Higher National Diploma in Accountancy (New)

First Year, Second Semester Examination - 20.18

1202: Statistical Analysis for Management- Answers

Instructions for Candidates: No. of questions: 06

Answer five (5) questions including question 01 No. ofpages : 04
All questions carry equal marks. Time : Three hours

Question 01

(i) State two types of statistics.

Descriptive and Inferential Statistics

(ii) Define the followings and give an example whichincludes both.

(a) Dependent variable - When we consider two( or more) variables and if one variable
changes according to the other variable( or variables) then the changing variable is
defined as the dependent variable

(b) Independent variable - When we consider two( or more) variables and if one variable
changes according to the other variable( or variables) then the influencing variable
(or variables) is(are) defined as the independent variable( or independent variables)

Consider production level and profit. Here we can consider production level as
independent variable and the profit as dependent variable( or similar examples)

(iii) What are the two major methods of sampling?

Probability and Non probability sampling

(iv) State two characteristics of negatively skewed distribution curve.

Most of the data very close to maximum of the data.

Mean< Median < Mode

Less data very close to minimum or similar answers

(v) Write two methods of collecting primary data.

Direct observation method, Interview method, Postal(letters emails ect) method,

• Telephone conversation method or similar methods
 (vi) What is meant by continuous data? Give an example.

Data that we can measure - Height of a production or similar answer

(vii) What is the difference between "parameters" and "statistics"?

Parameters - Measurements of characteristics of the population

Statistics - Measurements of characteristics of a sample

(viii) What are the factors involved in determining the size of a sample?
(ix) State four components of the time series.
(x) What is the difference between "in control" and "out of control" in the control process?

(10 x 2 = 20 Marks)

Question 02

Marks obtained by 25 students for the selection test are given below.

61 61 62 54 53
45 51 50 55 54·
58 60 62 59 64
48 53 66 52 57
. 53 60 60 50 51

(i) Prepare a frequency distribution for the above data by selecting classes as 45 - 49,

50-54, . (05 Marks)

Class Tally Marks Frequency

45-49 II 2
50-54 /J/111/1 10
55-59 Ill/ 4
60-64 /Jltl/1 8
65 ~69 I 1
HNDA - l Statistical Analysis for Management (N<-"W)2018 Second Semester
 (ii) Draw a frequency polygon for the distribution in part (i). (05 Marks)

Histogram
15

10

s
O·..,__---..J .,J . . . .,.
47 52 57 62 67

(iii)Find the average mark and median mark obtained by the students. (OS Marks)

Class
Boundaries Frcquencytr) x fx F
44.5-49.5 2 47 94 2
49.5- 54.5 10 52 520 12
54.5-59.5 4 57 228 16
59.5 - 64.5 8 62 496 24
64.5-69.5 1 67 67 25
25 1405

Mean= x=
- Ift 1405
LI =25 =56.2

Median= = L+[ ~; F ]x C = 54.5 +[ ·1~ ]x

12
12
5 = 54.75

(iv)Find the mark obtained by the majority of the students. (05 Marks)

Mode= =L+[ di di+d2 ]xC=49.5+[~]x5=52.3571

8+6

(Total 20 Marks)

HNDA ""'." l Statistical Analysis/or Management (New) 2018 Second Semester

.. ·:
Question 03

(i) According to the following data which player is more stable? (05 Marks)
~---------,., ,, _
Scores by Batsman Scores by Batsman
A B
3 6
·--------1------····· .... ····-··--
4 .)

_,
....
0
4 2
···---·- . . .·-·,-----..1--------
6 4
4 3

-·
· Batsman A Bals_ni_lill_B_l
x x2 x x2
-- -:-:;----i
3
4
9
16
6
3 !6~
3 9 0 0
4 16 2 4
6 36 4 16
,.,
4 16 .) 9
24 102 18 74

- ""x 24
Mean = x =_Li_= - =4 . _ - = --Ix = -1s = 3
Mean - x
n 6 n 6

• - 2 Ix2 -2 102 2 . 2
,2
Lix -2 74 2
Variance =S' =---x =--4 =1 Vanance=S =---x =--3 =3.3333
n 6 n 6
A player is more stable

(ii) Following table shows the marks obtained by 50 students for the first semester
examination.

