0511/Engg/SE/Pre_Pap/2011/AEM_Soln 1

Vidyalankar Vidyalankar Vidyalankar Vidyalankar

S.E. Sem. IV [ETRX]
Advanced Engineering Mathematics
Prelim Question Paper Solutions


Let X : is a random variable which denotes the no. of tosses required to get the first head.
P(H) =
1
2
; P(T) =
1
2


Event x P(x)
H 1
TH 2 .
TTH 3 . .
. . .
. . .
. . .




E(X) = x p(x)
=
2 3
1 1 1
1. 2 3 .............
2 2 2
| | | |
+ + +
| |
\ \

=
2
1 1 1
1 2 3 ..........
2 2 2

| | | |
+ + +
| |
\ \



=
2
1 1
1
2 2

| |

|
\
=
2
1 1
2 2

| |
|
\

=
1
(4)
2

E(X) = 2

To prove that R is on equivalence relation, we have to prove that R is reflexive, symmetric and
transistive.
(i) Reflexive :
x Z,
2x + 5x = 7x
and 7x is divisible by 7
x Rx x Z
R is Reflexive
(ii) Symmetric :
Let x Ry
2x + 5y is divisible by 7

2x 5y
k
7
+
=

7x 5x 7y 2y
k
7
+
=

( ) ( ) 7 x y 5x 2y
k
7
+ +
=
1. (a)

1. (b)

Vidyalankar : S.E. AEM
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 2
( )
2y 5x
k x y m(let)
7
+
= + + =

2y 5x
m
7
+
=
2y + 5x is divisible by 7.
y Rx
R is symmetric.

(iii) Transitive :
Let x Ry and y Rz
2x + 5y is divisible by 7 and 2y + 5z is divisible by 7.

2x 5y 2y 5z
m and n
7 7
+ +
= =

2x 7y 5z
m n
7
+ +
= +

2x 5z
y m n
7
+
+ = +
( )
2x 5z
m n y p let
7
+
= + =

2x 5z
p
7
+
=
2x + 5z is divisible by 7.
(x, z) R
R is transitive
R is an equivalence relation.

(i) Type I : When H
O
is true and we decide not to reject it or when H
O
is not true and we decide
to reject it, the decision is correct.
Type II : If we decide to reject H
O
when it is true or if we decide not to reject H
O
when it is
not true, the decision is not correct.

(ii) Critical Region : A critical region for a test is the region that corresponds to the subset of sample
space for which the null hypothesis H
O
is to be rejected if a sample point falls in the region.
Level of significance :
An upper bound on the probability of Type I error for a test is known as its level of significance,
denoted by (0 1).
Hence, level of significance = P (Type I error) if H
O
is simple
= Max P (Type I error) if H
O
is composite

Given : n = 6, N = 700
Prob (5 or 6) =
1 1
6 6
+ =
2
6
=
1
3

p =
1
(say)
3

p + q = 1
q =
2
3

P(x) =
n x n x
x
C (p) (q)

where x = 0, 1, 2, 3,
=
x 6 x
6
x
1 2
C
3 3

| | | |
| |
\ \
where x = 0, 1, 2, 3, 4, 5, 6
1. (c)

1. (d)

Prelim Question Paper Solutions
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 3
P(x 3) = p(3) + p(4) + p(5) + p(6)
= 1 [p(0) p(1) + p(2)]
=
0 6 1 5 2 4
6 6 6
0 1 2
1 2 1 2 1 2
1 C C C
3 3 3 3 3 3

| | | | | | | | | | | |
+ +
| | | | | |
\ \ \ \ \ \

=
6 5 2 4
2 1 2 1 2
1 1 6 15
3 3 3 3 3

| | | || | | | | |
+ +
| | | | |
\ \ \ \ \

P(x 3) =
233
729


Frequency Distribution
f(x) = N. P(x)
=
233
729
729
| |
|
\

f(x) = 233

Proof :
Let A be the square matrix. be the eigen value of A and X the corresponding eigen vector
of A.
Then, by definition
AX = X ... (1)
Premultiplying by (Adj A)
(Adj A) AX = (Adj A) X
[(Adj A)A]X = (Adj A) X
[ | A | I ] X = (Adj A) X [(Adj A) A = | A | I ]
| A | X = (Adj A) X

| A|

X = (Adj A) X [ 0]
or (adj A)X =
| A|

X.

| A|

is the eigen value of (adj A).

