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Chapter 8. 81. Confidence Intervals and Sample Size. Outline. 82. 81 Introduction 82 Confidence Intervals for the Mean [ Known or n 30] and Sample Size 83 Confidence Intervals for the Mean [ Unknown and n 30]. Outline. 83.
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Chapter 8
81
Confidence Intervals and Sample Size
82
83
84
85
86
A point estimate is a specific numerical value estimate of a parameter. The best estimate of the population mean is thesample mean .
X
87
88
89
810
811
812
813
s
s
<
<
z
z
m
æ
ö
æ
ö
+

X
X
ç
÷
ç
÷
n
n
ç
÷
ç
÷
a
/
2
a
/
2
ç
÷
ç
÷
ç
÷
ç
÷
è
ø
è
ø
814
815
Since
the
95%
confidence
interval
=
is
desired
,
z
1
.
96
.
Hence
,
a
2
substituti
ng
in
the
formula
s
s
æ
ö
æ
ö
<
m
<
ç
÷
ç
÷
X
–
z
X
+
z
è
ø
è
ø
a
a
n
n
2
2
one
gets
816
817
2
2
23
.
2
(1.96)
(
)
23.2
(1.96)
(
)
50
50
23
.
2
0
.
6
23
.
6
0
.
6
22
.
6
23
.
8
or 23.2 0.6 years.
Hence
,
the
president
can
say
,
with
95%
confidence
,
that
the
average
age
of
the
students
is
between
22
.
6
and
23
.
8
years
,
based
on
50
students
.
818
Since
the
99%
confidence
interval
is
desired
,
z
2
.
58
.
Hence
,
2
substituti
ng
in
the
formula
X
–
z
X
+
z
n
n
2
2
one
gets
819
820
5
5
104
(2.58)
.
(
)
104
(
)
(2.58)
30
30
104
2
.
4
104
2
.
4
101
.
6
106
.
4
.
Hence
,
one
can
say
,
with
99%
confidence
,
that
the
average
pulse
rate
is
between
101
.
6
and
106.4
beats per minute, based on 30 users.
821
×
2
z
æ
ö
s
=
ç
÷
n
ç
a
/
2
÷
ç
÷
E
ç
÷
ç
÷
ç
÷
è
ø
w
h
e
r
e
E
i
s
t
h
e
m
a
x
i
m
u
m
e
r
r
o
r
o
f
e
s
t
i
m
a
t
e
.
I
f
n
e
c
e
s
s
a
r
y
,
r
o
u
n
d
t
h
e
a
n
s
w
e
r
u
p
t
o
o
b
t
a
i
n
a
w
h
o
l
e
n
u
m
b
e
r
.
822
a
Since
=
0
.
01
(
or
1
–
0
.
99
),
z
=
2
.
58
,
and
E
=
1
,
substituti
ng
a
2
×
s
2
z
æ
ö
=
in
n
gives
ç
÷
a
2
è
ø
E
2
æ
(
2
.
58
)(
3
)
ö
=
»
ç
÷
n
=
59
.
9
60
.
è
ø
1
823
824
825
826
827
828
829
830
Thus
the
95%
confidence
interval
of
the
population
mean
is
found
by
substituti
ng
in
s
s
X
t
X
t
n
n
2
2
0.08
0.08
0.32
–
(2.262)
0
.
32
(2.262)
10
10
0
.
26
0
.
38
831
Symbols
Used
in
Proportion
Notation
p
population
proportion
p
(
read “p
hat”
)
sample
proportion
$
X
n
X
p
and
q
or
1
–
p
$
$
$
n
n
where
X
number
of
sample
units
that
possess
the
characteri
stic
of
interest
and
n
sample
size
.
832
ˆ
p
Since
X
54
and
n
150
,
then
X
54
p
=
=
0.36 = 36
%
$
n
150
n
X
150
54
and
q
=
$
n
150
0
.
64
64%
or
q
=
1
–
p
1
0
.
36
0
.
64
.
$
$
833
96
=
150
p
q
$
$
)
)
(z
(z
a
a
n
2
2
834
p
q
$
$

<
<
+
p
p
p
$
$
n
835
ˆ
p
p
q
z
z
2
2
836
Substituti
ng
in
p
q
$
$
$
$
$
p
p
p
$
n
n
we
get
(
0
.
12
)(
0
.
88
)
Lower limit
0
.
12
(
1
.
65
)
=
0
.
096
500
(
0
.
12
)(
0
.
88
)
Upper limit
0
.
12
(
1
.
65
)
=
0
.
144
500
Thus
,
0.096
<
p
<
0.144 or 9.6% < p < 14.4%.
837
w
h
e
r
e
E
i
s
t
h
e
m
a
x
i
m
u
m
e
r
r
o
r
o
f
e
s
t
i
m
a
t
e
.
I
f
n
e
c
e
s
s
a
r
y
,
r
o
u
n
d
t
h
e
a
n
s
w
e
r
u
p
t
o
o
b
t
a
i
n
a
w
h
o
l
e
n
u
m
b
e
r
.
838
=
1
.
96
,
E
=
0
.
02
839
z
Since
=
0
.
05
,
=
0
40
.
p
,
$
2
2
z
then
n
p
q
,
.
and
=
0
60
$
$
2
$
q
E
2
.
1
96
= (0.40)(0.60)
2304
.
96
.
0
02
Which, when rounded up is 2305 people to interview.
840
841
Formula
for
the
confidence
interval
for
a
variance
(
n
1
)
s
(
n
1
)
s
2
2
2
2
2
right
left
d
.
f
.
n
1
842
Formula
for
the
confidence
interval
for
a
standard
deviation
(
n
1
)
s
(
n
1
)
s
2
2
2
2
right
left
d.f.
n
1
843
844
The
95%
confidence
interval
for
the
variance
is
found
by
substituti
ng
in
(
n
1
)
s
(
n
1
)
s
2
2
2
2
2
right
left
2
(
20
1
)
(
20
1
)
(1.6)
(1.6)
2
2
32
.
852
8
.
907
1
.
5
5
.
5
2
845
The
95%
confidence
interval
for
the
standard
deviation
is
1
.
5
5
.
5
1
.
2
2
.
3