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No .of
3 5 10 13 8 5 4 2
Students

Cumulative
Class Frequency
Frequency
10-20 3 3
20-30 5 8
30-40 JO 18

..
HNDA -1 StatisticalAnalysisfor Management (New) 20 I 8 Second Semester
 40-50 13 31.
-·
50-60 8 39
..
60-70 5 44
70-80 4 48
80-90 2 50

The examination board decided to give grade A to the top 10% and to fail the lowest 30%.
Find the following;

(a) The minimwn mark for grade A. (05 Marks)

09orP9o=L+
[
..
lO
n
··X 9-F
f l 1

xC =70+[
45-44
4
Jx10=72.5

Therefore minimum mark for grade A is 72 or 73.

(h) The pass mark. (05 Marks)

Therefore pass mark is 37.

(c) Inter quartile range. (05 Marks)

Qi=L+

r
-xl-F
4n f l xC=30+[
125-8
·10 Jx10=34.5

-x.3-F
Q3 = L + 4n · f

r
l x C = 50 + [
37 5-31
.
8
J x 10 = 58.125
·

Inter QuartHe Range= Q3 - Q, = 58.125 - 34.5

(Total 20 Marks)

Question 04

Following table shows the relationship between students' performance on their mid-term
marks(x) and the final exam marks (v).

HNDA -{Statistical Analysis for rtanagement (New) 2018 Second Semester

 ···- .............

x
y
92
96
66
69
82
84
60
59 I
I
- · · i-·--·r--t ·-· · -
95 I I 00 ) 67 ' 50
95 I 98 63 48 ,
For the above
'
·(i) Draw a Scatter Plot for the above data. (05 Mai-ks)

Scatter Plot
; 120

100

80
; ..
60

40
j .....
I
·-~
i.
......

20
j ·. . . J·-·····
l
20 40 60 80 100

(ii) Find the regression line yon x by using least square method. (OS Ma1·ks)

x ·y xy x2 y2
.,____
92 96 8832 8464 9216
66 69. 4554 4356 4761
82 84 6888 6724 7056
60 59 3540 3600 3481
95 95 9025 9025 9025
100 98 9800 10000 9604
67 63 4221 4489 3969
50 48 2400 2500 2304
612 612 49260 49158 49416

LY=na+bL:X => 612=8a+612b~(l)

Lxy = a LX + b2>'2 => 49260 = 612 a+ 49158 b~(2)
By (1) and (2) a=- 3.3346 and b = 1.0436
Therefore the regression line is y = - 3.3346 + 1.0436 x.
(Note: similar method can apply to find a and h)

(iii) Find the correlation coefficient and interpret it. (05 Maries)

-x-----76.5and
I> 612 - "'y
y--LJ 612 -76.5
.
n 8 n 8
"''

..
HNDA -1 Statistical Analysis for Management (New) 2018 Second Semester
 r = I:xy-n-;y = 49260-8x76.5x76.5 =0.9904
)Ix -n-; x~Ly -n y
2 2 2 2 .J49158-8x76.52 x.J49416-8x76.5z

There is positive correlation between x and y. When a student perform well in mid-test
. that student perform well in final exam also.

(iv) If the student earns mid-term mark of 78, what is the predicted final exam mark?
(05 Marks)

Expected mark

y = -3.3346 + 1.0436 x

y =- 3.3346 + 1.0436 x 78 = 78.0662 = 78

(Total 20 Marks)

Question 05

(i) The demand for the production is observed for IO months and has been recorded as below.
:

Month Jan Feb Mar ·Apr May Jun Jul Aug Sep Oct

Demand 300 301 275 288 322 299 267 295 320 312

For the above, calculate three month moving averages. (06 Marks)

Three Month Three Month

Month Demand
Sum Average

Jan 300
.Feb 301 876 292
Mar 275 864 288
Apr 288 885 295
May 322 909 303
Jun · 299 888 296
Jul 267 861 287
Aug 295 882 294
Sep 320 927 309
Oct 312
(ii) Consider the following information.