The characteristic equation is |A I | = 0

y y
y y


= 0

2
2 y = 0
= 0, = 2y
Now when dividend = e

is divided by divisor =
2
2y
let the quotient be () and the remainder be a + b.
dividend = (quotient divisor) + remainder
e

= () (
2
2y) + a + b
put = 0 e
0
= (0) (0) + b
b = 1
put = 2y e
2y
= (2y) (0) + a (2y) + b
e
2y
= a (2y) + 1
a =
2y
e 1
2 y

1. (e)

2. (a)

Vidyalankar : S.E. AEM
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 4
e

= () (
2
2y) +
2y
e 1
2 y
| |

|
|
\
+ 1
e
A
= (A) (A
2
2Ay) +
2y
e 1
2 y
| |

|
|
\
A + I
e
A
=
2y
e 1
2 y
| |

|
|
\
y y
y y



+
1 0
0 1




=
2y 2y
2y 2y
e 1 e 1
1
2 2
e 1 e 1
1
2 2


+




+

=
2y 2y
2y 2y
e 1 e 1
2 2
e 1 e 1
2 2

+



+



= e
y

y y y y
y y y y
e e e e
2 2
e e e e
2 2



+



+


= e
y

cosh y sinh y
sinh y cosh y





Given : u = 662 , = 32, N = 1000
(i) (X > 700)
z =
Xu

z =
X 662
32

(X > 700) =
X 662 700 662
32 32
| |
>
|
\

= (z > 1.19)
= 0.5 P(0 < z < 1.19)
= 0.5 0.3830
(X > 700) = 0.117
No. of acre plots = 1000 0.117
n = 117
(ii) (x < 650) =
x 662 650 662
32 32
| |
<
|
\

= (z < 0.380)
= 0.5 P(0.380 < z < 0)
= 0.5 P(0 < z < 0.380)
= 0.5 0.1480
(x < 650) = 0.352
No. of acre plots = 1000 0.352
n = 352
(iii) Let the lowest yield of best 100 plots be x
1
kg
When X = x
1

z
1
=
1
x 662
32

P(X > x
1
) =
100
1000
= 0.1
P(X > x
1
) = 0.1
1.19
0.380
0.1
2. (b)

Prelim Question Paper Solutions
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 5

1
x 662 X 662
32 32
| |
>
|
\
= 0.1
P(z > z
1
) = 0.1
P(0< z < z
1
) = 0.4 (0.5 0.1)
z
1
= 1.28
1.28 =
1
x 662
32

x
1
= 702.96 kg

z
5
= {0, 1, 2, 3, 4}

(i) The composition table w.r.t. +
5
.
+
5
0 1 2 3 4
0 0 1 2 3 4
1 1 1 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3


All the elements belong to set z
5
in composition table.
(z
5
, +
5
) is an abelian group.

(ii) The composition table w.r.t. x
5
.

5
0 1 2 3 4
0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1

From the composition table
(z
5
,
5
) is a semigroup.

(iii) Distributive Law :
Now, we find, a, b, c z
5

a
5
(b +
5
c) = (a
5
b) +
5
(a
5
c)
2
5
(3 +
5
4) = (2
5
3) +
5
(2
5
4)
2
5
2 = 1 +
5
3
4 = 4
From table (i) and (ii)

5
is distributive over +
5
.
Hence (z
5
, +
5
,
5
) is a ring.

(i) f(x) =
2
x 4x
1. OneOne :

1
f (x ) =
2
f (x )

2
1 1
x 4x =
2
2 2
x 4x

1 1
x (x 4) =
2 2
x (x 4)
x
1
= 0 and x
2
= 4
x
1
x
2
F is not oneone (1)
2. (c)

3. (a)

Vidyalankar : S.E. AEM
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 6
2. Onto :
y = f(x)
y =
2
x 4x

2
x 4x y = 0
x =
2
4 16 4y
2


=
2
2 4 y
3 R (co domain)

But x = 2 4 9
= 2 5i R (domain)
f is not onto (2)
From (1) & (2)
f is not a bijective function.

(ii) Let
a 0 c 0 e 0
A , B , C
0 b 0 d 0 f

= = =



1. Closureness :
Consider,
A B =
a 0 c 0
0 b 0 d




A B =
ac 0
0 bd




A B R ; a, b, c, d R
G is closed under multiplication.
2. Associativity :
Multiplication is associative in
2 2
M (R)

and R is a of
2 2
M (R)

i.e. A (B C) = (A B) C a, b, c, d R
Multiplication is always associative in G.
3. Identify element :
I =
1 0
R
0 1

| I | 0 is the identity element with respect to multiplication.
A I = I A a, b R
4. Inverse element :
For any A =
1
1
a 0
R
0 b

( | A | 0)
Inverse exist
A
1
=
Adj A
| A|
=
1
1 1 1
b 0
1
0 a a b



=
1
1
1
0
a
R,
1
0
b

A R
G is a Group under multiplication.

