### 10/5

```© 1997 Prentice-Hall, Inc.
Importance of
Normal Distribution
n
Describes many random processes
or continuous phenomena
n
Can be used to approximate discrete
probability distributions
l
l
n
Binomial
Poisson
Basis for classical statistical
inference
5-1
Normal Distribution
n
n
n
n
‘Bell-shaped’ &
symmetrical
f(X )
Mean, median,
mode are equal
1.33 s
Random variable
has infinite range
5-2
X
Mean
Median
Mode
Standardize the
Normal Distribution
Z 
Normal
Distribution
X  
s
Standardized
Normal Distribution
s
s = 1

X
= 0
One table!
5-3
Z
Standardizing Example
Normal
Distribution
s = 10
= 5
5-4
6 .2
X
Standardizing Example
Z 
X  
s
Normal
Distribution

6 .2  5
10
 .1 2
Standardized
Normal Distribution
s = 10
= 5
5-5
s = 1
6 .2
X
= 0
.1 2
Z
Obtaining
the Probability
Standardized Normal
Probability Table (Portion)
Z
.0 0
.0 1
.02
s = 1
0 .0 .0 0 0 0 .0 0 4 0 .0 0 8 0
0.1
.0 3 9 8 .0 4 3 8
.0478
.0478
0 .2 .0 7 9 3 .0 8 3 2 .0 8 7 1
= 0
0 .3 .1 1 7 9 .1 2 1 7 .1 2 5 5
Probabilities
5-6
.1 2
Z
exaggerated
5-7
Example
P(3.8  X  5)
Example
P(3.8  X  5)
Z 
X  
s
Normal
Distribution

3 .8  5
10
  .1 2
Standardized
Normal Distribution
s = 10
s = 1
.0478
3 .8  = 5
5-8
X
-.1 2  = 0
Z
5-9
Example
P(2.9  X  7.1)
Example
P(2.9  X  7.1)
Z 
Normal
Distribution
Z 
X  
s
X  
s


2 .9  5
10
7 .1  5
10
  .2 1
 .2 1
Standardized
Normal Distribution
s = 10
s = 1
.1664
.0832 .0832
2 .9 5 7 .1
5 - 10
X
-.2 1 0 .2 1
Z
5 - 11
Example
P(X  8)
Example
P(X  8)
Z 
X  
s
Normal
Distribution

8 5
10
 .3 0
Standardized
Normal Distribution
s = 10
s = 1
.5000
.3821
.1179
 = 5
5 - 12
8
X
 = 0
.3 0 Z
Central Limit Theorem
As
sample
size gets
large
enough
( 30) ...
sampling
distribution
becomes
almost
normal.
XX
5 - 13
Introduction
to Estimation
5 - 14
Statistical Methods
SSta
tatis
tistic
ticaall
M
Meeth
thooddss
DDeessccrip
riptiv
tivee
SSta
tatis
tistic
ticss
In
Infe
fere
renntia
tiall
SSta
tatis
tistic
ticss
EEsstim
timaatio
tionn
5 - 15
HHyyppooth
theessis
is
TTeesstin
tingg
Estimation Process
Population
J
J
Mean, , is
unknown
J
J J
Sample
J
J
J J
5 - 16
Random Sample
Mean J
J`X = 50
I am 95%
confident that
 is between
40 & 60.
Population Parameters Are
Estimated
E
Esstim
tim aate
te ppooppuula
latio
tionn
ppaara
ram
m eete
ter...
r...
M
M eeaann

P
Pro
roppoortio
rtionn
pp
V
Vaaria
riannccee
D
Diffe
iffere
renncceess
5 - 17
ss
22
11

22
w
with
ith ssaam
m pple
le
ssta
tatis
tistic
tic
`x
`x
ppss
22
ss
`x
`x11 -`x
-`x22
Point Estimation
n
Provides single value
l
n
n
Based on observations from 1 sample
close value is to the unknown
population parameter
Example: Sample mean`X = 3 is
point estimate of unknown
population mean
5 - 18
Interval Estimation
n
Provides range of values
l
n
unknown population parameter
l
n
Based on observations from 1 sample
Stated in terms of probability
Example: Unknown population mean
lies between 50 & 70 with 95%
confidence
5 - 19
Key Elements of
Interval Estimation
A probability that the population parameter
falls somewhere within the interval.
Confidence
interval
Confidence
limit (lower)
5 - 20
Sample statistic
(point estimate)
Confidence
limit (upper)
Confidence Limits
for Population Mean
Parameter =
Statistic ± Error
(1)
  X  E rro r
(2 )
E rro r  X   o r X  
(3 )
T/Maker Co.
5 - 21
Z 
X  
s

xx
(4 )
E rro r  Z s
xx
(5 )
  X  Zs
xx
E rro r
s
xx
Many Samples Have
Same Interval
`X =  ± Zs`x
sx_
-1.65s`x

+1.65s`x
90% Samples
5 - 22
`X
Many Samples Have
Same Interval
`X =  ± Zs`x
sx_
-1.65s`x
-1.96s`x

+1.65s`x
+1.96s`x
90% Samples
95% Samples
5 - 23
`X
Many Samples Have
Same Interval
`X =  ± Zs`x
sx_
-2.58s`x
-1.65s`x
-1.96s`x

+1.65s`x
+2.58s`x
+1.96s`x
90% Samples
95% Samples
99% Samples
5 - 24
`X
Level of Confidence
n
n
Probability that the unknown
population parameter falls within
interval
Denoted (1 - a)%
l
n
ais probability that parameter is not
within interval
Typical values are 99%, 95%, 90%
5 - 25
Sampling
Distribution
of Mean
Intervals &
Level of Confidence
_
a /2
s x_
x
1 -a
a /2
 ``xx = 
X
(1 - a) % of
intervals
contain .
