YATES' CORRECTION
when the number of of freedom-1 (2xI table),
degrees
can get a better approximation for 2 test Statistic
(10i-Eil-0 5)2 .
XYates :
E -
Ei
e .
g response to a question :
test at 5 %
/Agree
Disagree
mum level if
CH-SQUARED
um
TESTS
PART 2 -
Female 53
there is any
association
Ho :
no association between gender & response to survey
Hi there is an association between gender &
response to survey
36
Ei : A
[
0 5163 841
.
.
2
do not reject Ho not sufficient
5
,
evidence to suggest an association 29151-0
XYates
.
=
a response to survey
between gender 48 025 .
5) 9751-0 5)
&
+ 129751-0 .
+
112 .
.
50 025
.
64 975 .
= 0 1657 . + 0 1276
.
+ 0 1225
. + 0 0943
.
XYates =
0 5 .
, combining discreted
EXPONENTIAL DISTRIBUTION if X & independant random
Y
f
continuous RV are
number of successes in a unit interval of time ,
X RV X can take values OX11 ,
1 or 2 variables following a normal distan
written as
X-exp(x) z aX by
f(x)
OxXI has pat =
KX = + + C
·
events must occur randomly &d independantly E(x) =
Sozx + ( z)
+
then z follows a normal dist.
·
mean rate events occur is constant K
E(z) =
aE(X) + bE(Y) + 2
pdf :
var(z) = a"var(x) + b2 var(y)
S
*x
Xe for X, 0
f(x) =
+ 2k + k 1 =
X N(5 4) & Y -N(12 26)
-
g Otherwise ,
,
of times 3 5k = 1
2x + 2y + 4
no
.
2/7 z
Thime
=
↓ CX : period k =
E(x) =
+ var(x) =
x find P(zc27)
NTNOs-
Po : no of
in
ifincome E(z)
times
VaN(z) 4x4 %
asman 2x12 + 4
"
(t)
** 2x5 +
: +
j8xxe
:
S. xxe var(x)
**
E(X)
= -
=
NORMAL DIST, = 38 = 120
C do not times hunkin
IBP
PBI twice want of z-N(38 120) ,
var(x)
Le P(z(27) 0 1577
Exe ] ? S: e
**
Exe ** e* * = .
* -
measurement likely to be within
*
rectangular distribution
Exe He ) ***e*) - n
*
:
** O .
Skg of true value
↓
-
large number of measurements of
a(X(b
f(x)
~ O
a =
b a
different object
O m
42kg
-
rounding
=
all to
Exe (1)
** *
-
Ye Y) -
= + -
are made & true value noted
↑
,
I Theight
f(x)
=
0 + x =
" al prob of 42kg value more than 0 3 kg .
away from 42
sneeze 2 times every hour
211 number of leaks
follows poisson dist
in 2 miles
mean > X tre mass
=
d b F(x) =
Find prob of going over 90 mins
prob first leak in first X= rec dist [41 5
.
,
42 . 5]
after waking up without sneezing
.
Sal
half mile
P(X(41 7) P(x42 3)
A
+
T time after waking up before sneezing X 3/2 X dis until first leak E(X) var(x)
. .
= : = =
=
S425
41 7
Tw exp(2)
Se Su it41 5
.
: =X
P(OcX0 5) +
-(
: =
Xa
.
1- PCTc1 5)
.
P(T71 5) = .
75
. c 42 3 .
E(X) = var(x) =
1-2-0
.
0 528 45
.
9:2e-2x
= :
1
=
-[a]-
.
= -
= 1(1 -
-
-
e
2x .
5)
if 5 mins since last call ,
b) O of
true value
diff between quoted & ·
Las]"
e
var(x) - ai a -
=
Find prob that next call var(42 -x) =
< 10mins later
Poisson dist mean s calls : e I (42 5 41 5)2
a
-
=
-Hub) b)
. .
per 10 mins .
unchanged
-
-
12
after a call find prob that P(x > 10) 1 - P(X(10) exp unaffected by
-
:
↳
12
/12
,
next call more than 10 mins time since last event =
a+ b
1-Soetox /es
(b)
: -
later =
:
-
↳
memorylessness
x =
7 O
2 =
when the number of of freedom-1 (2xI table),
degrees
can get a better approximation for 2 test Statistic
(10i-Eil-0 5)2 .
