistan during the third quarter of 1990.
Construct a pie chart to represent these data.
Ans.
To construct the pie chart, we will have to calculate the
percentage for each type of new business opened to the
total number of businesses opened during the third
quarter of 1990 and then angle of each representation will
determine in the following manner:
The total number of new businesses opened = 31728
Nature of
Business
No. of
Business
Retailing
Services
Construction
Wholesaling
Manufacturing
Other
Total:
10,724
4,886
4,315
3,776
2,760
5,267
31,728
Angle
33.8/100 x 360 =
15.4/100 x 360 =
13.6/100 x 360 =
11.9/100 x 360 =
8.7/100 x 360 =
16.6/100 x 360 =
122°
55°
49°
43°
31°
60°
360°
Now the pie chart will be constructed by dividing a
circular pie into six parts of proportional sizes as
indicated by the percentages representing them. The pie
chart would appear as:
16.6%
Retailing
33.8%
8.7%
Serv ices
Construction
Wholesaling
11.9%
Manuf acturing
13.6%
Ejaz Alam Khan - H 5279752
15.4%
Other
#1
Business Mathematics and Statistic (523)
2
First Assignment
(b)
Why does business statistics usually involve the use of
sample data instead of the entire population when
making a business decision.
Ans.
Business statistics usually involves the use of sample
data instead of the entire population, due to the following
reasons.
•
In most cases, it is impossible to obtain the data for
the entire population. For example, if a TV network
wants to obtain the views of its viewers about a
certain talk show, it would be practically impossible
for the network to reach all of its viewers.
•
Sampling saves time, cost, and effort. To gather
data about every member of the population is
usually very time consuming and costly. Usually, it
is so costly that the benefits expected to be
achieved from analyzing the data are not worth the
cost involved. Therefore, a random sample, which is
representative of the population, is chosen.
•
In some cases, inspection or testing of a good or
product destroys its usefulness. For example, if a
firm manufactures bullets, it will not be feasible to
test each and every bullet by firing it because then
they would become useless. Therefore, in such
cases, sampling is unavoidable.
Ejaz Alam Khan - H 5279752
#2
Business Mathematics and Statistic (523)
First Assignment
3
Q.2
A particular industrial product is shipped in lots of 20.
Testing to determine whether an item is defective is costly
hence the manufacturer samples production rather than
using a 100% inspection plan.
A sampling plan
constructed to minimize the number of defective units
shipped to customers calls for sampling 5 items from
each lot and rejecting the lot if more than one defective
units is observed. (If rejected, each item in the lot is then
tested). If a lot contains 4 defectives units, what is the
probability that it will be accepted.
Ans.
n=5
Probability of acceptance - p =
Probability of rejection - q
=
P(r ≤ 1) =
P(0)
1/5
4/5
P(0)+P(1)
= C n pr q n-r
r
5
0
5
= C 0 (.2) (.8)
= 0.3277
P(1)
= C n pr q n-r
r
5
1
4
= C1 (.2) (.8)
= 0. 4096
= 0.3277 + 4096
= 0.7373
Ejaz Alam Khan - H 5279752
#3
Business Mathematics and Statistic (523)
4
Q.3(a)
First Assignment
A decline in the demand for a specific brand of a product
may result from either a decline in demand for the
product, in general, or a reduction in the percentage of
the purchasers selecting this specific brand over its
competitors. Suppose that to remain competitive in the
marketplace, a company producing brand “A” must
maintain at least an equal share of the market with its
three major competitors. IF 100 consumers are randomly
selected and interviewed regarding their brand
preferences, what is the probability of observing sample
proportion preferring brand “A” as small as 0.15 or less if
in fact one-fourth or 25 per cent of the consumers prefer
brand “A”.
Ans.
n = 100
p 0   
q 0   
and q 0 = 1 - p 0
P(P ≤ 0.15) = ?
So
Z =
Z =
Z =
P - p0
p 0q 0
n
0.15 - 0.25
  

