Statistics and Numerical Methods Test Hypothesis Significance

MA8452 - STATISTICS AND NUMERICAL METHODS
UNIT I - TESTING OF HYPOTHESIS
PART A
1. Define Population, Sample and Sample Size.
Population:
The group of individuals under study is called population. The population may be finite or
infinite.
Sample:
A finite subset of statistical individuals in a population is called Sample.
Sample Size:
The number of individuals in a sample is called Sample Size (n).
2. Define Parameters and Statistics, Standard Error and Random Sampling.
(April/May 2018)
Parameter:
The statistical constants in population namely mean µ and variance 2
 which are called
parameters.
Statistic:
Statistical measures computed from sample observations alone, i.e. mean x and variance s2
are known as statistics.
Standard Error:
The standard deviation of the sampling distribution of a statistic is known as standard error.
Random Sampling:
Random Sampling is one in which each unit of the population has an equal chance of being
included in it.
3. What do you mean by critical region and acceptance region? (Nov/Dec 2017)
Critical region:
A region corresponding to a statistic, in the sample space S which amounts to rejection of
null hypothesis is called as critical region or region of rejection.
Acceptance region:
The region of the sample space S which amounts to the acceptance of null hypothesis
is called acceptance region.
4. Define Sampling of distribution.
The probability distribution of a sample statistic is called the sampling distribution.
5. Define one - tailed and two - tailed test.
One-tailed test:
A hypothesis test in which there is only one rejection region (either left or right side) is
called one – tailed test.
Two-tailed test:
A hypothesis test involving two rejection region is called two – tailed test.
6. Explain Null Hypothesis and Alternative Hypothesis.
Null Hypothesis:
For applying the tests of significance, we first set up a hypothesis which is a definite
statement about the population parameter. Usually, such a hypothesis is a hypothesis of no
MA8452/ Statistics and Numerical Methods II Year Mechanical Engineering 2021-2022

difference and it is denoted by 0
H .
Alternate Hypothesis:
Any hypothesis which is complementary to the null hypothesis is called an alternative
hypothesis, denoted by 1
H .
7. What are type I and type II errors? (April/May 2019) (Nov/Dec 2015, 2017)
Type I error: Reject Null hypothesis when it is true.
Example:
A test that shows a patient to have a disease when in fact the patient does not have the
disease.
Type II error: Accept Null hypothesis when it is wrong.
Example:
A blood test failing to detect the disease it was designed to detect, in a patient who
really has the disease.
8. Find the standard error of sample mean from the following data 16 16 5
= =
n , . ,

16 25
=
x . , 1 65
=
S .
Standard Error=
1.65
0.258
1 16 1
S
n
= =
− −
9. Define Level of Significance.
The probability that the value of the statistic lies in the critical region is called the level of
significance. It is denoted by α .
10. Write 95% confidence interval of the population mean.
95% confidence interval of the population mean is 0.05 0.05
1 1
S S
x t x t
n n

−   +
− −
11. For the following cases, specify which probability distribution to use in a hypothesis test
0 1
0 1
) : 25, : 25, 19.1, 6, 13
) : 96.6, : 96.6, 60, 13, 40
a H H X n
b H H X s n
  
 
=  = = =
=  = = =
a) student t-distribution b) Normal distribution
12. Write down the formula to test statistic ‘t’ to test the significance of difference between
the sample mean and population mean for small sample.
The test of significance of difference between the sample mean and population mean for
small sample is given by
/ 1
x
t
s n

−
=
−
Where Sample mean
Population mean
Sample Standard deviation
x
s
=
=
=
13. What are the assumptions of t-test?

➢ Parent populations, from which the samples have been drawn, are normally
distributed.
➢ The population variances are equal and unknown.
➢ The two samples are random and independent of each other.
14. Write down test statistic for paired t – test.
Test Statistic for paired t – test is
2
where .
d
t d x y
S
n
= = −
15. What are the uses of t – test?
➢ To test the significance of difference between mean of a random sample and the mean
of the population.
➢ To test the significance of the difference between two sample means.
16. What are the assumptions of F-test?
➢ Normality: The values in each group should be normally distributed.
➢ Independence of error: The variations of each value around its own group mean.
(i.e.) error should be independent of each value.
➢ Homogeneity: The variances within each group should be equal for all groups.
17. What are the uses of F – test?
➢ To test the equality of two sample variances.
➢ To test more than two population mean.
➢ To test the sample observation coming from normal population.
18. State the assumptions of Chi-square test.
➢ The sample observations should be independent.
➢ Constraints on the cell frequencies, if any, must be linear.
➢ The total frequency, should be at least 50
➢ No theoretical cell frequency should be less than 5
19. Write down the expected frequencies of 2
2 contingency table.
a B
c D
Expected frequency table:
N
c
a
b
a )
)(
( +
+
N
d
b
b
a )
)(
( +
+
N
d
c
c
a )
)(
( +
+
N
d
b
d
c )
)(
( +
+
where d
c
b
a
N +
+
+
=

20. State any two applications of 2
 - test. (Nov/Dec 2016 & 2017) (April/May 2019)
➢ To test the homogeneity of independent estimates of the population variances.
➢ To test the goodness of fit.
➢ To test for independence of attributes.
PART B
1. Random samples of 400 men and 600 women asked whether they would like to have a
flyover near their residence. 200 men and 325 women were in favour of the proposal.
Test the hypothesis that proportions of men and women in favour of the proposal are
same against that they are not, at 5% level. (May/June 2016)
Solution:
Given sample sizes n1=400, n2=600. Proportion of men p1=200/400=0.5
Proportion of women p2 = 325/600=0.541
Null Hypothesis H0:
Assume that there is no significant difference between the option of men and women
as far as proposal of flyover is concerned. i.e., p
p
p
:
H 2
1
0 =
=
Alternative Hypothesis H1:
2
1 p
p  (two tailed)
The Test Statistic is
1 2
1 2
1 1
p p
z
pq
n n
−
=
 
+
 
 
1 1 2 2
1 2
200 325
400 600
400 600
where 0.525
400 600
1 0.525 0.475
   
+
   
+    
= = =
+ +
= − =
n p n p
p
n n
q
( )( )
0.5 0.541 0.041
1.281
0.032
1 1
0.525 0.475
400 600
1.281
. ., 1.96
z
z
i e z
− −
= = = −
 
+
 
 
=

Therefore, we accept the null hypothesis Ho at 5% level of significance.
i.e., There is no difference between the option of men and women as far as proposal of
flyover are concerned.
2. In a random of 1000 people from city A, 400 are found to be consumers of rice. In a
sample of 800 from city B, 400 are found to be consumers of rice. Does this data give a
significant difference between the two cities as far as the proportion of rice consumers is
concerned? (Nov/Dec 2017)
Solution:
Given

1 2
1 2
1000 800
400 400
0.4 0.5
1000 800
n n
p p
= =
= = = =
0 1 2
1 1 2
1 2
1 2
1. :
2. :
3. 5% table value is 1.96
4.The teststatistic
1 1
H p p
H p p
p p
z
pq
n n

=

=
−
=
 
+
 
 
1 1 2 2
1 2
400 400
1000 800
4
1000 800 0.444
1000 800 9
1 0.444 0.556
n p n p
where p
n n
q
 + 
+
= = = =
+ +
= − =
0.4 0.5
4.167
1 1
0.444 0.556
1000 800
4.167
. ., 4.167 1.96
z
z
i e z
−
= = −
 
 +
 
 
=
= 
So, H0 is rejected.
Therefore, there is significant difference between the consumers of rice in the two cities A
and B.
3. The manufacturer of a medicine claimed that it was 90% effective in relieving an allergy
for a period of 8 hours. In a sample of 200 people who had the allergy, the medicine
provided relief for 160 people. Determine whether the manufacturer’s claim is
legitimate at 1% level of significance. (April/May 2019)
Solution:
H0: p = 0.9 and the claim is correct.
H1: p < 0.9, the claim is false where p: Probability of obtaining relief.
The claim is not legitimate if z is < -2.33 in which case we reject H0; Otherwise, it is
legitimate.
If H0 is true,
200 0.9 180,
200 0.9 0.1 4.23
160 180
4.78
4.23
np
npq
z


= =  =
= =   =
−
= = −
Which is much less than -2.33. Thus, we conclude that the claim is not legitimate.
4. A test of the breaking strengths of 6 ropes manufactured by a company showed a mean
breaking strength of 3515 kg and a standard deviation of 60 kg, whereas the
manufacturer claimed a mean breaking strength of 3630 kg. Can we support the
manufacturer’s claim at a level of significance of 0.05? (April/May 2019)

Solution:
Given: n = 6, sample mean = x =3515, population mean = 3630
 =
0
H : 3630 Kg
 = and the manufacturer’s claim is justified.
1
H : 3630 Kg
  and his claim is not justified.
A one tail test is required.
0
x- 3515-3630
Under H : t = 4.286
s 60
n-1 6-1

= = −
Calculated value of t = |t| = 4.286
Degrees of freedom = n – 1 = 5
Table value of t at 5% level of significance for 5 degrees of freedom in one tailed test = 2.01
Calculated t > Tabulated t
Hence, we reject H0.
5. A random sample of 100 bulbs from a company P shows a mean life 1300 hours and
standard deviation of 82 hours. Another random sample of 100 bulbs from company Q
showed a mean life 1248 hours and standard deviation of 93 hours. Are the bulbs of
company P superior to bulbs of company Q at 5% level of significance?
(Nov/Dec 2017)
Solution:
1 1 1
2 2 2
Given : 100 1300 82
100 1248 93
n x s
n x s
= = =
= = =
0 1 2
1 1 2
1. :
2. : (Useone-tailed right)
3. 5%
4.Theteststatistic
H
H
 
 

=

=
1 2
2 2 2 2
1 2
1 2
x x 1300 1248
Z=
s s (82) (93)
100 100
n n
− −
=
+
+
52
4.19
67.24 86.49
= =
+
5. Conclusion:
α 0 0
0
If Z<Z ,then weaccept H ;otherwise, wereject H .
Here,4.19>1.645
So,we reject H .
6. The sales manager of a large company conducted a sample survey in states A and B
taking 400 samples in each case. The results were (April/ May 2017)
State A State B
Average Sales Rs. 2,500 Rs. 2,200
S.D. Rs. 400 Rs. 550
Test whether the average sales is the same in the 2 states at 1% level of significance.
Solution:

1 1 1
Given : 400 2500 400
n x s
= = =
2 2 2
400 2200 550
n x s
= = =
0 1 2
1. :
H  
=
1 1 2
2. : (Use two-tailed test)
H  

3. 1%
 =
4.Theteststatistic
1 2
2 2 2 2
1 2
1 2
x x 2500 2200
Z=
s s (400) (550)
400 400
n n
− −
=
+
+
300
8.82
400 756.25
= =
+
5. Conclusion:
/2
/2 0 0
If Z < Z < Z ,then weaccept H ;otherwise,wereject H .


−
Here,8.82 2.58

0
So,we reject H .
7. Samples of two types tested gives the following results
Sample Size Mean Standard Deviation
I 32 72 8
II 36 74 6
Test if the means are significant.
Solution:
Null Hypothesis: There is no difference in two types.
Alternate Hypothesis: There is a significant difference in two types.
Level of significance: %
5
=

2
1
2 2
1 2
1 2
x x
z
s s
n n
−
=
+
72 74
1.55
64 36
32 36
−
= = −
+
1.155 1.96
z = 
Conclusion: Since Calculated value < Table value, We accept 0
H
8. The following data relate to the marks obtained by 11 students in two tests, one held at
the beginning of the year and other at the end of the year, after intensive coaching.
Test I: 19 23 16 24 17 18 20 18 21 19 20
Test II: 17 24 20 24 20 22 20 20 18 22 19
Do the data indicate that the students have benefited by coaching?
Solution:
The given data relate to the marks obtained in two tests by the same set of students.
Hence the marks in the two tests can be regarded as correlated.
Null Hypothesis H0: 1 2 The coaching is not useful
x x
=
Alternative Hypothesis H1: 1 2 The coaching is useful
x x


1
x 2
x 2
1 x
x
d −
= d
d −
2
)
( d
d −
19
23
16
24
17
18
20
18
21
19
20
17
24
20
24
20
22
20
20
18
22
19
2
-1
-4
0
-3
-4
0
-2
3
-3
1
3
0
-3
1
-2
-3
1
-1
4
-2
2
9
0
9
1
4
9
1
1
16
4
4
-11 58
2
2
( )
where
1
/
11
1
11
d d
d
t S
n
S n
d
d
n
−
= =
−
−
= = = −


41
.
2
8
.
5
1
11
58
1
)
( 2
2
=
=
−
=
−
−
=

S
n
d
d
S
38
.
1
38
.
1
11
/
41
.
2
1
/
=
−
=
−
=
=
t
n
S
d
t
Degrees of freedom = n-1 = 11-1 = 10
Tabulated value for 10 degrees of freedom at 5% level for one tail test is 1.812.
1.38<1.812
Calculated t < tabulated t
0
H
 is accepted.
 The course is not useful.
9. A random sample of 10 boys has the following I. Q’s: 70, 120, 110, 101, 88, 83, 95, 98,
107 and 100. Do these data support the assumption of a population mean I.Q of 100 at
5% level of significance? (May/June 2016) & (Nov/Dec 2017)
Solution:
100
:
H
&
100
:
H 1
0 

=

Level of Significance:
05
.
0
=

Test Statistic:
2
2
2
, where
1
x x
x
t s
S n n
n
  
−
= = − 
 
 
−
 
Analysis:
X X2
70 4900
120 14400

110 12100
101 10201
88 7744
83 6889
95 9025
98 9604
107 11449
100 10000
x
 = 972 2
x
 = 96312
2 2
2
2 96312 972
9631.2 9447.84 183.36 13.54
10 10
x x
s S
n n
   
= − = −  − =  =
   
   
 
 
97.2 100 2.8
0.62
13.54 4.513
1 9
x
t
S
n

− − −
= = = = −
−
0.62
t =
Table value:
, 1 5%,10 1 0.05,9 2.262
n
t t t
 − −
= = =
Conclusion:
The table value is greater than the calculated value; hence we accept the null
hypothesis and conclude that the data are consistent with the assumption of mean I.Q of 100
in the population.
10. The nicotine content in milligrams of 2 samples of tobacco were found to be as follows
Sample A 24 27 26 21 25 (April/May 2018)
Sample B 27 30 28 31 22 36
Can it be conclude that 2 samples come from normal population with the same mean.
Solution:
Null Hypothesis H0: 1 2
 