XYates :
E -
Ei
e .
g response to a question :
test at 5 %
/Agree
Disagree
mum level if
CH-SQUARED
um
TESTS
PART 2 -
Female 53
there is any
association
Ho :
no association between gender & response to survey
Hi there is an association between gender &
response to survey
36
Ei : A
[
0 5163 841
.
.
2
do not reject Ho not sufficient
5
,
evidence to suggest an association 29151-0
XYates
.
=
a response to survey
between gender 48 025 .
5) 9751-0 5)
&
+ 129751-0 .
+
112 .
.
50 025
.
64 975 .
= 0 1657 . + 0 1276
.
+ 0 1225
. + 0 0943
.
XYates =
0 5 .
, combining discreted
EXPONENTIAL DISTRIBUTION if X & independant random
Y
f
continuous RV are
number of successes in a unit interval of time ,
X RV X can take values OX11 ,
1 or 2 variables following a normal distan
written as
X-exp(x) z aX by
f(x)
OxXI has pat =
KX = + + C
·
events must occur randomly &d independantly E(x) =
Sozx + ( z)
+
then z follows a normal dist.
·
mean rate events occur is constant K
E(z) =
aE(X) + bE(Y) + 2
pdf :
var(z) = a"var(x) + b2 var(y)
S
*x
Xe for X, 0
f(x) =
+ 2k + k 1 =
X N(5 4) & Y -N(12 26)
-
g Otherwise ,
,
of times 3 5k = 1
2x + 2y + 4
no
.
2/7 z
Thime
=
↓ CX : period k =
E(x) =
+ var(x) =
x find P(zc27)
NTNOs-
Po : no of
in
ifincome E(z)
times
VaN(z) 4x4 %
asman 2x12 + 4
"
(t)
** 2x5 +
: +
j8xxe
:
S. xxe var(x)
**
E(X)
= -
=
NORMAL DIST, = 38 = 120
C do not times hunkin
IBP
PBI twice want of z-N(38 120) ,
var(x)
Le P(z(27) 0 1577
Exe ] ? S: e
**
Exe ** e* * = .
* -
measurement likely to be within
*
rectangular distribution
Exe He ) ***e*) - n
*
:
** O .
Skg of true value
↓
-
large number of measurements of
a(X(b
f(x)
~ O
a =
b a
different object
O m
42kg
-
rounding
=
all to
Exe (1)
** *
-
Ye Y) -
= + -
are made & true value noted
↑
,
I Theight
f(x)
=
0 + x =
" al prob of 42kg value more than 0 3 kg .
away from 42
sneeze 2 times every hour
211 number of leaks
follows poisson dist
in 2 miles
mean > X tre mass
=
d b F(x) =
Find prob of going over 90 mins
prob first leak in first X= rec dist [41 5
.
,
42 . 5]
after waking up without sneezing
.
Sal
half mile
P(X(41 7) P(x42 3)
A
+
T time after waking up before sneezing X 3/2 X dis until first leak E(X) var(x)
. .
= : = =
=
S425
41 7
Tw exp(2)
Se Su it41 5
.
: =X
P(OcX0 5) +
-(
: =
Xa
.
1- PCTc1 5)
.
P(T71 5) = .
75
. c 42 3 .
E(X) = var(x) =
1-2-0
.
0 528 45
.
9:2e-2x
= :
1
=
-[a]-
.
= -
= 1(1 -
-
-
e
2x .
5)
if 5 mins since last call ,
b) O of
true value
diff between quoted & ·
Las]"
e
var(x) - ai a -
=
Find prob that next call var(42 -x) =
< 10mins later
Poisson dist mean s calls : e I (42 5 41 5)2
a
-
=
-Hub) b)
. .
per 10 mins .
unchanged
-
-
12
after a call find prob that P(x > 10) 1 - P(X(10) exp unaffected by
-
:
↳
12
/12
,
next call more than 10 mins time since last event =
a+ b
1-Soetox /es
(b)
: -
later =
:
-
↳
memorylessness
x =
7 O
2 =