0.15 - 0.25
0.0433
P(P ≤ 0.15) = P(Z≤ -2.31)
= 0.5-P(-2.31≤ Z≤ 0)
= 0.5-0.4896
So
P(P ≤ 0.15) = 0.0104
Ejaz Alam Khan - H 5279752
#4
Business Mathematics and Statistic (523)
First Assignment
5
(b)
A sample of n=100 employees from a company was
selected and the annual salary for each was recorded.
The mean and standard deviation of their salaries were
found to be X = Rs. 7,750 and s = Rs. 900
Construct the 95 per cent confidence interval for the
population average salary µ.
Ans.
X - Z



n
7750 - Z 
7750 - Z 
<  X+Z
900
100
900



n
<   7 + Z 
900
100
100
7750 - 1.96)
900
100
7750 - 176. 4
= 7573.6
and
7750 + Z 
900
100
 7750 + 76.4
= 7926.4
7573.6 <   
Ejaz Alam Khan - H 5279752
#5
Business Mathematics and Statistic (523)
First Assignment
6
Q.4 (a)
The following data have been obtained for a random
sample of ten electronics firms, where X represents age of
the company in years and Y represents annual sales in
millions of dollars. Determine the correlation between age
of the firm and sales.
2
2
X
Y
XY
X
Y
3
10
5
6
12
15
9
2
9
7
78
2.5
6
2.5
3.5
6
6.5
6
1.5
5.5
5
45
9
100
25
36
144
225
81
4
81
49
754
6.25
36.00
6.25
12.25
36.00
42.25
36.00
2.25
30.25
25.00
232.50
7.5
60.0
12.5
21.0
72.0
97.5
54.0
3.0
49.5
35.0
412.0
The coefficient of correlation:
r
 X  XY  Y)
 ( X  X ) Y  Y 
2
2


 X  Y 
 
n
 X 

 X  XY  Y)   XY 

(X  X)
2
(Y  Y)
2
Ejaz Alam Khan - H 5279752
  X2
  Y2 
n
 
Y

n
#6
Business Mathematics and Statistic (523)
First Assignment
7

 X   Y 

XY 
 412
X
n
2
 

X

n
2
  
( Y  Y)   Y
r =
r =
2
 
 

n
78  45
10
= 61
= 145.6

Y
-

= 32.50 
 

= 30
61
145.6  30
61
66. 09
r = 0.923
Ejaz Alam Khan - H 5279752
#7
Business Mathematics and Statistic (523)
First Assignment
8
(b)
Describe a situation in your area of interest in which
simple correlation analysis would be useful. Use real or
realistic data and do a complete correlation analysis.
Ans.
The data given below corresponds to the months-wise
sales and marketing expenses of the ABM Data System
(Pvt) Limited which is renowned dealer in Computer
business.
In the following, the correlation coefficient analysis is
made between the total sales (X) and marketing expenses
(Y):
Month of
Year 1995
Total sales
(X)
Mktg.
Exp.
(Y)
X2
XY
Y2
January
820000
77900
672400000000
6068410000
63878000000
February
830000
82000
688900000000
6724000000
68060000000
March
720000
72000
518400000000
5184000000
51840000000
April
705000
69500
497025000000
4830250000
48997500000
May
980000
99275
960400000000
9855525625
97289500000
June
875000
86000
765625000000
7396000000
75250000000
1100000
98000
1210000000000
9604000000
107800000000
August
920000
99750
846400000000
9950062500
91770000000
September
700000
89000
490000000000
7921000000
62300000000
October
620000
70000
384400000000
4900000000
43400000000
November
860000
80000
739600000000
6400000000
68800000000
December
870000
76575
756900000000
5863730625
66620250000
10000000
1000000
8530050000000
84696978750
846005250000
July
r =
N  X
N  XY  (  x )(  y)
2
 (  X )2
N  Y
2
 (  Y )2