= . There is no significance difference between sample mean
and population mean
Alternative Hypothesis H1: 1 2
 
 . There is a significance difference between sample
mean and population mean.
x x x
−
( )
2
x x
−
y ( )
y y
− ( )
2
y y
−
24 -.6 0.36 27 -2 4
27 2.4 5.76 30 1 1
26 1.4 1.96 28 -1 1
21 -3.6 12.96 31 2 4
25 .4 0.16 22 -7 49
36 7 49
123 0 21.2 174 0 108

( ) ( )
( ) ( )
2 2
2 2
2
1 2
123 174
24.6 29
5 6
21.2 108
1
( )
2
1
(21.2 108) 14.356
5 6 2
3.78
X y
x x y y
s x x y y
n n
s
= = = =
− = − =
= − + −
+ −
= + =
+ −
=
 
 
The test statistic is
1 1
24.6 29
1.92
1 1 1 1
3.78
5 6
x y
t
s
n n
− −
= = = −
+ +
Calculated t=1.92
From table, ( )
0.5 9 2.262
t =
Calculated t < tabulated t
0
H
 is accepted.
11. A group of 10 rats fed on diet A and another group of 8 rats fed on diet B recorded the
following increase in weight
Diet A 5 6 8 1 12 4 3 9 6 10
Diet B 2 3 6 8 10 1 2 8
Does it show superiority of diet A over diet B.
Solution:
Given 1 2
10, 8
n n
= =
Null Hypothesis H0: 1 2
 
=
Alternative Hypothesis H1: 1 2
 
 (One tailed test)
To find 2
1
S and 2
2
S
Sample I Sample II
X1 X1
2
X2 X2
2
5
6
8
1
12
4
3
9
6
10
25
36
64
1
144
16
9
81
36
100
2
3
6
8
10
1
2
8
4
9
36
64
100
1
4
64
64 512 40 282

( ) ( )
( ) ( )
2
1 2 1 1
2
2 2
1 2
1 2
1 2
2
2 2
1
2
1 1
1
2
2 2
2
2
2 2
2
2 2
2 1 1 2 2
1 2
10, 8, 64, 512
40, 282
64 40
6.4, 5
10 8
512
6.4 10.24,
10
282
5 10.25
8
10(10.24) 8(10.25)
11.525
2 10 8 2
n n x x
x x
x x
x x
n n
x
s x
n
x
s x
n
n s n s
S
n n
= = = =
= =
= = = = = =
= − = − =
= − = − =
+ +
= = =
+ − + −
 
 
 


1 2
2
1 2
x x 6.4 5 1.4
t = 0.869
1.6103
1 1
1 1 11.525
10 8
S
n n
− −
= = =
   
+
+  
 
 
 
Conclusion:
Since the calculated value of t = 0.869 < the table value of t =1.75 , 0
H is accepted at
5% level of significance. Hence, the difference is not significant, so we cannot conclude the
diet A is superior to diet B.
12. Time taken by workers in performing a job is given below:
Method I: 20 16 26 27 23 22
Method II: 27 33 42 35 34 38
Test whether there is any significant difference between the variances of the time
distribution at 5% level of significance. (Nov/Dec 2017)
Solution:
Null Hypothesis H0: (Variances are equal)
Alternative Hypothesis H1: (Variances are not equal)
1 2
n =6, n =6,
Sample I Sample II
X1 X1
2
X2 X2
2
20
16
26
27
23
22
400
256
676
729
529
484
27
33
42
35
34
38
729
1089
1764
1225
1156
1444
134 3074 209 7407

( ) ( )
( ) ( )
2 2
2 2
1 1 2 2
1 2 1 2
1 2
1 2
1 2
1 2
2
2 2
1
2
1 1
1
2
2 2
2
2
2 2
2
n s n s
S = and S = ,n =n =6,
n -1 1
134 209
22.33, 34.833
6 6
3074
22.33 13.7,
6
7407
34.833 21.16
6
n
x x
x x
n n
x
s x
n
x
s x
n
−
= = = = = =
= − = − =
= − = − =
 


2
2 2
2
2 1
2
1
0
S
TestStatistic of Fis F= =1.5,sinceS >S
S
Tablevalueof Fat5%Los for (5,5) d.f =5.05
since1.5<5.05 H isaccepted.
thereisno significant difference between the two time distribution

13. Two independent samples of sizes 9 and 7 from a normal population had the following
values of the variables. Do the estimates of the population variance differ significantly
at 5% level of significance? (Apr/May 2017)
Sample I: 18 13 12 15 12 14 16 14 15
Sample II: 16 19 13 16 18 13 15
Solution:
Given 7
,
9 2
1 =
= n
n
Null Hypothesis H0: 2
2
2
1 
 = (Variances are equal)
Alternative Hypothesis H1: 2
2
2
1 
  (Variances are not equal)
To find 2
1
S and 2
2
S
Sample I Sample II
X1 X1
2
X2 X2
2
18
13
12
15
12
14
16
14
15
324
169
144
225
144
196
256
196
225
16
19
13
16
18
13
15
256
361
169
256
324
169
225
129 1879 110 1760
2
1 2 1 1
9, 7, 129, 1879
n n x x
= = = =
 
2
2 2
110, 1760
x x
= =
 
1 2
1 2
1 2
129 110
14.3333, 15.7143
9 7
x x
x x
n n
= = = = = =
 

( ) ( )
2
2 2
1
2
1 1
1
1879
14.3333 3.3342,
9
x
s x
n
= − = − =

( ) ( )
2
2 2
2
2
2 2
2
1760
15.7143 4.4894
7
x
s x
n
= − = − =

2 2
2 2
1 1 2 2
1 2
1 2
9 3.3342 7 4.4894
3.7510 , 5.2376
1 8 1 6
n s n s
S S
n n
 
= = = = = =
− −
3963
.
1
7510
.
3
2376
.
5
, 2
1
2
2
2
1
2
2 =
=
=

S
S
F
is
statistic
test
the
S
S
Since
Number of degrees of freedom )
8
,
6
(
)
1
,
1
(
)
,
( 1
2
2
1 =
−
−
= n
n
v
v
For )
,
( 2
1 v
v =(6,8), the table value of F at 5% level is 58
.
3
05
.
0 =
F
05
.
0
F
F 

Conclusion: Since the calculated value of F < the table value of F, 0
H is accepted at 5%
level of significance. The two samples are drawn from populations with same variances.
14. Two random samples gave the following results:
Sample Size Sample mean
Sum of squares of
deviations from the mean
I 10 15 90
II 12 14 108
Test whether the samples come from the same normal population at 5% level of
Significance.
Solution:
Null Hypothesis H0: The two samples have been drawn from the same normal population.
H0: 2 2
1 2 1 2
and
   
= =
Here we have to use two tests
(i) To test equality of variances by F-test
(ii) To test equality of means by t-test.
(i) F- test (equality of variances)
018
.
1
82
.
9
10
82
.
9
11
108
10
9
90
1
)
(
1
)
(
108
)
(
90
)
(
14
15
12
10
2
2
2
1
2
2
2
2
2
2
1
2
1
1
2
1
2
2
2
2
1
1
2
1
2
1
=
=
=
=
=
=
=
−
−
=
−
−
=
=
−
=
−
=
=
=
=




S
S
F
n
x
x
S
n
x
x
S
x
x
x
x
x
x
n
n
Given
Calculated F=1.018
Tabulated F at 5% level for (9, 11) degrees of freedom is 2.90.
90
.
2
)
11
,
9
(
05
.
0 =
F
Since calculated F < tabulated F, we accept the null hypothesis H0.
2
1 
 =
The samples come from the same normal population.

(ii) t-test (To test equality of means)
Null Hypothesis H0: 2
1 
 =
1 2
1 2
2 2
1 1 2 2
Given n 10 n 12
x 15 x 14
(x x ) 90 (x x ) 108
= =
= =
− = − =
 
 
2 2 2
1 1 2 2
1 2
1
( ) ( )
2
1
90 108
10 12 2
9.9
3.15
 
= − + −
 
+ −
= +
+ −
=
=
 
s x x x x
n n
s
1 2
1 2
15 14
The test statistic is 0.74
1 1 1 1
3.15
10 12
Calculated value 0.74
− −
= = =
+ +
 =
x x
t
s
n n
t
Tabulated value of t for 20 degrees of freedom )
2
( 2
1 −
+ n
n at 5% level of
significance is 2.086.
Since calculated value of t < tabulated value of t. We accept the null hypothesis
.
.,
.
. 2
1
0 
 =
e
i
H
Hence from (i) and (ii), the given samples have been drawn from the same normal
population.
15. It is believed that the precision of an instrument is no more than 0.16. Write down the
null and alternative hypotheses for testing this belief. Carry out the test at 1% level of
significance, given 11 measurements of the same subject on the instrument. 2.5, 2.3, 2.4,
2.3, 2.5, 2.7, 2.5, 2.6, 2.6, 2.7, 2.5. (April / May 2017)
Solution:
2
: 0.16
27.6
X = 2.51
11
Given
x
n
 =
= =

X X X
− ( )
2
X X
−

2.5
2.3
2.4
2.3
2.5
2.7
2.5
2.6
2.6
2.7
2.5
-0.01
-0.21
-0.11
-0.21
-0.01
0.19
-0.01
0.09
0.09
0.19
-0.01
0.0001
0.0441
0.0121
0.0441
0.0001
0.0361
0.0001
0.0081
0.0081
0.0361
0.0001
0.1891
( )
2
0
2
1
2
2
2
2
2 2
1. : 0.16
2. : 0.16
3. 1%,d.f =n-1=11-1=10
4.Table valueχ 23.2
5.Theteststatistic
X X
ns
χ =
σ σ
0.1891
1.182
0.16
=

=
=
−
=
= =

H
H



6. Conclusion:
2 2
0 0
0
If χ < table χ ,then weaccept H ;otherwise,wereject H .
Here,1.182 23.2
So,we accept H .

We conclude that the data are consistent with the hypothesis that the precision of the
instrument is 0.16.
16. Fit a binomial distribution for the following data and also test the goodness of fit.
x : 0 1 2 3 4 5 6 Total
f(x) : 5 18 28 12 7 6 4 80
Solution:
Null Hypothesis H0: Binomial fit is a good to the data.
x 0 1 2 3 4 5 6 Total
f 5 18 28 12 7 6 4 80
f*x 0 18 56 36 28 30 24 192

192
2.4
80
2.4
6 2.4 ( 6)
0.4
1 1 0.4 0.6
80
= = =
= =
= =
=
= − = − =
= =



fx
x
f
x np
p n
p
q p
N f
n r n r
r
6 0 6 0
0
6 1 6 1
1
6 2 6 2
2
6 3 6 3
3
N(r) c p q N
N(0) c (0.4) (0.6) 80 3.73 4
N(1) c (0.4) (0.6) 80 14.93 15
N(2) c (0.4) (0.6) 80 24.88 25
N(3) c (0.4) (0.6) 80 22.12 22
−
−
−
−
−
= 
=  = 
=  = 
=  = 
=  = 
Expected frequencies are given by ( for r successes)
6 4 6 4
4
6 5 6 5
5
6 6 6 6
6
i
i
N4) c (0.4) (0.6) 80 11.06 11
N(5) c (0.4) (0.6) 80 2.95 3
N(6) c (0.4) (0.6) 80 0.13 0
o : 5 18 28 12 7 6 4
E : 4 15 25 22 11 3 0
−
−
−
=  = 
=  = 
=  = 
Since the first two frequencies and last three frequencies are small, we pool them
together
2
2
: 23 28 12 17
: 19 25 22 14
( )
i
i
i i
i
o
E
O E
E

−
= 
2 2 2 2
2
2
0.05
2 2
0.05
(23 19) (28 25) (12 22) (17 14)
19 25 22 14
6.39
2 , 5.59 from table
− − − −
= + + +
=
= =
 
for n


 
Hence reject the binomial fit.
17. The following data gives the number of aircraft accidents that occurred during the
various days of a week. Find whether the accidents are uniformly distributed over the
week.
Days: Sun Mon Tue Wed Thu Fri Sat
No. of accidents: 14 16 8 12 11 9 14
Solution:
Null Hypothesis H0: The accidents are uniformly distributed over the week.
Alternative Hypothesis H1: The accidents are not uniformly distributed over the week.
2
=0.05,d.f = n-1 =7-1=6
Table value 12.592

 =

St. Joseph’s College of Engineering 17
( )
2
2 84
12
7
x
O E
whereE x
E n

−
= = = = =


O E
( )
2
O E
E
−
14
16
8
12
11
9
14
12
12
12
12
12
12
12
0.333
1.333
1.333
0
0.083
0.75
0.333
TOTAL 4.165
( )
2
2
4.165 12.592
O E
E

−
 = = 

Hence, we accept H0.
18. Using the data given in the following table to test at 1% level of significance whether a
person’s ability in Mathematics is independent of his/her interest in Statistics.
(Nov/Dec 2017)
Interest in
Statistics
Ability in Mathematics
Low Average High
Low 63 42 15
Average 58 61 31
High 14 47 29
Solution:
The expected frequencies are
Ability in
Mathematics
Interest
in Statistics
Low Average High Total
Low
135 120
45
360

=
120 150
50
360

=
120 75
25
360

= 120
Average
135 150
56.25
360

=
150 150
62.5
360

=
75 150
31.25
360

= 150
High
135 90
33.75
360

=
150 90
37.5
360

=
90 75
18.75
360

= 90
Total 135 150 75 360
0
1
1. : AbilityinMathematicsandinterestinStatisticsare independent
2. : AbilityinMathematicsandinterestinStatistics are not independent
H
H

2
2
2
0
( )
32.14
ablevalueof χ at1% Losfor4 d.f =13.277
Since32.14>13.2777,H isrejected.
i i
i i
O E
E
T

−
= =

Therefore there is a relationship between a person’s ability in Mathematics and his/her
interest in Statistics.
19. In an investigation into the health and nutrition of two groups of children of different
social status, the following results are got and their economic conditions. What
conclusion can you draw from the following data?
Health Social Status Total
Poor Rich
Below normal 130 20 150
Normal 102 108 210
Above normal 24 96 120
Total 256 224 480
Discuss the relation between the Health and their Social Status.
Solution:
Expected frequencies are:
Social
status
Health
Poor Rich Total
Below normal
80
480
150
256
=