HNDA - J Statistical Analysisfor Management(New) 2018 Second Semester

•
 Price Quantityl
Item ._, .......
q,
I
J
po p1 qo
A 5 8
...
20
.
I
I
28
·-·
R 4 3 15 20
.......
....
c 8 .) 10 14
-··
n 5 7 12
_18
__... ,_ ,
22 . 21 57 80

By taking 1991 as the base year, calculate the following;

21
(a) Simple aggregate price index= ~ Pi x 100 ;:: x 100 = 95 .45% (02 Marks)
LJPo 22

(b) Simple aggregate quantity index =~ q1 x 100 = 80 x 100 = 140.35% (02 Marks)
LJ% 57

(iii) The following data are related to a set of commodities used in a particular process.

By taking 1996 as the base year, calculate the followings;

p() qo p1 q1 poqo poql p1qo p1q1

275 8 325 12 2200 3300 2600 3900

40 15 85 24 600 960 1275 2040

100 12 130 18 1200 1800 1560 2340

4000 6060 5435 8280

(a) Laspeyears' price and quantity index. (05 Marks)

. , · · d
1.. .aspeyears pnce m ex="'°'
%LP1 5435 · ggo;
x 100. =--x 100 = 135. 10
LJPo % 4000 .

· · • . . d
Laspeycars quantity m ex === '"'
I A) q,
x 100
6060
= -- x 1 0 0 = 151.5~0
11

L,Po % 4000

(b) Paasches' price and quantity index. (05 Marks)

TfNDA - I StatisticalAnalysisfor Management(New) 2018.Second Semester

•
 Paasches' price index = f P, q, x I 00
Po s;
= 8280 x 100 = 136.63%
6060

. . .
Paasches' quantity mdcx ="
LP1 q, 8280
x 100 =--x 100 = 152.35%
· L.iPi % 5435

(Total 20 Marks)

Question 06

(i) Consider the following probability distribution.

x -1 0 1 2 3

P(X) 0.1 0.26 4a 0.14 a

For the above find;

(a) The value of a. (02 Marks)

Since it's a probability distribution LP(X) =1

0.1 + 0.26 + 4a + 0.14 +a= 1 =>a= 0.1

(b) Expected value. (02 Marks)

E(X) = :Z:xx P(X)= (- 1 xO.l)+ (0 x0.26)+ (1 x0.4)+ (2 x0.14)+ (3 x0.1)

= (-0.1) + (0) + ( 0.4) + (0.28) +(0.3)

1:::: 0.88

(c) Variance. (02 Marl{S)

E(X2) = LX2 x P(X) = ((- 1)2 xO.l)+ (02 x0.26)+ (12 x0.4)+ 02 xO. 14)+ (32 xO.l)
= (0.1) + (0) + ( 0.4) + (0.56) +(0.9)

= 1.96

Variance=E(X2)-E(X)2= 1.96-0.882 = l.1856

(ii) The mean height of 40 students is 145 cm and the standard deviation is 12 cm. Assuming
that the height is· normally distributed, find how many students have height between 11 Ocm
cm and 148 cm? (04 Marks)

P { students have height between 110 cm and 148 cm}

HNDA -I StatisticalAnalysisfor. Management (New) 20.18 Second Semester

 =P(llO:S X:Sl48)

(iii) If 5% of pens manufactured by a company are defective. find the probability that a box
with 10 pens contains;

(a) Exactly 4 defective pens. (02 Marks)

· ( b) At least 2 defective pens. (03 Marks)

(iv) Researchers are interested in the mean age of a certain population. A random sample of 16
individuals ·drawn from the population of interest has a mean of 27.Assuming that the
population is approximately normally distributed with variance 25, can we conclude that
the mean age of the population is different from 30 years.(significance level is 0.05)?

(05 Marks)

(Total 20 Marks)

..
!TNDA··· I Statistical Ancdysisfor Management (New) 2018 Second Semester