12  846005250000 - 10000000  1000000
r =

12  8530050000000  10000000

2
 12  846005250000
- (1000000)2
r = 0.77
Ejaz Alam Khan - H 5279752
#8
Business Mathematics and Statistic (523)
First Assignment
9
Q. 5 (a)
A penal of potential cereal product consumers was asked
to rate one of four potential new products. Each member
of the panel rated only one of the products on a 100 point
scale, as compared to a standard, existing cereal. The
data is given below:
I
II
III
IV
16,
30,
43,
29,
31,
35,
51,
39,
57,
52,
53,
46,
62,
60,
54,
50,
67,
64,
56,
59,
71,
65,
58,
61,
73, 75
65, 67, 70, 71, 75, 82
61, 64, 64, 67, 68, 70, 71, 75, 79
62
Test the hypothesis of equal mean product scores by an
analysis of variance. First use
 = 0.05, then find a - value.
Ans.
H0          
H1 : All means are not equal
 =
0.05
I
II
16
31
57
62
67
71
73
75
30
35
52
60
64
65
65
67
70
71
75
82
C1 = 452 C2 = 736 C3
Ejaz Alam Khan - H 5279752
III
IV
43
29
51
39
53
46
54
50
56
59
58
61
61
62
64
64
67
68
70
71
75
79
= 934 C4 = 346
#9
Business Mathematics and Statistic (523)
First Assignment
10
452
X1 =
X2 =
X3 =
X4 =
8
736
12
934
15
346
7
= 56.5
= 61.33
= 62.27
= 49.43
Sum of square = SS total =

 X
2


2
 X  -

( X )2
n




 = (16 )           
= 154080
X
= 452 + 736 + 934 + 346
= 2468
2
 X)
SS(T)
2
= (2468 )  6091024
=
2
 X  -
=  
( X )2
n


= 154080 - 145024.381
= 9055.62
SS (Factors) + SS (Error) = SS (Total)
SS (Products) =
Ejaz Alam Khan - H 5279752
# 10
Business Mathematics and Statistic (523)
First Assignment
11
 2
2 C2
2 
C
C
C
1
2
3
 

 4 
K1 K2 K 3 K4 





 
+
 

 
 X 

n
 
 +  +    







= 145938.69 - 145024.381
= 914.305
SS (Error ) =
X
2



   
 

 
C2 C2 C2 C2 
-  1  2  3  4 
K1 K 2 K3 K 4 



 
 +  +  







 154080 - 145938 .69
= 8141.31
SS (Factor) + SS (Error) = SS (Total)
914.305 + 8141.31 = 9055.62
ANOVA TABLE
SOURCE
S.S.
d.f.
M.S.
Product
914.305
C-1=3
914.305
3
= 304.768
Error
8141.31
n-c=38
8141.31
38
= 214.245
Total
9055.62
n-1 = 41
Ejaz Alam Khan - H 5279752
# 11
Business Mathematics and Statistic (523)
12
F =
304.768
214.245
F = 1.422
First Assignment

0.05
F (3,38, 0.05 = 2.85)
F  = 0.05 is 2.85
Result
The calculated value F = 1422 is less than F 0.05 = 2.85,
so we do not reject H0 at 0.05 level of significance
Ejaz Alam Khan - H 5279752
# 12
Business Mathematics and Statistic (523)
First Assignment
13
(b)
A production superintendent claims that there is no
difference in the employee accident rates for the day and
evening shifts in a large manufacturing plant.
The
number of accidents per day is recorded for both the day
and the evening shifts for n=100 days. It is found that
the number of accidents
y
E
per day for the evening shift
exceeded the corresponding number of accidents y D on
the day shift on 63 of the 100 days. Do these results
provide sufficient evidence to indicate that more accidents
tend to occur on one shift than on the other, or
1
P(y E  y D) 
2?
equivalently, that
Ans.
H0 : p ( Y E  Y D  
H0 : p ( Y E  Y D  




n  100 X = 63
so that
P = 63
100
also
p 0  0.50
q 0 = 0.50
Z =
P - p0
p 0q 0
n
Z =
0.63 - 0.50
(0.50)(0.50 )

 = 2.6
Ejaz Alam Khan - H 5279752
# 13