70
480
150
224
=
 150
Normal
112
480
210
256
=

98
480
210
224
=
 210
Above normal
64
480
120
256
=

56
480
120
224
=
 120
Total 256 224 480
2 2 2 2 2 2
2
2
2 2
0.05
(130 80) (20 70) (102 112) (108 98) (24 64) (96 56)
80 70 112 98 64 56
31.25 35.71 0.89 1.02 25 28.57 122.44
. . (3 1) (2 1) 2 ( )
2 5 5.991
(
d f class total are kept same
For v at level value from table
v


 
− − − − − −
= + + + + +
= + + + + + =
= −  − =
= =
 = 2)
H0 is rejected. Hypothesis is not acceptable.
That is, social status and health are associated (dependent).
UNIT – II - DESIGN OF EXPERIMENTS
PART A
1. What is the aim of the design of experiment? (Nov/Dec2017) (April/May2019)
Aim is to control the extraneous variables so that the result could be attributed only to the
experimental variables.
2. What are the three essential steps to plan Design of experiment?
To plan an experiment, the following three are essential

➢ A Statement of the objective. Statement should clearly mention the hypothesis to be
tested
➢ A description of the experiment. Description should include the type of experimental
material, size of the experiment and the number of replications.
➢ The outline of the method of analysis. The outline of the method consists of analysis
of variance
3. What are the basic principles of design of experiments? (April/May2017, 2018)
Basic Principle of Design of Experiments are
(i) Randomization (ii) Replication and (iii) Local Control
4. Define Experimental error. ( Nov /Dec 2016)
Factors beyond the control of the experiment are known as experimental error.
5. Define Analysis of Variance. (Nov /Dec 2016)
Analysis of Variance is a statistical method used to test the difference between 2 or more
means.
6. What are uses of ANOVA (April/ May2017)
➢To test the homogeneity of several means.
➢It is now frequently used in testing the linearity of the fitted regression line or in the
significance of the correlation ratio
7. What are the assumptions of analysis of variance? (April/ May2014)
➢ The sample observations are independent
➢ The environmental effects are additive in nature
➢ Sample observation are coming from normal distribution/population
8. Define completely randomized design. (Nov/Dec2017)
In completely randomized design the treatments are given to the experimental units by a
procedure of random allocation. It is used when the units are homogeneous.
9. What are the advantages of Completely Randomized Block Design?
The advantages of Completely Randomized Experimental Design as follows:
(a) Easy to lay out. (b) Allow flexibility. (c) Simple Statistical Analysis.
(d) Lots of information due to missing data is smaller than with any other design.
10. Find A, B, C, D from the ANOVA table
S. V D.F S. S M.S. S F cal
Treatments 4 A 6.86 2.37
Error B C 16.26 -
Total 24 D - -
A = 4  6.86 = 27.44, B = Total D.F – Treatments = 24 – 4 = 20, C = 20  16.26 = 325.20,
D = A + C = 27.44 + 325.20 = 352.64

11. Find A, B, C, D from the ANOVA table
S. V D.F S. S M. S. S F cal
Treatment 2 A 89.5 7.10
Error B C 12.6 -
Total 11 D - -
A = 2  89.5 = 179, B = Total D.F – Treatments = 11 – 2 = 9, C = 9  12.6 = 113.4,
D = A + C = 179 + 113.4 = 292.4
12. Explain the situation in which RBD is considered an improvement over CRD.
RBD CRD
It is flexible and so any number of treatments
and any number of replications may be used
There is complete flexibility as the number
of replications is not fixed
The analysis of design is simple as it results in a
two-way classification analysis of variance
The analysis of design is simple as it
results in a one-way classification analysis
of variance
13. What do you mean by two-way classification in ANOVA?
When data are classified according to two factors one classification is taken column wise and
the other row wise. Such a classification is called two-way classification
14. Write the ANOVA table for randomized block design.
Source of
Variation
D.F S.S M.S.S F cal
Column
Treatments
c – 1 SSC
1
SSC
MSC
c
=
−
C
MSC
F
MSE
=
Row
Treatments
r – 1 SSR
1
SSR
MSR
r
=
−
R
MSR
F
MSE
=
Error (c – 1) (r – 1) SSE
( )( )
1 1
SSE
MSE
r c
=
− −
-
15. Write the advantages of Latin Square Design.
➢ With the Two-way stratification, Latin square controls more of the variation than the
CRD or RBD
➢ The analysis is simple it is only slightly more complicated than that of the randomized
complete block design.
16. Compare RBD, LSD, CRD.

CRD RBD LSD
To influence one
factor
To influence two factors To influence more than two factors
No restriction
further treatments
No restriction on treatment
& replications
The number of replications of each
treatment is equal to the number of
treatments
-
Use only rectangular or
Square field
Use only Square field
17. Construct 5  5 Latin Square Design.
Answer: Each Treatments appears only once in each row and each column
(5  5 LSD)
A B C D E
B C D E A
C D E A B
D E A B C
E A B C D
18. Find A, B, C from LSD ANOVA.
S.V D.F S.S M.S.S F cal
Treatment 3 A 96.06 20.68
Row 3 19.19 6.397 1.377
Column 3 19.69 C 1.413
Error B 27.87 4.645 -
Total 15 - -
A = 3  96.06 = 288.18, B = 15 – 3 – 3 – 3 = 6, C = 19.69 / 3 = 6.563
19. Why a 2  2 Latin Square is not possible? Explain. (Nov/Dec2015) (May/June2016)
Consider a n  n Latin square design,
2
2
2
The degrees of freedom for SSE ( 1) ( 1) ( 1) ( 1)
1 3 3
3 2
( 1)( 2)
n n n n
n n
n n
n n
= − − − − − − −
= − − +
= − +
= − −
For n = 2, d.f. of SSE = 0 and hence MSE is not defined.
Comparisons are not possible.
Hence a 2  2 Latin Square Design is not possible.
20. Write the main effect and interaction effect of 22
factorial design.

Answer: Main Effect of A =  
)
1
(
)
b
(
)
ab
(
)
a
(
n
2
1
−
−
+
Main Effect of B =  
)
1
(
)
a
(
)
ab
(
)
b
(
n
2
1
−
−
+
Interaction Effect of A and B =  
1
( ) (1) ( ) ( )
2
ab a b
n
+ − −
PART B
1. As head of the department of a consumer’s research organization you have the
responsibility of testing and comparing life times of 4 brands of electric bulbs. suppose
you test the life time of 3 electric bulbs each of 4 brands, the data is given below, each
entry representing the life time of an electric bulb, measured in hundreds of hours.
(Nov/Dec 2017)
A B C D
20 25 24 23
19 23 20 20
21 21 22 20
Solution:
H0: Here the population means are equal .
H1: The population mean are not equal.
X 1 X2 X3 X4 X1
2
X2
2
X3
2
X4
2
20 25 24 23 400 625 576 529
19 23 20 20 361 529 400 400
21 21 22 20 441 441 484 400
TOTAL 60 69 66 63 1202 1595 1460 1329
Step 1: Grand Total T = 258
Step 2: N = Total No of Observations = 12
Step 3: Correction Factor =
ns
Observatio
of
No
Total
)
total
Grand
( 2
= 5547
Step 4: 39
2
2
3
2
2
2
1 =
−
+
+
= 

 N
T
X
X
X
TSS
Step 5:
( ) ( ) ( )
15
2
1
2
3
1
2
2
1
2
1
=
−
+
+
=



N
T
N
X
N
X
N
X
SSC
( 1
N = No of element in each column)
Step 6: SSE = TSS – SSC = 39 – 15 = 24
ANOVA TABLE
Source of
Variation
Sum of
Squares
Degree of
freedom
Mean Square F- Ratio

Between
Samples
SSC = 15 C – 1 = 4 – 1 = 3
1
C
SSC
MSC
−
= = 5
MSE
MSC
FC =
=1.67
Within
Samples
SSE = 24
N – C = 12 – 4
= 8 C
N
SSE
MSE
−
= = 3
Total TSS = 39
Cal FC = 1.67 & Tab FC (3,8) = 4.07
Conclusion: Cal FC < Tab FC  Hence we accept H0
2. A farmer wishes to test the effects of four different fertilizers A, B, C, D on the yield of
wheat. In order to eliminate sources of error due to variability in soil fertility he uses
the fertilizers in a Latin square arrangement as indicated below where the number
indicate yields in Kilograms per unit area. Perform an analysis of variance to
determine if there is a significant difference between the fertilizers at 0.01 level of
significance
(April/May 2019)
A 18 C 21 D 25 B 11
D 22 B 12 A 15 C 19
B 15 A 20 C 23 D 24
C 22 D 21 B 10 A 17
Solution:
H0: There is no significant difference between fertilizers and yield.
H1: There is a significant difference between them.
C1 C2 C3 C4 Ri
2
4
i
R
R1 18 21 25 11 75 1406.25
R2 22 12 15 19 68 1156.00
R3 15 20 23 24 82 1681.00
R4 22 21 10 17 70 1225.00
j
C 77 74 73 71 295
2
4
i
R
=
 5468.25
2
4
j
C
1482.25 1369 1332.25 1260.25
2
4
j
C
=
 5443.75

1 2 3 4 Ti
2
4
i
T
A 18 20 15 17 70 1225.00
B 15 12 10 11 48 576.00
C 22 21 23 19 85 1806.25
D 22 21 25 24 92 2116.00
2
4
i
T
=
 5723.25
Step 1: Grand Total T = 295
Step 2: Correction Factor (C. F) =
( )
2
2
295
5439.06
16
T
N
= =
Step 3: TSS = Sum of squares of individual observations - Correction Factor
2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2
18 21 25 11 22 12 15 19 15
5439.06
20 23 24 22 21 10 17
329.94
 
+ + + + + + + + +
 
= −
 
+ + + + + + +
 
 
=
Step 4: SSR
2
. 5468.25 5439.06 29.19
4
i
R
C F
= − = − =

Step 5: SSC
2
. 5443.75 5439.06 4.69
4
j
C
C F
= − = − =

Step 6: SST
2
. 5723.25 5439.06 284.19
4
i
T
C F
= − = − =

Step 7: ( ) ( )
SSE = TSS SSR+SSC+SST 329.94 29.19 4.69 284.19 11.87
− = − + + =
ANOVA Table
Source of
Variation
Sum of
Squares
Degree of
freedom
Mean
Square
F- Ratio
Rows SSR = 29.19 4 – 1 = 3 MSR = 9.73 FR = 4.91
Columns SSC = 4.69 4 – 1 = 3 MSC = 1.56
FC = 1.27
Treatments SST = 284.19 4 – 1 = 3 MST = 94.73 FT = 47.84
Error SSE = 11.87 (4 – 1)(4 – 2) = 6 MSE = 1.98

Total 22 15
F0.99,3,6 = 9.78
We conclude that there is a difference between fertilizers and yield at 1% level of
significance
3. Analyze the following RBD and find the conclusion
Treatments T1 T2 T3 T4
Blocks B1 12 14 20 22
B2 17 27 19 15
B3 15 14 17 12
B4 18 16 22 12
B5 19 15 20 14
Solution
H0: There is no significant difference between blocks and treatment
H1 : There is no significant difference between blocks and treatment
We subtract 15 from the given value
T1 T2 T3 T4 Total=Ti [Ti
2
]/k Xij
2
B1 -3 -1 5 7 8 16 84
B2 2 12 4 0 18 81 164
B3 0 -1 2 -3 -2 1 14
B4 4 0 5 -1 8 16 42
B5 4 0 5 -1 8 16 42
Total=Tj 6 11 23 0 40 130 372
[Tj
2
]/h 7.2 24.2 105.8 0 137.2

2
ij
y 38 147 119 68 372
N = 20
T=Grand Total = 40 ;
Correction Factor =
2 2
( ) (40)
80
20
Grand total
Total No of Observations
= =
 =
−

= 292
2
2
N
T
X
TSS ij
2
.
57
.
2
=
−
=
 F
C
h
T
SSC
J

SSR =  =
−
 50
2
2
N
T
Yij
SSE = TSS – SSC – SSR = 184.
ANOVA Table
Source of
Variation
Sum of
Squares
Degree of
freedom
Mean Square F- Ratio FTab Ratio
Between
Rows
(Blocks)
SSR = 50 r - 1= 3 MSR=12.5 FR = 1.232
FC= 1.238
F5%(12,4) =
5.91
F5%(3, 12) =
3.49
Between
Columns
(Treatmen
ts)
SSC= 57.2 c – 1=4 MSC = 19.07
Residual
SSE =
184.8
(h – 1)( k – 1)
= 12
MSE = 15.4
Total 292
Conclusion : Cal FC < Tab FC and Cal FR < Tab FR  hence the difference between the
blocks and that treatments are not significant.
4. The accompanying data results from an experiment comparing the degree of soiling for
fabric co-polymerized with the three different mixtures of methacrylic acid. Analysis is
the given classification (April /May 2017)
Mixture 1 0.56 1.12 0.90 1.07 0.94
Mixture 2 0.72 0.69 0.87 0.78 0.91
Mixture 3 0.62 1.08 1.07 0.99 0.93
Solution:
H0: The true average degree of soiling is identical for 3 mixtures.
H1 : The true average degree of soiling is not identical for 3 mixtures.
We shift the origin
Total
X 1 X2 X3 TOTAL X1
2
X2
2
X3
2
0.56 0.72 0.62 1.9 0.3136 0.5184 0.3844
1.12 0.69 1.08 2.89 1.2544 0.4761 1.1664
0.90 0.87 1.07 2.84 0.8100 0.7569 1.1449
1.07 0.78 0.99 2.84 1.1449 0.6084 0.9801
0.94 0.91 0.93 2.78 0.8836 0.8281 0.8649
4.59 3.97 4.69 13.25 4.4065 3.1879 4.5407
N = Total No of Observations = 15
T= Grand Total = 13.25
Correction Factor =
ns
Observatio
of
No
Total
)
total
Grand
( 2
= 11.7042

4309
.
0
N
T
X
X
X
TSS
2
2
3
2
2
2
1 =
−
+
+
= 


( ) ( ) ( )
0608
.
0
N
T
N
X
N
X
N
X
SSC
2
1
2
3
1
2
2
1
2
1
=
−
+
+
=


 ( 1
N = No of element in each column )
SSE = TSS – SSC = 0.4309 – 0.0608 = 0.3701
ANOVA TABLE
Source of
Variation
Sum of
Squares
Degree of
freedom
Between
Samples
SSC=0.0608 C-1=3-1=2 1
C
SSC
MSC
−
= =0.030
4
MSC
MSE
FC =
=10.144
Within
Samples
SSE=0.3701 N-C=15-3=12 C
N
SSE
MSE
−
= =0.30
84
Cal FC = 10.144 & Tab FC (12,2)=19.41
Conclusion : Cal FC < Tab FC  Hence we accept H0
5. Table below shows the seeds of 4 different types of corns planted in 3 blocks. Test at
0.05 level of significance whether the yields in kilograms per unit area vary significantly
with different types of corns. (April /May 2019)
Types of Corns
Blocks
I II III IV
A 4.5 6.4 7.2 6.7
B 8.8 7.8 9.6 7.0
C 5.9 6.8 5.7 5.2
Solution:
H0: There is no significant difference between types of corns and blocks.
H1 There is a significant difference between them.
Solution:
X1 X2 X3 X4 Tot
al
A 4.5 6.4 7.2 6.7 24.8
B 8.8 7.8 9.6 7.0 33.2
C 5.9 6.8 5.7 5.2 23.6
Total 19.
2
21 22.
5
18.9 81.6
H0 : There is no significant difference between rows, columns & treatments.