Prelim Question Paper Solutions
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 7

2
(z 2z 1) 4 + + + = 0

2 2
(z 1) (2i) + = 0
z + 1 + 2i = 0 , z + 1 2i = 0
z = 1 2i , z = 1 + 2i are simple poles.

(i) Since both pole lie outside the circle z 1 = .
By Cauchys Integral Theorem

2
C
z 4
. dz 0
z 2z 5
+
=
+ +

where C is z 1 = .
(ii) The centre C of the circle z 1 i 2 + + = is 1i and radius is 2. If A is z = 12i then
CA = ( ) ( ) 1 2i 1 i i 1 2 = = <
Hence, A lies inside the circle.
Residue (at z = 12i)
=
( ) z 1 2i
z 4
lim
z 1 2i

+
+

=
1 2i 4 3 2i
1 2i 1 2i 4i
+
=
+


2
C
z 4 3 2i i
dz 2 i
4i z 2z 5
+ 3+ 2 | |
= =
|

2 \ + +



2
C
z 4 i
dz
z 2z 5
+ 3+ 2
=

2 + +



(iii) The centre C of the circle z 1 i 2 is 1 i + = + and
radius is 2. If B is z = 1 + 2i then
CB = 1 2i 1 i i 1 2 + + = = <
Hence, B lies inside the circle
Residue (at z = 1 + 2i)
=
( ) z 1 2i
z 4
lim
z 1 2i
+
+
+ +

=
1 2i 4 3 2i
1 2i 1 2i 4i
+ + +
=
+ + +


2
C
z 4 3 2i i
. dz 2 i
4i z 2z 5
+ + 3+ 2 | |
= =
|

2 \ + +



2
C
z 4 i
. dz
z 2z 5
+ 3+ 2
=

2 + +



The characteristics matrix is
A I =
6 2 2 1 0 0
2 3 1 0 1 0
2 1 3 0 0 1






=
6 2 2
2 3 1
2 1 3







The characteristic polynomial is
| A I | =
6 2 2
2 3 1
2 1 3







=
3 3 2
( 1) 12 36 32 + +
3. (c)

C
A
1i
12i
1+ 2i B
y
x
0
y
B
C
1+2i
1+i
x
0
A
12i
3. (b)

Vidyalankar : S.E. AEM
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 8
Characteristic equation is

2 2
12 36 32 0 + + = or
3 2
12 36 32 0 + =
on factorization we get
1
= 8 and
2, 3
= 2, 2 are the eigen values.
The characteristic vector has to satisfy the equation (A I)X = 0 which for the given matrix A is

1
2
3
6 2 2 x
2 3 1 x
2 1 3 x






=
0
0
0






Corresponding to
1
= 8 the eigen vector X
1
is given by

1
2
3
2 2 2 x
2 5 1 x
2 1 5 x






=
0
0
0







3 3 2
R R R + =
2 2 2
2 5 1
0 6 6








2 2 1
R R R + =
2 2 2
0 3 3
0 6 6








3 3 2
R R 2R =
2 2 2
0 3 3
0 0 0








2 2
1
R R
3
=
2 2 2
0 1 1
0 0 0







2 1
1
R R
2
=
1 1 1
0 1 1
0 0 0







1 2 3
x x x + = 0

2 3
x x + = 0

1
x
1 1
1 1

=
2
x
1 1
0 1

=
3
x
1 1
0 1



1
x
2

=
2
x
1

=
3
x
1


1
X =
2
1
1






Corresponding to
2, 3
= 2 the eigen vector X
2
and X
3
is given by

1
2
3
4 2 2 x
2 1 1 x
2 1 1 x






=
0
0
0







3 3 2
R R R + gives
4 2 2
2 1 1
2 1 1







Prelim Question Paper Solutions
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 9
Also
1
2 2
R
R R
2
+
4 2 2
0 0 0
0 0 0







