LOYOLA COLLEGE (AUTONOMOUS), CHENNAI –600 034 M.Sc., DEGREE EXAMINATION - STATISTICS

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13.11.2003
1.00 - 4.00
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI –600 034
M.Sc., DEGREE EXAMINATION - STATISTICS
FIRST SEMESTER – NOVEMBER 2003
ST-1804/S719 - COMPUTATIONAL STATISTICS - I
Max:100 marks
SECTION-A
Answer any THREE questions.
All questions carry equal marks
1. a) Compare the exact mean square error of ratio estimator with the variance of HartelyRoss unbiased estimator using the following population data assuming simple random
sampling is done with sample size 3.
Votes polled in 1991
Votes polled in 1996
Booth No.
election
election
1
1024
964
2
886
931
3
950
1014
4
1251
895
5
684
768
b) Consider a finite population of size 30 where the population units are arranged in
ascending order with respect to the levels,
Label : 1
2
3 …………..30
Y : 13
21
29………….245
YI = 5+8i, i = 1, 2, ……. 30.
Draw a linear systematic sample of size 6 and estimate the population total using Yates
corrected estimator and give your comments.
(20+14)
2. a) To estimate the volume of timer available in forest area consisting of 36 geographical
regions, a sample of size 6 is drawn using random group method. The following table
gives the data collected.
Volume of timber
Total no. of
S.No.
No.of trees
(in cubic units)
trees
1
46
112.6
243
2
73
143.6
180
3
50
121.2
140
4
48
98.7
70
5
36
76.3
107
6
20
42.3
90
Estimate the total volume of timber assuming that there are 2800 trees in the forest
area and also estimate the variance of your estimator. (Here it is assumed that the
random groups are all of equal size).
b) A sample survey is conducted with the aim of estimating the total yield of paddy, the
area is divided into 5 strata and from each stratum 4 plots are selected by the method of
simple random sampling without replacement (SRSWOR). Using the data given below
obtain the estimate of total yield along with the variance of the estimator.
Stratum No. Total No. of plots
Yield of paddy for 4 plots
1
105
104 182 148
87
2
87
108
64
132 156
3
76
110
281 120 114
4
98
96
102 141 111
5
64
112 128
124 118
(17+17)
3. a) Show that (A-1)T = (AT)-1 for the following matrix.
 2 2 3 3
 2 3 3 2

A
5 3 7 9 


3 2 4 7 
b) Find g inverse for the given rectangular matrix.
 6 7 8 11
A   2 4 3 2 
14 7 9 3 
c) Find the rank of the given matrix.
2 3 4 9 
1
1
0  1 1 1 

 3  1 1 0  1


0 2 9
 1 1
 3 1
0 3 9 
(14+10+10)
4. a) Find BT A-1 B without finding A-1 for given matrices A and B.
38 47 63 21
51 62 37 18 

A
;
B T  26 38 43 24
 71 37 46 36


23 32 71 24
b) A population containing 420 birds of same weight and age were taken for a stimulus
study. The birds were divided into 60 equal groups. They were then given the stimulus
to increase the growth. The following data gives the frequency distribution of those
with significance increase in weight at the end of 6 weeks. Fit a truncated binomial
distribution an test the goodness of fit at 5% level of significance.
No.of weeks: 1
2
3
4
5
6
No. of groups: 5
12
18
15
7
3
(14+20)
5. a) For the following distribution, fit a negative binomial distribution and test the goodness
of fit at 5% level of significance.
X:
0
1
2
3
4
5
F:
212
128
37
18
3
2
1
 p1 ( x)  p 2 ( x) where p1 (x) = pqx-1 , x = 1,2,…
b) Fit a distribution of the type p(x) =
2

x
e 
, x  0,1, 2, ......,   0. Also test the goodness of fit at 5%
0 ≤ p ≤ 1 and p 2 ( x) 
x!
level.
X:
0
1
2
3
4
5
6
7
8
F:
72
110
119
58
28
12
4
2
2
(14+20)
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