H1 : There is significant difference between rows, columns & treatments.
N = 12 T = 81.6
C.F =
2
554.88
T
N
=
2
2 2 2 2
1 2 3 4
18.9 40.96 51.84 44.89 77.44 60.84
92.16 49 34.81 46.24 32.49 27.04 554.88
21.73
T
TSS X X X X
N
= + + −
= + + + + +
+ + + + + + −
=
   
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 2 2 2
2
1 2 3 4
2 2 2 2
19.2 21 22.5 18.9
554.88
3 3 3 3
2.82
X X X X T
SSC
n n n n N
= + + + −
= + + + −
=
   
( n = No of element in each column )
( ) ( ) ( )
( ) ( ) ( )
2 2 2
2
2 2 2
24.8 33.2 23.6
554.88
4 4 4
13.68
A B C T
SSR
n n n N
= + + −
= + + −
=
  
( n = No of element in each row )
SSE = TSS – SSC - SSR = 554.88 – 2.82 – 13.68 = 538.38
ANOVA TABLE
S.V DF SS MSS F cal F
tab
Between
Columns
c-1=3 SSC = 2.82 0.94
.
SSC
MSC
d f
= =
95.46
C
MSE
F
MSC
= =
5.14
Between
Rows
r-1=2 SSR = 13.68 6.84
.
SSR
MSR
d f
= =
13.12
R
MSE
F
MSR
= = 5.14
Error
(c-1)(r-
1) = 6
SSE =
538.38
89.73
.
SSE
MSE
d f
= =
Conclusion: Cal 
c
F Tab c
F Reject 0
H
Cal R
F  Tab R
F Reject 0
H
We conclude that there is a difference between the corns and blocks.

6. The following data represent a certain person to work from Monday to Friday by 4
different roués.
DAYS
MON TUE WED THU FRI
ROUTES
1 22 26 25 25 31
2 25 27 28 26 29
3 26 29 33 30 33
4 26 28 27 30 30
Test at 5% level of significance whether the difference among the means obtained for
the different routes are significant and also whether the differences among the means
obtained from the different days of the week are significant. (Nov/ Dec 2017)
Solution:
DAYS
MON TUE WED THU FRI TOT
ROUTES
1 22 26 25 25 31 129
2 25 27 28 26 29 135
3 26 29 33 30 33 151
4 26 28 27 30 30 141
TOT 99 110 113 111 123 556
N = 20 T = 556
C.F =
2
15456.8
T
N
=
2
2 2 2 2
1 2 3 4 153.2
T
TSS X X X X
N
= + + − =
   
( ) ( ) ( ) ( )
2 2 2 2
2
1 2 3 4
1 1 1 1
52.8
X X X X T
SSC
N N N N N
= + + + − =
   
( 1
( ) ( ) ( ) ( ) ( )
2 2 2 2 2
2
1 2 3 4 5
2 2 2 2 2
73.2
Y Y Y Y Y T
SSR
N N N N N N
= + + + + − =
    
( 2
N = No of element in each row )
SSE = TSS – SSC-SSR = 37793.75 – 1613.5 – 26234.95 = 27.2
ANOVA TABLE
tab
Column
treatment
C-1=
5-1 =3
SSC=52.8 17.6
1
SSC
MSC
C
= =
−
7.75
C
MSC
F
MSE
= =
3.49
Between
Row
R-1=4-
1=4
SSR=73.2 18.3
1
SSR
MSR
R
= =
−
8.06
C
MCR
F
MSE
= =
3.26
Error
N-c-
R+1=1
2
SSE=27.2 2.27
12
SSE
MSE = =
Conclusion: Cal c
F Tab c
F , Reject H0

Cal 
R
F Tab R
F , Reject H0
7. A set of data involving four “four tropical feed stuffs A, B, C, D” tried on 20 chicks is
given below. All the twenty chicks are treated alike in all respects except the feeding
treatments and each feeding treatment is given to 5 chicks. Analyze the data.
(April/May 2017)
A 55 49 42 21 52
B 61 112 30 89 63
C 42 97 81 95 92
D 169 137 169 85 154
Solution:
H0 : There is no significant difference between column means as well as row means
H1 : There is no significant difference between column means as well as row means
X1 X2 X3 X4 X5 Total X2
1 X2
2 X2
3 X2
4 X2
5
A 5 -1 -8 -29 2 -31 25 1 64 841 4
B 11 62 -20 39 13 105 121 3844 400 1521 169
C -8 47 31 45 42 157 64 2209 961 2025 1764
D 119 87 119 35 104 464 14161 7569 14161 1225 10816
Total 127 195 122 90 161 695 14371 13623 15586 5612 12753
N = 20 T = 695
C.F = 25
.
24151
2
=
N
T
75
.
37793
2
2
5
2
4
2
3
2
2
2
1 =
−
+
+
+
=  


 N
T
X
X
X
X
X
TSS
( ) ( ) ( ) ( ) ( )
50
.
1613
2
1
2
5
1
2
4
1
2
3
1
2
2
1
2
1
=
−
+
+
+
+
=





N
T
N
X
N
X
N
X
N
X
N
X
SSC
( 1
( ) ( ) ( ) ( ) ( )
95
.
26234
2
2
2
5
2
2
4
2
2
3
2
2
2
2
2
1
=
−
+
+
+
+
=





N
T
N
Y
N
Y
N
Y
N
Y
N
Y
SSR
( 2
SSE = TSS – SSC-SSR = 37793.75 – 1613.5 – 26234.95 = =9945.3
ANOVA TABLE
S.V SS DF SS MSS F cal F
tab
Column
treatme
nt
SSC=1613
.5
c-1=
5-1
=4
SSC=1613.5
375
.
403
1
=
−
=
C
SSC
MSC 055
.
2
=
=
MSE
MSC
FC
3.26
Between
Row
SSR=2623
4.95
r-1=4-
1=3
SSR=26234.9
5
98
.
8744
1
=
−
=
R
SSR
MSR 552
.
10
=
=
MSE
MCR
FC
3.49
Error
SSE=9945
.3
N-c-
r+1=1
2
SSE=9945.3
775
.
828
12
=
=
SSE
MSE

Conclusion: Cal 
c
F Tab c
F , Accept H0
Cal 
R
F Tab R
F , Reject H0
8. Three varieties of coal were analyzed by 4 chemists and the ash content is given below.
Perform an ANOVA Table (May/June 2016)
Chemists
A B C D
COAL
I 8 5 5 7
II 7 6 4 4
III 3 6 5 4
Solution:
Chemists
A B C D TOT
COAL
I 8 5 5 7 25
II 7 6 4 4 21
III 3 6 5 4 18
TOT 18 17 14 15 64
N = 12
T = 64
C.F =
2
341.33
T
N
=
2
2 2 2 2
1 2 3 4 24.67
T
TSS X X X X
N
= + + − =
   
( ) ( ) ( )
2 2 2
2
1 2 3
1 1 1
3.34
X X X T
SSC
N N N N
= + + − =
  
( 1
( ) ( ) ( ) ( )
2 2 2 2
2
1 2 3 4
2 2 2 2
6.17
Y Y Y Y T
SSR
N N N N N
= + + + − =
   
( 2
SSE = TSS – SSC - SSR = 24.67 – 3.34 – 6.17 = 15.16
ANOVA TABLE
tab
Column C-1= 4- SSC=3.34 1.11
1
SSC
MSC
C
= =
−
2.28
C
MSE
F
MSC
= = 3.49

treatment 1 =3
Between
Row
R-1=3-
1=2
SSR=6.17 3.09
1
SSR
MSR
R
= =
−
1.22
C
MCR
F
MSE
= = 3.26
Error
N-c-
R+1=6
SSE=15.16 2.53
12
SSE
MSE = =
Conclusion: Cal c
F Tab c
F , Reject H0
Cal 
R
F Tab R
F , Reject H0
9. A company appoints 4 salesmen A, B, C and D and observes their sales in 3 seasons,
summer, winter and monsoon. The figures are given in the following table:
(Nov/Dec 2016)
Salesmen A B C D
Season
Summer 45 40 28 37
Winter 43 41 45 38
Monsoon 39 39 43 41
Carry out an analysis of variances.
Solution:
H0 : There is no significant difference between salesmen and between seasons.
H1 : There is significant difference between salesmen or between seasons.
Season X1 X2 X3 X4 Total X2
1 X2
X2
13 X2
1
Summer 5 0 -2 -3 0 25 0 4 9
Winter 3 1 5 -2 7 9 1 25 4
Monsoon -1 -1 1 1 0 1 1 1 1
Total 7 0 4 -4 7 35 2 30 14
N = 12 T = 7
C.F =
2
7
4.083
12
=
( ) ( ) ( )
2 2 2
2 2
TSS 5 0 2 3 ... 1 4.083 76.917
= + + − + − + + − =
2 2 2 2
7 0 4 4
SSC 4.083 22.917
3 3 3 3
−
= + + + − =
2 2 2
0 7 0
SS 4.083 8.167
4 4 4
R = + + − =
SSE TSS - SSC-SSR
=
SSE 76.917 22.917 8.167 45.833
= − − =
ANOVA TABLE
S.V DF SS MSS F cal F tab
salesmen 3 SSC = 22.917 MSC=7
.639
1
C
MSE
F
MSC
= = 4.76

seasons 2 SSR = 8.167 MSR=4
.0835
1.87
C
MCR
F
MSE
= = 19.33
Error 6 SSE = 45.833 MSE=7
.6388
Total 11 TSS = 76.917
Conclusion: Cal c
F  Tab c
F , Accept H0
Cal R
F  Tab R
F , Accept H0
10. The following is the Latin Square of a design when 4 varieties of seed are being tested.
Set up the analysis of variance table and state your conclusion. You can carry out the
suitable change of origin and scale.
(A) 110 (B) 100 (C)130 (D) 120
(C) 120 (D) 130 (A) 110 (B) 110
(D) 120 (C) 100 (B) 110 (A) 120
(B) 100 (A) 140 (D) 100 (C) 120
Solution:
Subtracting 100 and dividing by 10
1 2 3 4 Total=Ti* [Ti*
2
]/n X*ij
2
1 A 1 B 0 C 3 D 2 6 9 14
2 C 2 D 3 A 1 B 1 7 12.25 15
3 D 2 C 0 B 1 A 2 5 6.25 9
4 B 0 A 4 D 0 C 2 6 9 20
Total=T*j 5 7 5 7 24 36.5 58
[T*j
2
]/n 6.25 12.25 6.25 12.25 37
2
ij
y
 9 25 11 13 58
Letters Total=TK [TK
2
]/n
A 1 1 2 4 8 16
B 0 1 1 0 2 1
C 3 2 0 2 7 12.25
D 2 3 2 0 7 12.25
Total 24 41.5
2
2
ij
T
TSS Y 22
N
=  − =
2
2
i
1 T
SSR T 0.5
n N
= − =

2
2
j
1 T
SSC T 1
n N
= − =


2
2
K
1 T
SSK T 5.5
n N
SSE TSS SSC SSR SSK 15
= − =
= − − − =

ANOVA TABLE
Source of
Variation
Sum of
Squares
Degree of
freedom
FTab Ratio
( 5% level)
Between
Rows
SSR = 0.5 3 MSR = 0.167 FR =14.97
FC =7.508
FK =1.364
FR(6,36)=8.94
Fc(6,3)=8.94
FK(6 ,3)=8.94
Between
Columns
SSC = 1 3 MSC = 0.333
Between
Letters
SSK = 5.5 3 MSK = 1.833
Residual SSE = 15 6 MSE = 2.5
Total 22 15
Conclusion:
R
R F
Tab
F
Cal  , C
C F
Tab
F
Cal  , K K
Cal F Tab F
 There is a significant
difference between rows and no significant difference between column and also between
letters.
11. In a Latin square experiment given below are the yield in quintals per acre on the
paddy crop carried out for testing the effect of five fertilizers A, B, C, D, E. Analyze the
data for variations. (Nov / Dec 2017)
B 25 A 18 E 27 D 30 C 27
A 19 D 31 C 29 E 26 B 23
C 28 B 22 D 33 A 18 E 27
E 28 C 26 A 20 B 25 D 33
D 32 E 25 B 23 C 28 A 20
Solution:
Subtract 20 from all the items
X1 X2 X3 X4 X5 Total X2
1 X2
2 X2
3 X2
4 X2
5
1
Y 0 -7 2 5 2 2 0 49 4 25 4
2
Y -6 6 4 1 -2 3 36 36 16 1 4
3
Y 3 -3 8 -7 2 3 9 9 64 49 4
4
Y 3 1 -5 0 8 7 9 1 25 0 64
5
Y 7 0 -2 3 -5 3 49 0 4 9 25
Total 7 -3 7 2 5 18 103 95 113 84 101
N = 25 T = 18