2 1
1
R R
2

2 1 1
0 0 0
0 0 0







1 2 3
2x x x + = 0
and
1
X =
2
1
1






X
2
=
0
1
1






X
3
=
1
0
2







3
A 2I + =
6 2 2 6 2 2 6 2 2 1 0 0
2 3 1 2 3 1 2 3 1 2 0 1 0
2 1 3 2 1 3 2 1 3 0 0 1


+




=
344 168 168 2 0 0
168 92 84 0 2 0
168 84 92 0 0 2


+




=
3 2
534 (1780 4300 4300) 51400 + + + +
Eigen values of
3
A 2I + are

1
514 = and
2 3
10 = =


1
n = 40
2
n = 40

1
x = 42
2
x = 45

1
= 8
2
= 7
Let,
H
0
= u
A
= u
B

H
1
= u
A
u
B
(Two tailed)

LOS = 0.01
z
cal
=
1 2
2 2
1 2
1 2
x x
n n

+
=
42 45
64 49
40

+

| z
cal
| = 1.78
| z
tab
| at LOS = 0.01 is
| z
tab
| = 2.575
| z
cal
| < | z
tab
|
H
0
is accepted.

Hence there is no significant difference between sales with the two types of displays.
4. (a)

Vidyalankar : S.E. AEM
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 10
D
42
= {1, 2, 3, 6, 7, 14, 21, 42}
D
30
= {1, 2, 3, 5, 6, 10, 15, 30}








(i) f : (D
42
, |) (D
30
, |)
f(1) 1
f(2) 2
f(3) 3
f(6) 6
f(7) 5
f(14) 10
f(21) 15
f(42) 30
Clearly f is oneone and onto

(ii) f(a b) = f(a) f(b)
Take a = 7 and b = 21
f(7 21) = f(7) f(21)
L.H.S. = f(7 21) = f(21) = 15
R.H.S. = f(7) f(21)
= 5 15
= 15
L.H.S. = R.H.S.
Thus, f(a b) = f(a) f(b)
Also, f(a b) = f(a) f(b) a, b D
42
and D
30

Hence, D
42
is isomorphic to D
30
under divisibility.

In degrees : Let R is a relation on A and a A then the indegree of a A is the number of b
such that b R a i.e. number of b | (b, a) R

Out degrees : Let R is a relation on A and a A then the outdegree of a A is the number of
b such that a R b i.e. number of b | (a, b) R.

R
a b c d
a 1 0 0 0
b 0 0 1 0
M
c 0 1 0 1
d 1 0 1 0



=





Here R = {(a, b) | a R b}
R = {(a, a), (b, c)(c, b), (c, d), (d, a), (d, c)}








10
2
1
3
5
6
15
30 42
14
2
1
3
7
6 21
(D
30
, |) (D
42
, |)
a
b
b
c
d

a b c d
Indegrees 1 1 2 1
Outdegrees 1 1 2 2

4. (b)

4. (c)

Prelim Question Paper Solutions
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 11
Given Mean = 2
Variance = 3
Mean = m = 2
E(X) = mean = 2
By formula
E(X
2
) = m
2
+ m = 4 + 2
E(X
2
) = 6
and Variance V(X) =
2 2
E(X ) [E(X)]
=
2
6 (2) = 6 4
V(X) = 2
The statement given is false.

(i) x (x y) x (by definition) (1)
x (x y) (by definition) and x x (by reflexive)
x x (x y) (2) z x, z y z x y
From (1) & (2)
x (x y) = x (by antisymmetric relation)

(ii) x x (x y) (1)
x x (by reflexive) x y x (by definition)
x (x y) x (2) x z, y z x y z
From (1) & (2)
x (y x) = x ( is antisymmetric)

Raw moments about origin :

1
u =
3
0
x f (x)dx

=
3
2 2
0
4x (9 x )
.dx
81

=
3
2 4
0
4
(9x x ) dx
81

=
3 3
3 5
0 0
4 x x
9
81 3 5

| | | |

| |
\ \

=
4 27 243
9
81 3 5
| |

|

\



1
u =
8
5


2
u =
3
2
0
x .f (x) dx

=
3
3 5
0
4
(9x x ) dx
81

=
3 3
4 6
0 0
4 x x
9
81 4 6

| | | |

| |
\ \



2
u = 3

3
u =
3
3
0
x .f (x) dx

=
3
2
3
0
4x{9 x }
x dx
81
| |

|
\

=
3
4 6
0
4
(9x x ) dx
81

=
3 3
5 7
0 0
4 x x
9
81 5 7

| | | |

| |
\ \

=
216
35

4. (d)

5. (b)