C.F = 96
.
12
2
=
N
T
04
.
483
2
2
5
2
4
2
3
2
2
2
1 =
−
+
+
+
=  


 N
T
X
X
X
X
X
TSS
( ) ( ) ( ) ( ) ( )
24
.
14
2
1
2
5
1
2
4
1
2
3
1
2
2
1
2
1
=
−
+
+
+
+
=





N
T
N
X
N
X
N
X
N
X
N
X
SSC
( 1
( ) ( ) ( ) ( ) ( )
04
.
3
2
2
2
5
2
2
4
2
2
3
2
2
2
2
2
1
=
−
+
+
+
+
=





N
T
N
Y
N
Y
N
Y
N
Y
N
Y
SSR
( 2
SSK
A B C D E
-7 0 2 5 2
-6 -2 4 6 1
-7 -3 3 8 2
-5 0 1 8 3
-5 -2 3 7 0
TOT -30 -7 13 34 8
SSK =
( ) ( ) ( ) ( ) ( ) 64
.
454
5
8
5
34
5
13
5
7
5
30 2
2
2
2
2
2
=
−
+
+
+
−
+
−
N
T
SSE = TSS – SSC - SSR = 483.04 – 14.24 – 3.04 – 454.64 = 11.12
ANOVA TABLE
tab
Column
treatment
k-1=4 SSC=14.24 56
.
3
1
=
−
=
K
SSC
MSC 84
.
3
=
=
MSE
MSC
FC 3.26
Between
Row
k-1=4 SSR=3.04 76
.
0
1
=
−
=
K
SSR
MSR 22
.
1
=
=
MSR
MSE
FR
3.26
Between
Treatment
k-1=4 SSK=11.12 66
.
113
1
=
−
=
K
SSK
MSK
61
.
122
=
=
MSE
MSK
FT
3.26
Error
(k-1)(k-
2) = 12
SSE=483.04
927
.
0
)
2
)(
1
(
=
−
−
=
K
K
SSE
MSE
Conclusion: Cal 
c
F Tab c
F Reject 0
H
Cal 
R
F Tab R
F Accept 0
H
Cal 
T
F Tab T
F Reject 0
H

12. A variable trial was conducted on wheat with 4 varieties in a Latin Square Design. The
plan of the experiment and the per plot yield are given below:
(Apr/May 2017)( Dec 2016)
C 25 B 23 A 20 D 20
A 19 D 19 C 21 B 18
B 19 A 14 D 17 C 20
D 17 C 20 B 21 A 15
Solution:
Subtract 20 from all the items
X1 X2 X3 X4 Total X2
1 X2
2 X2
3 X2
4
1
Y 5 3 0 0 8 25 9 0 0
2
Y -1 -1 1 -2 -3 1 1 1 4
3
Y -1 -6 -3 0 10 1 36 9 0
4
Y -3 0 1 -5 -7 9 0 1 25
Total 0 -4 -1 -7 -12 36 46 11 29
N = 16 T = -12
C.F = 9
2
=
N
T
113
2
2
5
2
4
2
3
2
2
2
1 =
−
+
+
+
=  


 N
T
X
X
X
X
X
TSS
( ) ( ) ( ) ( ) ( )
5
.
7
2
1
2
5
1
2
4
1
2
3
1
2
2
1
2
1
=
−
+
+
+
+
=





N
T
N
X
N
X
N
X
N
X
N
X
SSC
( 1
( ) ( ) ( ) ( ) ( )
5
.
46
2
2
2
5
2
2
4
2
2
3
2
2
2
2
2
1
=
−
+
+
+
+
=





N
T
N
Y
N
Y
N
Y
N
Y
N
Y
SSR
( 2
SSK
T
A 0 -1 -6 -5 -12
B 3 -2 -1 1 1
C 5 1 0 0 6
D 0 -1 -3 -3 -7
SSK =
( ) ( ) ( ) ( ) 5
.
48
4
7
4
6
4
1
4
12 2
2
2
2
2
=
−
−
+
+
+
−
N
T
SSE = TSS – SSC - SSR = 113-7.5-46.5-48.5 = 10.5
ANOVA TABLE

tab
Column
treatment
k-1=3 SSC=7.5 5
.
2
1
=
−
=
K
SSC
MSC 43
.
1
=
=
MSE
MSC
FC 4.76
Between
Row
k-1=3 SSR=46.5 5
.
15
1
=
−
=
K
SSR
MSR 86
.
8
=
=
MSE
MCR
FR 4.76
Between
Treatment
k-1=3 SSK=48.5 17
.
16
1
=
−
=
K
SSK
MSK 24
.
9
=
=
MSE
MSK
FT 4.76
Error
(k-1)(k-
2) = 6
SSE=10.5
Conclusion: Cal 
c
F Tab c
F
Cal 
R
F Tab R
F
Cal 
T
F Tab T
F
There is significant difference between treatment and rows but there is no significant
difference between columns.
13. Given the following observation for the 2 factors A & B at two levels, Compute (i) the
main effect (ii) make an analysis of variance.
Treatment
Combination
Replication
I
Replication
II
Replication
III
(1) 10 14 9
A 21 19 23
B 17 15 16
AB 20 24 25
Solution:
H0 : No difference in the Mean effect.
H1 :There is a difference in the Mean effect.
We code the data by subtracting 20
Treatment Replication Total X2
1 X2
2 X2
3
(l) -10 -6 -11 -27 100 36 121
(a) 1 -1 3 3 1 1 9
(b) -3 -5 -4 -12 9 25 16
(ab) 0 4 5 9 0 16 25
Total -27 110 78 171
T= Grand Total = -27 N =12
Correction Factor =
2 2
( ) ( 27)
60.75
12
Grand total
Total No of Observations
−
= =
A Contract = a + ab – b – (1) = 3+9+-(12)-(-27) = 51
B Contract = b + ab – a – (1) = -12+9-3-(-27) = 21
A Contract = (1) + ab – a – b = - 27+9-3-(-12) = -9
(i) Main effects of A = A Contract / 2n = 51/6 = 8.5

Main effects of B = B Contract / 2n = 21/6 = 3.5
Main effects of AB = AB Contract / 2n = -9 / 6 = -1.5
2
2 2 2
1 2 3 110 78 171 60.75 298.25
T
TSS X X X
N
= + + − = + + − =
  
( )
2
2
51
( contract)
216.75
4 12
A
SSA
n
= = =
( )
2
2
21
( contract)
36.75
4 12
B
SSB
n
= = =
( )
2
2
9
( contract)
6.75
4 12
AB
SSAB
n
−
= = =
SSE = TSS – SSA – SSB - SSAB = 298.25 – 216.75 – 36.75 – 6.75 = 38
ANOVA Table
Source of
Variation
Sum of
Squares
Degree
of
freedom
Mean
Square
F- Ratio FTab Ratio
A SSA=216.75 1
MSA=216.
75
FA = 45.63
FB = 7.74
FAB = 1.42
F5%(1, 8) = 5.32
F5%(1, 8) = 5.32
F5%(1, 8) = 5.32
B SSB=36.75 1
MSB=36.7
5
AB SSAB=6.75 1
MSAB=6.7
5
Error SSE=38 4(n-1)=8 MSE=4.75
Total TSS=298.25 4n-1=11
Conclusion : Cal FA > Tab FA , Reject H0
Cal FB > Tab FB Reject H0
Cal FAB < Tab FAB Accept H0
UNIT – III SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS
PART – A
1. State Newton’s iterative formula.
)
(
'
)
(
1
n
n
n
n
x
f
x
f
x
x −
=
+
2. State the condition for convergence of Newton-Raphson Method.
The order of convergence of Newton-Raphson method is 2 and the convergence condition is
2
( ) ''( ) | '( )
f x f x f x


3. Write down the order of convergence and condition for convergence of fixed point
iteration method x = g(x).
The Order of convergence is One and condition for convergence is g'(x) 1
 , for
x I
 where I is the interval containing the root of x = g(x).
4. Find the interval in which the root lies for x
x e−
=
( ) ; ( ) 1
(0) 1( )
(1) 0.632( )
The root lies between 0 and 1.
x x
Let f x x e f x e
f ve
f ve
− −

= − = +
= − −
= +

5. What are the merits of Newton Raphson method? (Nov/Dec 2017)
Newton’s method is successfully used to improve the result obtained by other methods. It is
applicable to find the solution of equations involving algebraic functions as well as
transcendental functions.
6.
State Newton’s algorithm for finding square root of 12.
2 2
x = 12 x -12 = 0. Let f(x) = x -12; f'(x) = 2x
2
f(x ) x -12 x 12 1 12
n n n
x = x - = x - = x - + = x +
n+1 n n n n
f'(x ) 2x 2 2x 2 x
n n n n
1 12
x = x + is the iterative formula for 12
n+1 n
2 x
n

 
 
 
 
 
 

 
 
7. Obtain the iterative formula to find
1
N
using Newton Raphson method.
(Nov/Dec 2015)
Let
1 1 1
( )
x N f x N
N x x
=  =  = −
1 1
( )
1
The iterative formula to find is (2 )
( )
i
i i i
i i
i
f x
x x x x Nx
N f x
+ +
= −  = −

8. Find the smallest positive root of 3
2 5 0
x x
− − =
Let 3
( ) 2 5
f x x x
= − +
f (0) = - 5 = -ve; f (1) = –6 = – ve, f (2) = –1 = – ve, f (3) = 16 = + ve
Therefore, a root lies between 2 and 3.
The Newton-Raphson formula is given by
1
( )
( )
n
n n
n
f x
x x
f x
+ = −

3
2
( ) 2 5
( ) 3 2
f x x x
f x x
= − −
 = −

Take x0 = 0. Putting n = 1,
3
0
1 0 2
0
( ) (2) 2 2(2) 5
2 2 2.1
( ) (2) 3(2 ) 2
f x f
x x
f x f
− −
= − = − = − =
  −
1
2 1
1
( ) (2.1)
2 2.1 2.0946
( ) (2.1)
f x f
n x x
f x f
=  = − = − =
 
2
3 2
2
( ) (2.0946)
3 2.0946 2.0946
( ) (2.0946)
f x f
n x x
f x f
=  = − = − =
 
The root is 2.0946.
9. Define a direct and indirect methods of solving systems of simultaneous linear equations.
(April / May 2019)
A direct method is one which involve a certain amount of fixed computation whereas in an
indirect method the amount of computation depends on the degree of accuracy and the
solution is obtained by successive approximations.
10. Distinguish between direct and iterative methods of solving simultaneous equations.
S.No Direct Method Iterative Method
1. We get exact solution. Approximate solution.
2. Simple, take less time. Laborious, time consuming.
11. For solving a linear system of equations, compare Gauss Elimination method and Gauss
Jordan method.
S.No. Gauss-Elimination method Gauss – Jordan method
1.
2.
Augmented matrix is transformed into
upper triangular matrix
We obtain the solutions by back
substitution method
Augmented matrix is transformed
into diagonal matrix
No need of back substitution method
12. What is the procedure of Gauss-Jordan method to solve the simultaneous equations?
In this method, the augmented matrix is reduced to a diagonal matrix (or even a unit matrix)
by elementary row operations. Here we get the solutions directly, without using back
substitution method.
13. Using Gauss elimination method solve 2, 2 3 1
x y x y
+ = − = −
The given system equations can be written as AX = B 
1 1 2
2 3 1
x
y
    
=
    
− −
    
  2 2 1
1 1 2 1 1 2
, 2
2 3 1 0 5 5
~
A B R R R
   
= → −
   
− − − −
   
- 5y = - 5 y=1
2 1 2 1
x y x x
 
+ =  + =  =

14. Solve the linear system of equations 4x-3y=11, 3x+2y=4 by Gauss-Jordan method.
The given system of equations can be written as AX = B 
4 3 x 11
3 2 y 4
−
    
=
    
    
2 2 1
4 3 11 4 3 11
A,B R 4R 3R
3 2 4 0 17 17
   
     
 
   
− −
 =  → −
−
2
2 1 2 1
4 3 11 4 0 8
R
R R 3R R
0 1 1 0 1 1
17
   
   
   
−
 →  → +
− −
n
Bybacksubstutio
4x 8& y 1
x 2&y 1
= = −
 = = −
15. What is meant by Diagonally Dominant?
A matrix is diagonally dominant if the numerical value of the leading diagonal element in
each row is greater than or equal to the sum of the numerical values of the other element in
that row. Consider the system of equations,
1 1 1 1
2 2 2 2
3 3 3 3
a x b y c z d
a x b y c z d
a x b y c z d
+ + =
+ + =
+ + =
i.e. 1 1 1 2 2 2 3 3 3
; ;
a b c b a c c a b
 +  +  +
16. Solve the following system of equations by Gauss-Seidel method 8x+11y-4z = 95,
7x+52y+13z = 104, 3x+8y+29z = 71
Given equations are 8x +11y-4z = 95, 7x+52y+13z = 104 , 3x+8y+29z = 71
The given system of equation does not satisfy the diagonally dominant condition
i.e., 8 isnotgreater than 11 4
+ −
Rearrangement above equation fails.
Hence the given system of equation does not converge by Gauss Seidel method
17. Write down the condition for the convergence and the iterative formula of Gauss Seidel
technique. (April/May 2018,2019)
Consider the system of equations,
, ,
1 1 1 1 2 2 2 2 3 3 3 3
a x b y c z d a x b y c z d a x b y c z d
+ + = + + = + + =
The given system of equation should be diagonally dominant.
1 1 1 2 2 2 3 3 3
i.e. a b c ; b a c ; c a b
 +  +  +
Then the iterative formula is given by,
(n 1) (n) (n)
1 1 1
1
(n 1) (n 1) (n)
2 2 2
2
(n 1) (n 1) (n 1)
3 3 3
3
1
x [d b y c z ]
a
1
y [d a x c z ]
b
1
z [d a x b y ]
c
+
+ +
+ + +
= − −
= − −
= − −

18 Distinguish between Gauss elimination and Gauss seidel methods. (Nov/Dec 2017)
Gauss Elimination Gauss Seidel
a) Exact solution obtained
b) No initial values are needed
c) Direct method
a) Approximate solution obtained
b) Initial values are needed
c) Iterative method.
19
Write the procedure to find all the eigen values of a 3x3 matrix A by using power
method.
Step 1: Find the dominant Eigen value of A, say 1
 , by using power method.
Step 2: Consider the matrix B = A – 1
 I
Step 3: Again, by applying power method, find the dominant Eigen value of B. Then the
smallest eigen value of A is equal to 1
 + the dominant eigen value of B
Step 4: The remaining eigen value is found by using the property.
20 Determine the largest eigen value and the corresponding eigen vector of the matrix






1
1
1
1
correct to two decimal places using power method.
The initial Eigen vector Let 1
1
0
X
 
=  
 
AX1 = 2
1
1
0
1
1
1
1
1
X
=






=












AX2= 3
X
2
1
1
2
1
1
1
1
1
1
=






=












; AX3= 4
2
1
1
2
1
1
1
1
1
1
X
=






=












The largest Eigen value is 2 & the corresponding Eigen vector is
1
1
 
 
 
.
PART – B
1. Find Newton’s iterative formula to find the reciprocal of a given number N and hence
find the value of
1
19
Solution:
Let
x
N
N
x
1
1
=

=
)
2
(
;
)
(
)
(
;
1
)
( 1
1 i
i
i
i
i
i
i Nx
x
x
x
f
x
f
x
x
N
x
x
f −
=

−
=
−
= +
+
To find
19
1
,
take x0= 05
.
0
20
1
=
x1=0.0525; x2=0.05263; x3 =0.05263 ; x4=0.05263.

05263
.
0
19
1
=

2. Find a root of 10
log 1.2 0
x x − = Newton’s method correct to three decimal places.
(May/June 2014)
Solution:
10
( ) log 1.2 ; (2) 0.5980 ( ); (3) 0.2313( )
f x x x f ve f ve
= − = − − = +
The root lies between 2 and 3
Hence the root is 2.74065
3. Find a positive root of the following equations by the method of fixed point iteration.
(a) cos 3 1 0
x x
− + = (b)
Solution:
(a) f(x) = cos x-3x+1=0
The root lies between 0 and
Take
........
(b) The root lies between 2 & 3
take
,

.....
4. Solve the following system of equations by Gauss Elimination method
2 3 5, 4 4 3 3, 2 3 2 2
x y z x y z x y z
+ − = + − = − + =
Solution:
Given
2 3 5,
4 4 3 3,
2 3 2 2;
x y z
x y z
x y z
+ − =
+ − =
− + =
2 2 1
2 3 1 5 2 3 1 5
(A,B) 4 4 3 3 0 2 1 7 R R 2R
2 3 4 2 0 6 3 3
 −   − 
   
= − − − − → −
   
   
− − −
   
3 3 2
2 3 1 5
0 2 1 7 R R 3R
0 0 6 18
Hence 2x 3y z 5
2y z 7
6z 18
z 3 y 2 x 1
 − 
 
− − − → −
 
 
 
+ − =
− − = −
=
 = = =
5. Solve the following system of equations by Gauss Elimination method
1 2 3 4
5 4,
x x x x
+ + + = 1 2 3 4
7 12,
x x x x
+ + + = 1 2 3 4
6 5,
x x x x
+ + + = −
1 2 3 4
4 6
x x x x
+ + + = −
Solution:
1 2 3 4 1 2 3 4
1 2 3 4 1 2 3 4
5 4, 7 12,
6 5, 4 6
x x x x x x x x
x x x x x x x x
+ + + = + + + =
+ + + = − + + + = −
2 2 1 3 3 1 4 4 1
5 1 1 1 4 5 1 1 1 4
1 7 1 1 12 0 34 4 4 56
(A,B) R 5R R , R 5R R , R 5R R
1 1 6 1 5 0 4 29 4 29
1 1 1 4 6 0 4 4 19 34
   
   
   
= → − → − → −
   
− −
   
− −
   
   
3 3 2 4 4 2
5 1 1 1 4
0 34 4 4 56
R 34R 4R , R 34R 4R
0 0 970 120 1210
0 0 120 630 1380
 
 
 
= → − → −
 
−
 
−
 
 

4 4 3
5 1 1 1 4
0 17 2 2 28
R 970R 120R
0 0 97 12 121
0 0 0 5967 11934
 
 
 
= → −
 
−
 
−
 
 
4
4
3 4
3
2 3 4
2
1 2 3 4
1
1 2 3 4
Hence 5967x 11934
x 2
97x 12x 121
x 1
17x 2x 2x 28
x 2
5x x x x 4
x 1
Hence the solution is x 1;x 2; x 1; x 2
= −
= −
+ = −
= −
+ + =
=
+ + + =
=
= = = − −
6. Apply Gauss Jordan method to find the solution of the following system:
3 2 12, 2 3 11, 2 2 2
x y z x y z x y z
− + = + + = − − =
Solution:
3 1 2 12
( , ) 1 2 3 11
2 2 1 2
A B
 − 
 
=  
 
− −
 
2 2 1
3 3 1
3 1 2 12
0 7 7 21 3
0 4 7 18 3 2
R R R
R R R
 − 
 
= → −
 
 
− − − → −
 
3 3 2
3 1 2 12
0 7 7 21
0 0 21 42 7 4
R R R
 − 
 
=  
 
− − → +
 
2
3
3 1 2 12
0 1 1 3
7
0 0 1 2
21
R
R
 − 
 
=  
 
 
−
1 1 2
3 0 3 15
0 1 1 3
0 0 1 2
R R R
  = +
 
=  
 
 
1 1 3
2 2 3
3 0 0 9 3
0 1 0 1
0 0 1 2
R R R
R R R
  = −
 
= = −
 
 
 

1
1 0 0 3
0 1 0 1
3
0 0 1 2
R
 
 
=  
 
 
Therefore x = 3, y = 1, z = 2.
7. Solve the system of the following equations using Gauss Jordan method correct to two
decimals places 1 2 3 4
2 2 4,
x x x x
+ − + = 1 2 3 4
4 3 2 6,
x x x x
+ − + =
1 2 3 4
8 5 3 4 12
x x x x
+ − + = and 1 2 3 4
3 3 2 2 6
x x x x
+ − + = (April/May 2017)
Solution:
1 2 3 4 1 2 3 4 1 2 3 4
2 2 4, 4 3 2 6,8 5 3 4 12
x x x x x x x x x x x x
+ − + = + − + = + − + = and
1 2 3 4
3 3 2 2 6
x x x x
+ − + =
2 2 1 1 4
4 3 1 2 6
( , )
8 5 3 4 12
3 3 2 2 6
A B
−
 
 
−
 
=
 
−
 
−
 
1 1
1 1 0.5 0.5 2
4 3 1 2 6
/ 2
8 5 3 4 12
3 3 2 2 6
R R
−
 
 
−
 
= →
 
−
 
−
 
2 2 1
3 3 1
4 4 1
1 1 0.5 0.5 2
4
0 1 1 0 2
8
0 3 1 0 4
3
0 0 0.5 0.5 0
R R R
R R R
R R R
−
 
→ −
 
− −
 
= → −
 
− −
→ −
 
−
 
2 2
1 1 0.5 0.5 2
( 1)
0 1 1 0 2
0 3 1 0 4
0 0 0.5 0.5 0
R R
−
 
→ −
 
−
 
=
 
− −
 
−
 
3 3 2
2 1 2
3
1 0 0.5 0.5 0
0 1 1 0 2
0 0 2 0 2
0 0 0.5 0.5 0
R R R
R R R
→ +
−
 
→ −
 
−
 
=
 
−
 
−
 
3 3 /( 2)
1 0 0.5 0.5 0
0 1 1 0 2
0 0 1 0 1
0 0 0.5 0.5 0
R R
→ −
 
 
−
 
=
 
−
 
−
 
1 1 3
2 2 3
4 4 3
1 0 0 0.5 0.5
0.5
0 1 0 0 1
0 0 1 0 1
0.5
0 0 0 0.5 0.5
R R R
R R R
R R R
 
→ −
 
 
= → +
 
−
→ +
 
−
 
1 1 4
4 4
1 0 0 0.5 0.5
0.5
0 1 0 0 1
0 0 1 0 1
/ 0.5
0 0 0 1 1
R R R
R R
 
→ −
 
 
=
 
−
→
 
−
 
1
2
3
4
1 0 0 0 1
0 1 0 0 1
0 0 1 0 1
0 0 0 1 1
x
x
x
x
 
   
 
   
 
   
=
 
   
−
 
   
−
 
   
 
1 2 3 4
1 , 1 , 1 , 1
x x x x
 = = = − = −
8. Solve the following system by Gauss Jacobi method
20 2 17, 3 20 18,
+ − = + − = −
x y z x y z 2 3 20 25
x y z
− + =

Solution:
Clearly the given equation satisfies the diagonally dominant condition.
( ) ( ) ( )
1 1 1
17 2 18 3 25 2 3
20 20 20
x y z , y x z & z x y
= − + = − − + = − +
First Iteration : Put
17 18 25
0 0 0 0 85 0 9 1 25
20 20 20
x ,y & z x . ; y . & z .
−
= = =  = = = = − = =
Second Iteration: 0 85 0 9 1 25
x . , y . & z .
= = − =
   
 
1 1
17 0 9 2 1 25 1 02 18 3 0 85 1 25 0 965
20 20
1
25 2 0 85 3 0 965 1 03
20
x . ( . ) . ; y ( . ) . . &
z ( . ) ( . ) .
 = + + = = − − + = −
= − + − =
The remaining iteration is
Iteration ( )
1
17 2
20
x y z
= − +
( )
1
18 3
20
y x z
= − − + ( )
1
25 2 3
20
z x y
= − +
0 0 0 0
1 0.85 –0.90 1.250
2 1.02 –0.965 1.03
3 1.00125 –1.0015 1.0032
4 1.000 –1.0003 0.9965
5 1.000 –1.000 1.000
 The approximate root is 1 1 1
x ,y , z
= = − =
9. Using Gauss- Seidel method, solve x + y + 54z = 110, 27x + 6y – z = 85,
6x + 15y + 2z = 72. (April/May 2018,2019) (Nov/Dec 2017)
Solution:
As the coefficient matrix is not diagonally dominant we rewrite the equation as
27 16 85,
6 15 2 72,
54 110;
+ − =
+ + =
+ + =
x y z
x y z
x y z
 
 
 
1
85 6
27
1
72 6 2
15
1
110
54
0, 0, 0
x y z
y x z
z x y
Let the initial value be x y z
= − +
= − −
= − −
= = =
 
 
 
(1)
(1)
(1)
1
: 85 6(0) (0) 3.148
27
1
72 6(3.148) 2(0) 3.5408
15
1
110 3.148 3.5408 1.913
54
First Iteration x
y
z
= − + =
= − − =
= − − =

 
 
 
(2)
(2)
(2)
1
: 85 6(3.5408) (1.913) 2.4322
27
1
72 6(2.4322) 2(1.913) 3.572
15
1
110 2.4322 3.572 1.9259
54
Second Iteration x
y
z
= − + =
= − − =
= − − =
Proceeding like this we get
(3) (3) (3)
(4) (4) (4)
(5) (5) (5)
x 2.4257, y 3.5729,z 1.926
x 2.4255, y 3.573,z 1.926
x 2.4255, y 3.573,z 1.926
= = =
= = =
= = =
4th
and 5th
iterations are same.
Hence x= 2.426 , y= 3.573 , z=1.926
10. Solve the following system of equations by Gauss Seidel method 9 15,
x y z
+ + =
17 2 48, 30 2 3 75
x y z x y z
+ − = − + = (May/June 2016)
Solution:
The system of equations can be written as,
30 2 3 75
17 2 48
9 15
x y z
x y z
x y z
− + =
+ − =
+ + =
Now, the system is diagonally dominant.
Therefore
Let the initial value be y=0,z=0
Iterations 1
x (75 2y 3z)
30
= + −
1
y (48 x 2z)
17
= − +
1
z (15 x y)
9
= − −
1 2.5 2.6765 1.0915
2 2.5693 2.8008 1.07
3 2.5797 2.7977 1.0692
4 2.5796 2.7976 1.0692
5 2.5796 2.7976 1.0692
Hence the solution is x = 2.5796, y = 2.7976, z = 1.0692
11. Solve the following system by Gauss Jacobi method 4 2 14,
x y z
+ + = 5 10,
x y z
+ − =
8 20
x y z
+ + =
Solution:
Clearly the given equation satisfies the diagonally dominant condition.
( ) ( ) ( )
1 1 1
14 2 10 20
4 5 8
x y z , y x z & z x y
= − − = − + = − −
First Iteration : Put
14 10 20
0 0 0 3 5 2 2 5
4 5 8
x ,y & z x . ; y & z .
= = =  = = = = = =
Second Iteration:

( )
( )
( )
( )
1
14 2 2 2 5 1 875
4
1
10 3 5 2 5 1 8
5
1
20 3 5 2 1 8125
8
x . .
y . . .
z . .
= − − =
= − + =
= − − =
The remaining iteration is
Iteration ( )
1
14 2
4
x y z
= − − ( )
1
10
5
y x z
= − + ( )
1
20
8
z x y
= − −
0 0 0 0
1 3.5 2 2.5
2 1.875 1.8 1.8125
3 2.1469 1.9875 2.0406
4 1.9961 1.9787 1.9832
5 2.0149 1.9974 2.003
6 2.0006 1.9976 1.9985
7 2.0016 1.9996 2.0002
8 2.0002 1.9997 1.9998
9 2.0002 1.9999 2.0000
10 2.0001 2.0000 2.0000
11 2 2 2
12 2 2 2
Hence the root is 2 2 2
x ,y , z
= = =
12. Using Jacobi method, find the eigen values of
1 3
3 4
A
 
=  
 
11 22
12
2
2 2
Solution:
a a 1 4 1
cot 2
2a 6 2
1 5
cot 1 1/ 2 1 1/ 4
2
5 1 1 2 5 1
take 0 sin and cos
2 1 1
10 2 5 10 2 5
− − −
 =  = = =
− 
 =  =   +  = −  + =
−  −
 =   = =  = =
+  + 
+ +
2 2
11 11 12 22
5 3 5
b a cos a sin 2 a sin
2
+
=  +  +  =
22 11 22 11
5 3 5
b a a b
2
5 3 5 5 3 5
hence eigen values are and
2 2
−
= + − =
+ −

1 2
cos sin
R
sin cos
2
5 1
Theeigen vectors are X and X
2 5 1
 − 
 
=  
 
 
−
   
−
= =
   
−
   
13.
Using Jacobi method, find the eigen values and eigen vectors of the matrix
2 0 1
0 2 0
1 0 2
A
−
 
 
=  
 
−
 
Solution:
Given
2 0 1
0 2 0
1 0 2
A
−
 
 
=  
 
−
 
is a symmetric matrix,
Here, the numerically largest off-diagonal element is 13 31 1
a a
= = −
and 11 33 2
a a
= =
.
The rotation matrix is
cos 0 sin
0 2 0
sin 0 cos
P
 
 
 
 
=  
 
−
 
The angle is
1 13
11 33
1
2
1
tan
2
1 2( 1)
tan
2 2 2 4
a
a a


−
−
 
=  
−
 
−
 
= =
 
−
 
The rotation matrix
1 1
0
cos 0 sin
2 2
4 4
0 2 0 0 2 0
1 1
sin 0 cos 0
4 4 2 2
P
 
 
 
 
 
 
 
 
= =  
 
 
 
 
 
− −
   
   
1 1 1 1
0 0
2 0 1
2 2 2 2
0 2 0 0 2 0 0 2 0
1 1 1 0 2 1 1
0 0
2 2 2 2
T
D P AP
   
−
   
−
 
   
 
= =    
 
   
 