5. (a)

Vidyalankar : S.E. AEM
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 12

4
u =
3
4
0
x .f (x) dx

=
3
5 7
0
4
(9x x ) dx
81

=
3 3
6 8
0 0
4 x x
9
81 6 8

| | | |

| |
\ \



4
u =
27
2

Row moments (origin) =
8 216 27
, 3, ,
5 35 2

Central moments about mean :

1
u = 0

2
u =
2
2 1
( ) u u

2
u =
11
25


3
u =
3 2 1 1
3 2( )
3
u u u + u
=
3
216 8 8
3 3 2
35 5 5
| | | |
+
| |
\ \


3
u = 0.0365

4
u =
4 3 1 2 1 1
4 6 3
2 4
u u u + u (u ) (u )
=
2 4
27 216 8 8 8
4 6(3) 3
2 35 5 5 5
| || | | | | |
+
| | | |
\ \ \ \


4
u = 0.422
Central moments (mean) =
11
0, , 0.0365, 0.422
25


x : 0 1 2 3 4 5
f
0
: 2 20 48 70 46 6

Mean =
f . x
f

=
0 20 96 210 184 30
192
+ + + + +
,where N = f = 192
=
540
192


Mean = np = 2.8125
n = 5
5p = 2.8125
p = 0.435
q = 0.565

Expected frequency (f
e
) :
f(x) = N. P(x)
=
n x n x
x
N. C p q

; x = 0, 1, 2, 3, 4, 5,
N = f

= 192
f(x) =
5 x 5 x
x
192 C (0.435) (0.565)

; x = 0, 1, 2, 3, 4, 5

5. (c)

Prelim Question Paper Solutions
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 13
x f(x)
0
f(0) =
5 0 5
0
192 C (0.435) (0.565)
= 11.05 11
1
f(1) =
5 1 4
1
192 C (0.435) (0.565)
= 42.55 43
2
f(2) =
5 2 3
2
192 C (0.435) (0.565)
= 65.52 66
3
f(3) =
5 3 2
3
192 C (0.435) (0.565)
= 50.45 50
4
f(4) =
5 4 1
4
192 C (0.435) (0.565)
= 19.42 19
5
f(5) =
5 5 0
5
192 C (0.435) (0.565)
= 2.99 3
Total 192

Total
x : 0 1 2 3 4 5
f
0
: 2 20 48 70 46 6 192
f
e
: 11 43 66 50 19 3 192



2
2
0
sin
.d
5 4cos

Let z =
2 2
i
dz z 1 z 1
e ; d ; cos ; sin
iz 2z 2iz

+
= = =
I =
2
2
2
C
1
z
dz 2iz
.
iz
z 1
5 4
2z
| |

|
\
| |
+
+
|
\

=
2 2
2
2
C
(z 1) dz
.
iz
10z 4z 4
( 4z )
2z

| |
+ +

|
\

=
2 2
2 2
C
1 (z 1)
. dz
2i
(10z 4z 4) z

+ +

=
2 2
2 2
C
1 (z 1)
.dz
2i
z (4z 10z 4)

+ +

where C is the circle | z | = 1
z
1
= 0.5 and z
2
= 2
z
1
lies inside the circle | z | = 1
Residue of f (z) (at z = 0)
=
2 2
2
2 2
z 0
d (z 1)
lim z .
dz z (4z 10z 4)

+ +


=
2 2
2
z 0
d (z 1)
lim
dz 4z 10z 4

+ +


=
2 2 2 2
2 2
z 0
(4z 10z 4).2(z 1).2z (z 1) (8z 10)
lim
(4z 10z 4)

+ + +
+ +

=
10
16


2
4z 10z 4 + + = 4(z + 0.5) (z + 2)

6. (a)

Vidyalankar : S.E. AEM
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 14
Residue at f(z) at z = 0.5 is
=
2 2
2
z 0.5
(z 1)
lim (z 0.5) .
z .4(z 0.5)(z 2)

+
+ +

=
2 2
2
z 0.5
(z 1)
lim
4.z (z 2)

+
=
1 0.5625
4 0.375
| |
|
\

=
6
16

I =
1 10 6
2 i
2i 16 16
| |
+
|

\

=
1 4
2 i
2i 16
| |

|

\

I =
4

L = {1, 2, 3, 4, 6, 24} under divisible by

Elements : 1 2 3 4 6 24
Complements : 24 NC 4 3 NC 1






Complement of 2 and 6 does not exist. Hence it is not a complemented lattice.
Complimented Lattice :
A lattice L is called complemented if it is bounded and every element of L has a complement.