−
 
   
−
   
   
1 1 1 1
0 0
2 0 1
2 2 2 2
0 2 0 0 2 0 0 2 0
1 1 1 0 2 1 1
0 0
2 2 2 2
T
D P AP
   
−
   
−
 
   
 
= =    
 
   
 
−
 
   
−
   
   
1 1 1 1
0 0
1 0 0
2 2 2 2
0 2 0 0 2 0 0 2 0
3 3 1 1 0 0 3
0 0
2 2 2 2
T
D P AP
  
−
    
    
= = =
    
    
 
  
−
  
  

The eigen values are 1, 2, 3 and eigen vectors are
1 0 1
0 , 1 , 0 .
1 0 1
−
     
     
     
     
     
14. Find the dominant eigen value of A =
25 1 2
1 3 0
2 0 4
 
 
 
 
−
 
and the corresponding eigen vector,
by power method. (May/June 2016, Nov/Dec 2017)
Solution:
1
1
0 .
0
Let x be an arbitraryinitial eigen vector
 
 
=  
 
 
1 2
25 1 2 1 25 1
1 3 0 0 1 25 0.04 25
2 0 4 0 2 0.08
AX X
      
      
= = = =
      
      
−
      
2 3
25 1 2 1 25.2 1
1 3 0 0.04 1.12 25.2 0.0444 25.2
2 0 4 0.08 1.68 0.0667
AX X
      
      
= = = =
      
      
−
      
3 4
25 1 2 1 25.1778 1
1 3 0 0.0444 1.1332 25.1778 0.0450 25.1778
2 0 4 0.0667 1.7337 0.06888
AX X
      
      
= = = =
      
      
−
      
4 5
25 1 2 1 25.1826 1
1 3 0 0.0450 1.135 25.1826 0.0451 25.1826
2 0 4 0.06888 1.7248 0.0685
AX X
      
      
= = = =
      
      
−
      
5 6
25 1 2 1 25.1821 1
1 3 0 0.0451 1.1353 25.1821 0.0451 25.1821
2 0 4 0.0685 1.7260 0.0685
AX X
      
      
= = = =
      
      
−
      
6 7
25 1 2 1 25.1821 1
1 3 0 0.0451 1.1353 25.1821 0.0451 25.1821
2 0 4 0.0685 1.7260 0.0685
AX X
      
      
= = = =
      
      
−
      
7 8
25 1 2 1 25.182 1
1 3 0 0.0451 1.1353 25.182 0.0451 25.182
2 0 4 0.0685 1.7260 0.0685
AX X
      
      
= = = =
      
      
−
      
We have reached the limit
1
1
25.182 and the corresponding eigen vector is 0.0451
0.0685

 
 
 =  
 
 

15. Find the largest Eigen value and Eigen vector of
1 6 1
1 2 0
0 0 3
A
 
 
=  
 
 
, by using power method.
Also find the least latent root and hence find the third Eigen value also.
Solution:
1
1
Let X = 0 beanapproximateeigen vector.
0
 
 
 
 
 
1 2
1 6 1 1 1 1
1 2 0 0 1 1 1 1
0 0 3 0 0 0
AX X
      
      
= = = =
      
      
      
2 3
1 6 1 1 7 1
1 2 0 1 3 7 0.4286 7
0 0 3 0 0 0
AX X
      
      
= = = =
      
      
      
3 4
1 6 1 1 3.5714 1
1 2 0 0.4286 1.8572 3.5714 0.52 3.5714
0 0 3 0 0 0
AX X
      
      
= = = =
      
      
      
4 5
1 6 1 1 4.12 1
1 2 0 0.52 2.04 4.12 0.4951 4.12
0 0 3 0 0 0
AX X
      
      
= = = =
      
      
      
5 6
1 6 1 1 3.9706 1
1 2 0 0.4951 1.9902 3.9706 0.5012 3.9706
0 0 3 0 0 0
AX X
      
      
= = = =
      
      
      
6 7
1 6 1 1 4.0072 1
1 2 0 0.5012 2.0024 4.0072 0.4997 4.0072
0 0 3 0 0 0
AX X
      
      
= = = =
      
      
      
7 8
1 6 1 1 3.9982 1
1 2 0 0.4997 1.9994 3.9982 0.5000 3.9982
0 0 3 0 0 0
AX X
      
      
= = = =
      
      
      
8 9
1 6 1 1 4 1
1 2 0 0.5000 2 4 0.5 4
0 0 3 0 0 0
AX X
      
      
= = = =
      
      
      
9
1 6 1 1 4 1
1 2 0 0.5 2 4 0.5
0 0 3 0 0 0
AX
       
       
= = =
       
       
       
Dominant eigen value 4;corresponding eigen vector is (1,0.5,0)
 =
1
To find the least eigen value,let B A-4I since 4

= =

3 6 1
1 2 2
0 0 2
B
−
 
 
 = −
 
 
−
 
We will find the dominant eigen value of B
1
1
Let Y 0 .
0
be the initial vector
 
 
=  
 
 
1 2
3 6 1 1 3 1
1 2 2 0 1 3 0.3333 3
0 0 2 0 0 0
BY Y
− −
      
     
 
= − = = − − = −
     
 
     
 
−
      
2 3
3 6 1 1 5 1
1 2 2 0.3333 1.6666 5 0.3333 5
0 0 2 0 0 0
BY Y
− −
      
     
 
= − − = = − − = −
     
 
     
 
−
      
3
3 6 1 1 5 1
1 2 2 0.3333 1.6666 5 0.3333
0 0 2 0 0 0
BY
− −
      
     
 
= − − = = − −
     
 
     
 
−
      
Dominant eigen value of B is -5

Adding smallest eigen value of A 5 4 1
= − + = −
Sum of eigen values Trace of A 1 2 3 6
= = + + =
sum of eigen value of A= Trace of A
Here 1 2
4, 1
 
= = −
1 2 3
3
3
1 2 3 6
4 1 6
3
  


+ + = + + =
− + =
=
3
6 4 8

+ + =
3 2
 = −
Hence the eigenvalues are 1 2 3
4, 1, 3
  
= = − =
UNIT – IV INTERPOLATION, NUMERICAL DIFFERENTIATION AND NUMERICAL
INTEGRATION
PART – A
1. What is meant by Interpolation and Extrapolation? (Nov/Dec 2017)
Interpolation is the process of finding the values of f(x) for intermediate values of x in
the given interval. Extrapolation is the process of finding the values of f(x) for outside the
given interval.
2. Write down Inverse Lagrange’s Interpolation formula.
Inverse Lagrange’s Interpolation formula is

1 2 2
1
1 2 1 1 2 1
0 1 1
1 1
( )( ) ( ) ( )( ) ( )
( )( ) ( ) ( )( ) ( )
( )( ) ( )
( )( ) ( )
n o n
o
o o o n o n
n
n
n o n n n
y y y y y y y y y y y y
x x x
y y y y y y
x
y y y y y y
−
−
− − − − − −
= + +
− − − − − −
− − −
+
− − −
3. Obtain Lagrangian interpolation polynomial from the data.
x: -1 1 2
f(x) : 7 5 15
2
1
2
2
1
0
1
2
1
1
2
2
1
2
1
)
)(
(
)
)(
(
)
)(
(
)
)(
(
)
)(
(
)
)(
(
)
( y
x
x
x
x
x
x
x
x
y
x
x
x
x
x
x
x
x
y
x
x
x
x
x
x
x
x
x
f
o
o
o
o
o
o −
−
−
−
+
−
−
−
−
+
−
−
−
−
=
2
1
11x 3x 7
3
(x 1)(x 2) (x 1)(x 2) (x 1)(x 1)
y f(x) (7) (5) (15)
( 1 1)( 1 1) (1 1)(1 32) (2 1)(2 1)
 
− +
 
− − + − + −
= =  +  +  =
− − − − + − + −
4. Evaluate 1 2 3
 + + +
[ ( )( )( )]
x x x x (April/ May 2017)
We Know that [ ( )] ( 1) ( )
f x f x f x
 = + −
[ ( 1)( 2)( 3)] ( 1)( 2)( 3)( 4) ( 1)( 2)( 3)
( 1)( 2)( 3)[ 4 ] 4( 1)( 2)( 3)
x x x x x x x x
 + + + = + + + + − + + +
= + + + + − = + + +
5. State Newton’s Backward formula.
n
n
n
3
n
2
n
n y
!
n
))
1
n
(
v
)....(
1
v
(
v
.....
y
!
3
)
2
v
)(
1
v
(
v
y
!
2
)
1
v
(
v
y
v
y
)
x
(
y 
−
+
+
+
+

+
+
+

+
+

+
=
where
h
x
x
v n
−
=
6. Find the polynomial which takes the following values given f (0) = - 1, f (1) = 1 and
f (2) = 4 using Newton’s interpolating formula. (Nov/Dec 2016)
X y = f(x) ∆y ∆2
y
0 -1
2
1 1 1
3
2 4
There are only 3 data given. hence, the polynomial of degree 2
2 3
( 1) ( 1)( 2) ( 1)...( ( 1))
( )
2! 3! !
n
o o o o o
u u u u u u u u n
y x y u y y y y
n
− − − − − −
= + + + + + +
   
where 0
x x
u
h
−
=
( )
0
2
0 1
( 1) 1
( ) 1 (2) (1) 3 2
2! 2
x h u x
x x
y x x x x
= =  =
−
= − + + = + −
7. Define Numerical Differentiation.
Numerical differentiation is the process of computing the values of
2
2
,
dy d y
dx dx
…. for some
particular values of x from the given data (xi,yi) where y is not known explicitly.

8. Define Numerical Integration
The process of computing the values of definite integrals
b
a
y dx
 from a set of (n + 1) paired
values (xi,yi), i = 0,1,2,…,n, where xo = a, xn = b of the function y = f(x) is called Numerical
integration.
9. Write down Newton’s divided difference formula.
Newton’s divided difference formula is
0 0 0 0 1 0 1 2
0 1 0 1 2
1
)
1
( ) ( ) ( ) ( , ) ( )( ) ( , , )
( )( ) ( ( , , ,.... )
n n
f x f x x x f x x x x x x f x x x
x x x x x x f x x x x
−
= + − + − − +
+ − − −
10. State any two properties of Divided difference. (Nov/Dec 2016)
(i) The divided differences are symmetrical in all their arguments. i.e. the value of any
difference is independent of the order of the arguments.
(ii) The divided differences of the sum or difference of two functions is equal to the sum or
difference of the corresponding separate divided differences.
11. When do we use the divided difference methods and Newton’s forward and backward
interpolation methods? (April/May 2019)
When the values of x are at equal intervals the Newton’s forward and backward formulae can
be applied. But when the values of x given at unequal intervals then we use the divided
difference formula. Further Newton’s forward formula can be used when we want to
interpolate the value of the beginning of the table and backward at the end of the table.
12. Show that the second order divided difference [x0, x1, x2] is independent of the order of
the arguments.
0 1 2
0 1 2
0 1 0 2 1 0 1 2 2 0 2 1
( ) ( ) ( )
( , , )
( )( ) ( )( ) ( )( )
f x f x f x
f x x x
= + +
− − − − − −
 0 1 2
( , , )
f x x x = 1 2 0
( , , )
f x x x = 2 0 1
( , , )
f x x x
This shows that 0 1 2
( , , )
f x x x is independent of the order of the arguments.
13. Show that .
abcd
1
a
1
3
bcd
−
=







abcd
1
a
d
abc
1
bcd
1
a
d
)
c
,
b
,
a
(
f
)
d
,
c
,
b
(
f
)
d
,
c
,
b
,
a
(
f
a
1
3
bcd
−
=
−
−
=
−
−
=
=







14. State Newton’s formula to find the derivatives 2
2
,
dx
y
d
dx
dy
at x = x0 using forward
differences.







−

+

−

=
=
=






−

+

−

=
=
=
.......
12
11
1
)
(
'
'
'
'
........
3
1
2
1
1
)
(
'
'
4
3
2
2
2
2
3
2
o
o
o
o
o
o
y
y
y
h
dx
y
d
x
f
y
y
y
y
h
dx
dy
x
f
y
15. Write down Newton’s formula to find the derivatives 2
2
,
dx
y
d
dx
dy
using backward
differences. (April / May 2016)
2
2 3
1 (2 1) (3 6 2)
' '( ) ........
2! 3!
n n n
dy v v v
y f x y y y
dx h
+ + +
= = =  +  +  +
 
 
 
2 2
2 3 4
2 2
1 6 18 11
'' ''( ) ( 1) ...
12
n
n n n
x x
d y v v
y f x y v y y where v
dx h h
−
+ +
= = =  + +  +  + =
 
 
 
 
 
 
16. Write the formula for trapezoidal and Simpson’s 1/3rd
rules. (April/May 2019)
Trapezoidal rule:
0
0 1 2 1
( ) [( ) 2( ... )]
2
n
x
I n n
x
h
f x dx y y y y y
= −
= + + + + +

Simpson’s 1/3rd
rule:
0
0 2 4 2 1 3 1
( ) [( ) 2( ... ) 4( ... )]
3
n
x
I n n n
x
h
f x dx y y y y y y y y
= − −
= + + + + + + + +

17. What is the order of error in Trapezoidal and Simpson’s 1/3rd
rule?
(April/May 2016, Nov/Dec 2017)
Trapezoidal rule: 2
2
h
error
the
of
Order
;
b
x
a
,
)
(
'
'
y
h
12
)
a
b
(
Error =



−
−
=
Simpson’s rule: 4
iv
4 h
error
the
of
Order
;
b
x
a
,
)
(
y
h
180
)
a
b
(
Error =



−
−
=
18. Why Simpson’s one third rule is called a closed formula?
Since the end point ordinates yo and yn are included in the Simpson’s one third rule, so it is
called closed formula.
19. Given the following data, find
4
x
0
e dx
 by using Simpson’s
1
3
rule with h = 1
: 0 1 2 3 4
: 1 2.72 7.39 20.09 54.60
x
x
y e
=
( ) ( ) ( )
4
0
1
1 54.60 2 7.39 4 2.72 20.09 53.8733
3
x
e dx = + + + + =
 
 

20. Write the Trapezoidal rule to evaluate ( , )
f x y dxdy


Trapezoidal rule for Double Integration is
( )
( )
( )
sumof valuesof f(x,y)at the remainingnodes
sumof valuesof f x,y at the 4corners 2
on theboundary
4 sumof valuesof f(x,y)at the interior nodes
hk
I
4
+
+
 
 
 
 
=  
 
 
 