Bounded Lattice :
A lattice L is said to be bounded if it has a greatest element 1 and a least element 0.

Proof :
Consider the function
f (z)
z a
which is analytic at every point within C except at z = a.
Draw a circle C with centre at z = a and radius 'r' such that C/lies entirely inside C.
Then
f (z)
z a
is analytic in the region between C and C ( z = a does not lie in this region).

By Cauchy's Integral Theorem, we have

'
C
C
f (z) f (z)
dz dz
z a z a
=


(1)
Now, the equation of circle C is | z a | = r
any point on this circle is given by
z a = r e
i


dz = i r e
i

d
Limits for : = 0 to = 2
Then,
C'
f (z)
dz
z a

=
2 i
i
i
0
f (a re )
ire d
re

=
2
i
0
i f (a re )d

4
2
1
3
6
24
C
C'
r
a
6. (b)

6. (c)

Prelim Question Paper Solutions
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 15
In the limiting case as the circle C shrinks to the point 'a', then , r 0
Hence,
C'
f (z)
dz
z a

= i
2
0
f (a) d

= i f(a)
2
0
d

= i f(a) [2] = 2i f(a)
Substituting in (1) we get,

C
f (z)
dz
z a

= 2if(a)
f (a) =
C
1 f (z)
2 i (z a)

dz

Mean =
f . x
f

=
0 59 28 9 4
200
+ + + +

= 0.5 ,where N = f = 200
= 0.5
Expected frequency
f(x) = N . P(x)
=
x
e .
200
x!

| |

|
\

f(x) =
0.5 x
e .(0.5)
200.
x!

=
x
(121.3)(0.5)
x!
where x = 0, 1, 2, 3, 4

Total
x : 0 1 2 3 4
O
i
: 123 59 14 3 1 200
E
i
: 121 60 15 3 1 200

All the expected frequencies must be greater than 5. Therefore the above table becomes


x

O
i


E
i

2
cal
x =
2 n
i i
i i 1
(O E )
E
=

0 123 121 0.033
1 59 60 0.016
2 or 3 or 4 18 19 0.067

Test of goodness of Fit :
H
0
: Poisson fit is good
H
i
: Poisson fit is not good
LOS = 5% (let)

2
cal
x =
2 n
i i
i i 1
(O E )
E
=

= 0.033 + 0.016 + 0.067 + 0

2
cal
x = 0.116
d.f. = k s 1 where, k = no. of new classes = 3
= 3 1 1 s = no. of parameters = 1
d.f. = 1

7. (a)

Vidyalankar : S.E. AEM
0511/Engg/SE/Pre_Pap/2011/AEM_Soln 16

2
tab
x (0.05, 1 df) = 3.84

2 2
cal tab
x x <
H
0
is accepted.
Hence poisson fit is good.

H
0
: There is association between darkness of eye colour in father and son
H
i
: There is no association between darkness of eye colour in father and son
LOS = 1%

Dark Not Dark Total
Dark 48 90 138
Not Dark 80 782 862
Total 128 872 1000

d.f. = (2 1) (2 1) = 1

2
cal
x =
2
N(ad bc)
(a b)(a c)(b d)(c d)

+ + + +
=
2
1000(48 782 90 80)
(48 90)(48 80)(90 782)(80 782)

+ + + +


2
cal
x = 69.311

2
tab(0.05,1df )
x = 3.84

2 2
cal tab
x x >
Reject H
0
.
Hence there is no association between darkness of eye colour in father and son.

Type I Type II
n
1
= 8 n
2
= 7
1
x 1234 =
2
x 1036 =
S
1
= 36 S
2
= 40

Let, H
0
: Type I = Type II
H
1
: Type I > Type II (one tailed test)
LOS = 5%
S
2
=
2 2
1 1 2 2
1 2
n s n s
n n 2
+
+
=
2 2
8(36) 7(40)
8 7 2
+
+

S
2
= 1659.07
S = 40.73
t
cal
=
1 2
1 2
x x
1 1
S
n n

+
=
1234 1036
1 1
40.73
8 7

+

| t
cal
| = 9.39
d.f. =
1 2
n n 2 +
d.f. = 13
t
(0.05, 13df)
= 1.771
| t
observed
| = 1.771
| t
cal
| > | t
obs
|
Reject H
0

Type I is superior to Type II.


7. (c)

7. (b)