PART B
1. Using Lagrange’s interpolation formula, find the polynomial f(x) from the following
data
x 0 1 4 5
f(x) 4 3 24 39
Solution:
1 2 3 0 2 3
0 1
0 1 0 2 0 3 1 01 1 2 1 3
0 1 3 0 1 2
2 3
2 0 2 1 2 3 3 0 3 1 3 2
0 1 2 3
0 1
( )( )( ) ( )( )( )
( )
( )( )( ) ( )( )( )
( )( )( ) ( )( )( )
( )( )( ) ( )( )( )
0 1 4 5
4 3
y f x y y
y y
x x x x
y y y
− − − − − −
= = + +
− − − − − −
− − − − − −
+
− − − − − −
= = = =
= =
( ) ( )
( ) ( )
2 3
3 2 3 2
3 2 3 2
2
24 39
4 40 116 80 5 45 100
1
( )
20 4 10 60 50 39 195 156
( ) 2 3 4
y
x x x x x x
f x
x x x x x x
f x x x
= =
 
− − + − + − +
 
=  
− − + + − +
 
 
 = − +
2. Using Lagrange’s interpolation calculate the profit in the year 2000 from the following
data:
Year: 1997 1999 2001 2002
Profit (in Lakhs): 43 65 159 248
Solution:
100
)
2000
(
248
159
65
43
2002
2001
1999
1997
)
)(
)(
(
)
)(
)(
(
)
)(
)(
(
)
)(
)(
(
)
)(
)(
(
)
)(
)(
(
)
)(
)(
(
)
)(
)(
(
)
(
3
2
1
0
3
2
1
0
3
2
3
1
3
0
3
2
1
0
2
3
2
1
2
0
2
3
1
0
1
3
1
2
1
01
1
3
2
0
0
3
0
2
0
1
0
3
2
1
=
=
=
=
=
=
=
=
=
−
−
−
−
−
−
+
−
−
−
−
−
−
+
−
−
−
−
−
−
+
−
−
−
−
−
−
=
=
f
y
y
y
y
x
x
x
x
y
x
x
x
x
x
x
x
x
x
x
x
x
y
x
x
x
x
x
x
x
x
x
x
x
x
y
x
x
x
x
x
x
x
x
x
x
x
x
y
x
x
x
x
x
x
x
x
x
x
x
x
x
f
y
3. Using Lagrange’s inverse interpolation formula, find the value of x when y=0 from the
given data ( ) ( ) ( ) ( )
30 30, 34 13, 38 3, 42 18
f f f f
= − = − = =
Solution:

1 2 3 2 3
1
1 2 3 1 1 2 1 3
0 1 3 0 1 2
2 3
2 2 1 2 3 3 3 1 3 2
( )( )( ) ( )( )( )
( )( )( ) ( )( )( )
( )( )( ) ( )( )( )
( )( )( ) ( )( )( )
o
o
o o o o
o o
x x x
x x
− − − − − −
= + +
− − − − − −
− − − − − −
+ +
− − − − − −
0 1 2 3
0 1 2 3
0 30 13 3 18
30 34 38 42
0.78208 6.53225 33.681818 2.201612
37.2304
y y y y y
x x x x
x
x
= = − = − = =
= = = =
= − + + −
=
4. Using Lagrange’s inverse interpolation formula, find the value of x when y = 13.5 from
the given data
x 93.0 96.2 100.0 104.2 108.7
y 11.38 12.80 14.70 17.07 19.91
Solution:
The Lagrange’s inverse interpolation formula is,
1 2 3 4 0 2 3 4
1
1 2 0 3 0 4 1 0 1 2 1 3 1 4
0 1 3 4 0 1 2 4
2
2 0 2 1 2 3 2 4 3 0
( )( )( )( ) ( )( )( )( )
( )
( )( )( )( ) ( )( )( )( )
( )( )( )( ) ( )( )( )( )
( )( )( )( ) (
− − − − − − − −
= +
− − − − − − − −
− − − − − − − −
+ +
− − − − −
o
o o
y y y y y y y y y y y y y y y y
x y x x
x
y y y y y y y y y y
3
3 1 3 2 3 4
0 1 2 3
4
4 0 4 1 4 2 4 3
)( )( )( )
( )( )( )( )
( )( )( )( )
− − −
− − − −
+
− − − −
x
y y y y y y
y y y y y y y y
x
y y y y y y y y
0 1 2 3 4
0 1 2 3 4
Here x 93.0 ; x 96.2 ; x 100; x 104.2 ; x 108.7 ;
y 13.5; y 11.38; y 12.80 ; y 14.70 ; y 17.07 ; y 19.91
= = = = =
= = = = = =
Therefore x(13.5)=97.65567455
5. Using Newton’s forward interpolation formula, find the cubic polynomial which takes
the following values:
x 0 1 2 3
f(x) 1 2 1 10
Solution:
The forward difference table is given below:
x y y
 2
 y 3
 y
0
1
2
3
1
2
1
10
1
-1
9
-2
10
12

2 3
0 0 0 0
0
0 0
3 2
u(u 1) u(u 1)(u 2)
( ) y
1! 2! 3!
0 ; 1; 1;
( 1) ( 1)( 2)
( ) 1 (1) ( 2) (12)
1 2 6
( ) 2 7 6 1 ( )
4 ( )
(4) 41
u
y x y y y
x x
here x y h u x
h
x x x x x x
y x
y x x x x f x
Put x in f x
f
− − −
= +  +  +  +
−
= = = = =
− − −
= + + − +
 = − + + =
=
 =
6. From the given data, find  at x= 43 and x = 84 (Nov/Dec 2016)
X 40 50 60 70 80 90
 184 204 226 250 276 304
Solution:
x  Δ Δ2
 Δ3
 Δ4

40 184
20
50 204 2
22 0
60 216 2 0
24 0
70 250 2 0
26 0
80 276 2
28
90 304
2 3 0
o o o o
x x
u u(u 1) u(u 1)(u 2) 43 40
(x) ... where u 0.3
1! 2! 3! h 10
(0.3)( 0.7)
184 (0.3)20 (2)
2
(43) 189.79
−
− − − −
 =  +  +   +   + = = =
−
= + +
 =
2 3 n
n n n n
x x
v(v 1) v(v 1)(v 2) 84 90
(x) v where v 0.6
2! 3! h 10
( 0.6)(0.4)
304 ( 0.6)28 (2)
2
(84) 286.96
−
+ + + −
 =  +  +   +   + = = = −
−
= + − +
 =

7. The table gives the distances in nautical miles of the visible horizon for the given heights
in feet above the earth’s surface.
X 100 150 200 250 300 350 400
Y 10.63 13.03 15.04 16.81 18.42 19.90 21.27
Find the values of y when x = 218 ft and 410 ft. (April/ May 2015)
Solution:
X y ∆y ∆2
y ∆3
y ∆4
y
100 10.63
2.40
150 13.03 -0.39
2.01 0.15
200 15.04 -0.24 -0.07
1.77 0.08
250 16.81 -0.16 -0.05
1.61 0.03
300 18.42 -0.13 -0.01
1.48 0.02
350 19.90 -0.11
1.37
400 21.27
0
2 3 4
0 0 0 0 0
choose x 200 (Nearer to 218)
Newton's forward formula :
u u(u -1) u(u -1)(u -2) u(u -1)(u -2)(u -3)
y(x) y y y y y ......
1! 2! 3! 4!
=
= +  +  +  +  +
0
2 3 4
0 0 0 0 0 0
218 200
here 0.36
50
200, 15.04, 1.77, 0.16, 0.03, 0.01
x x
u
h
x y y y y y
− −
= = =
= =  =  = −  =  = −
0.36 0.36(0.36-1) 0.36(0.36-1)(0.36- 2)
(218) 15.04 (1.77) ( 0.16) (0.03)
1 2 6
0.36(0.36-1)(0.36- 2)(0.36-3)
( 0.01)
24
15.696
hence (218) 15.7
y
y
= + + − +
+ −
=
=
n
2 3 4
choose x =400 (Nearer to 410)
Newton's backward formula :
v(v 1) v(v 1)(v 2) v(v 1)(v 2)(v 3)
( ) y .....
1! 2! 3! 4!
n n n n n
v
y x y y y y
+ + + + + +
= +  +  +  +  +

2 3 4
410 400
here 0.2
50
400, 21.27, 1.37, 0.11, 0.02, 0.01
0.2 0.2(0.2 1) 0.2(0.2 1)(0.2 2)
(410) 21.27 (1.37) ( 0.11) (0.02)
1 2 6
0.2(0.2 1)(0.2 2)(0.2 3)
( 0.01) 21.53
24
hence (410)
n
n n n n n n
x x
v
h
x y y y y y
y
y
− −
= = =
= =  =  = −  =  = −
+ + +
= + + − +
+ + +
+ − =
= 21.53
8. From the given data, find the number of students whose weight is between 60 and 70.
Weight in lbs. : 0 – 40 40 – 60 60 – 80 80 – 100 100 - 120
No of students : 250 120 100 70 50
Solution :
x y Cumulative
frequency
y
 2
 y 3
 y 4
 y
Below 40
Below 60
Below 80
Below 100
Below 120
250
120
100
70
50
250
370
470
540
590
120
100
70
50
-20
-30
-20
-10
10
20
2 3 4
0 0 0 0 0
0
u u(u 1) u(u 1)(u 2) u(u 1)(u 2)(u 3)
y(x) y y y y y
1! 2! 3! 4!
x x 70 40
here u 1.5
h 20
hence y(70) 423.5937
now we take x 60
60 40
u 1
20
y(60) 370
− − − − − −
= +  +  +  + 
− −
= = =
=
=
−
= =
=
Therefore, the no.of students whose weight below 60=370
The no.of students whose weight below 70=423.5937
No. Of students whose weight is between 60 &70=423.5937-370=53.5937=54 approximately.
9. From the following values, find f(x) and hence find f (6) by Newton’s divided difference
formula. (Nov/Dec 2017)
x 1 2 7 8
f(x) 1 5 5 4

Solution:
X f(x) ( )
f x
 2
( )
f x
 3
( )
f x

1
2
7
8
1
5
5
4
4
0
-1
-0.6667
-0.1667
0.0714
0 0 0 1 0 1 0 1 2
0 1 2 0 1 2 3
( ) ( ) ( ) ( , ) ( )( ) ( , , )
( )( )( ) ( , , , )
= + − + − −
+ − − − +
0 1 2 3
0
0 1
0 1 2
0 1 2 3
2 3 2
1, 2, 7, 8
( ) 1
( , ) 4
( , , ) 0.6667
( , , , ) 0.0714
( ) 1 ( 1)4 ( 1)( 2)( 0.6667) ( 1)( 2)( 7)(0.0714)
( ) 1 4 ( 3 2)( 0.6667) ( 10 23 14)(0.0714)
( ) 0.071
here x x x x
f x
f x x
f x x x
f x x x x
f x x x x x x x
f x x x x x x x
f x
= = = =
=
=
= −
=
= + − + − − − + − − −
= + + − + − + − + −
= 3 2
4 1.3809 7.6428 5.3333
6,
(6) 6.2381
x x x
Putting x we get
f
− + −
=
=
10. Using Newton’s divided difference formula compute f (2) from the data
x -4 -1 0 5
y 1245 33 5 1335
Solution:
X f(x) ( )
f x
 2
( )
f x
 3
( )
f x

-4
-1
0
5
1245
33
5
1335
-404
-28
266
94
49
-5
0 0 0 1 0 1 0 1 2
0 1 2 0 1 2 3
( ) ( ) ( ) ( , ) ( )( ) ( , , )
( )( )( ) ( , , , )
= + − + − −
+ − − − +

0 1 2 3
0
0 1
0 1 2
0 1 2 3
4, 1, 0, 5
( ) 1245
( , ) 404
( , , ) 94
( , , , ) 5
( ) 1245 ( 4)( 404) ( 4)( 1)(94) ( 4)( 1)( 0)( 5)
2,
(2) 1245 (2 4)( 404) (2 4)(2 1)(94) (2 4)(2 1)(2
here x x x x
f x
f x x
f x x x
f x x x x
f x x x x x x x
Putting x we get
f
= − = − = =
=
= −
=
= −
= + + − + + + + + + − −
=
= + + − + + + + + + )( 5) 333
− =
11. Find the missingg values from the following table
x 0 5 10 15 20 25
y 6 10 - 17 - 31
Solution:
x y Δ Δ2
Δ3
Δ4
0 6
4
5 10 a-14
a-4 41-3a
10 a 27-2a -102+6a+b
17-a -61+3a+b
15 17 -34+b+a 143-4b-4a
b-17 82-3b-a
20 b 48-2b
31-b
25 31
Since only four values are given, we have fourth difference is zero
Therefore Δ4
y = 0
6a+b=102, 4a+b=143
Solving this we get a=13.25, b= 22.5
12. The following table gives the population of a town during the last six censuses. Estimate,
using Newton’s interpolation formula, the increase in the population during the period
1946 to 1948.
Year 1911 1921 1931 1941 1951 1961
Population(in thousands) 12 13 20 27 39 52
Hints:
X y y
 2
 y 3
 y 4
 y 5
 y
1911
1921
1931
1941
12
13
20
27
1
7
7
6
0
5
-6
5
-4
11
-9
-
20

1951
1961
39
52
12
13
1
2 3
0 0 0 0 0
0 0
0
n n(n 1) n(n 1)(n 2)
y(x nh) y y y y ...
1! 2! 3!
here x 1911,h 10 , y 12
To find y(1946)
x nh 1946
i.e.,1911 n.10 1946
i.e., n 3.5
− − −
+ = +  +  +  +
= = =
+ =
+ =
=
0
Therefore, y(1946) 32.18
To find y(1948)
x nh 1948
i.e.,1911 n.10 1948
i.e., n 3.7
Therefore, y(1948) 34.73
Increasein the population during the period1946 to1948
Population in 1948 population in 1946
2.55 Thousands
=
+ =
+ =
=
=
= −
=
13. Find the first derivative of f(x) at x = 0.4 from the following table. (April/May 2018)
x 0.1 0.2 0.3 0.4
f(x) 1.10517 1.2214 1.34986 1.49182
Solution: Since x = 0.4 is the final value of the given table, we use Newton’s backward
formula.
n
2 3
n n n
x x v 0
The Newton 's backward formula for first derivative is given by
dy dy 1 1 1
[ y y y .........)
dx dx h 2 3
= =
   
= =  +  +  +
   
   
x y ∆y ∆2
y ∆3
y
0.1 1.10517
0.11623
0.2 1.2214 0.01223
0.12846 0.00127
0.3 1.34986 0.0135
0.14196
0.4 1.49182

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