Intervals
extend from
`X - Zs`X to
`X + Zs`X
a % do not.
Large number of intervals
5 - 26
_
n
Data dispersion
l
n
Measured by s
Intervals extend from
`X - Zs`X to`X + Zs`X
Sample size
l
n
Factors Affecting
Interval Width
s`X = s / n
Level of confidence
(1 - a)
l
Affects Z
5 - 27
Confidence Interval
Estimates
CCoonnfid
fideennccee
In
ls
rvaals
terv
Inte
M
Meeaann
ssK
Knnoowwnn
5 - 28
PPro
rtionn
roppoortio
ss UUnnkknnoowwnn
VVaaria
riannccee
FFin
ite
inite
PPooppuula
tionn
latio
Confidence Interval Estimate
Mean
(s Known)
5 - 29
Confidence Interval
Estimates
CCoonnfid
fideennccee
In
ls
rvaals
terv
Inte
M
Meeaann
ss KKnnoowwnn
5 - 30
PPro
rtionn
roppoortio
ss UUnnkknnoowwnn
VVaaria
riannccee
FFin
ite
inite
PPooppuula
tionn
latio
n
Confidence Interval
Mean (s Known)
Assumptions
l
l
l
Population standard deviation is known
Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)
5 - 31
Confidence Interval
Mean (s Known)
n
Assumptions
l
l
l
n
Population standard deviation is known
Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)
Confidence interval estimate
X  Z aa //22 
5 - 32
s
n
   X  Z aa //22 
s
n
Estimation Example
Mean (s Known)
The mean of a random sample of n = 25
is`X = 50. Set up a 95% confidence
interval estimate for  if s = 10.
5 - 33
Estimation Example
Mean (s Known)
The mean of a random sample of n = 25
is`X = 50. Set up a 95% confidence
interval estimate for  if s = 10.
X  Z aa //22 
5 0  1.9 6 
s
n
10
   X  Z aa //22 
   5 0  1.9 6 
25
4 6 .0 8    5 3 .9 2
5 - 34
s
n
10
25
Confidence Interval
Solution*
X  Z aa //22 
1.9 9  1.6 4 5 
s
n
.0 5
100
   X  Z aa //22 
n
   1.9 9  1.6 4 5 
1.9 8 2    1.9 9 8
5 - 35
s
.0 5
100
Confidence Interval Estimate
Mean
(s Unknown)
5 - 36
Confidence Interval
Estimates
CCoonnfid
fideennccee
In
ls
rvaals
terv
Inte
M
Meeaann
ss KKnnoowwnn
5 - 37
PPro
rtionn
roppoortio
ssU
Unnkknnoowwnn
VVaaria
riannccee
FFin
ite
inite
PPooppuula
tionn
latio
n
Confidence Interval
Mean (s Unknown)
Assumptions
l
l
Population standard deviation is
unknown
Population must be normally distributed
n
Use Student’s t distribution
n
Confidence interval estimate
X  t aa //22,, nn 11 
5 - 38
S
n
   X  t aa //22,, nn 11 
S
n
Student’s t Distribution
Standard
Bellnormal
shaped
Symmetric
t (df = 13)
‘Fatter’ tails
t (df = 5)
0
5 - 39
Z
t
Student’s t Table
Assume:
n=3
df
=n-1
=2
a = .10
a/2 =.05
a/2
U p p e r T a il A re a
df
.2 5
.1 0
.0 5
1
1 .0 0 0 3 .0 7 8 6 .3 1 4
2
0 .8 1 7 1 .8 8 6 2 .9 2 0
3
0 .7 6 5 1 .6 3 8 2 .3 5 3
.05
0
t values
5 - 40
t
Student’s t Table
Assume:
n=3
df
=n-1
=2
a = .10
a/2 =.05
a/2
U p p e r T a il A re a
df
.2 5
.1 0
.0 5
1
1 .0 0 0 3 .0 7 8 6 .3 1 4
2
0 .8 1 7 1 .8 8 6 2 .9 2 0
3
0 .7 6 5 1 .6 3 8 2 .3 5 3
.05
0
t values
5 - 41
2.920
t
Estimation Example
Mean (s Unknown)
A random sample of n = 25 has`X = 50
& S = 8. Set up a 95% confidence
interval estimate for .
X  t aa //22,, nn 11 
5 0  2 .0 6 3 9 
S
n
8
   X  t aa //22,, nn 11 
   5 0  2 .0 6 3 9 
25
4 6 .6 9    5 3 .3 0
5 - 42
S
n
8
25
Thinking Challenge
You’re a time study
analyst in manufacturing.
You’ve recorded the
3.6, 4.2, 4.0, 3.5, 3.8, 3.1.
What is the 90%
confidence interval
estimate of the population
5 - 43
Alone
Group Class
Confidence Interval
Solution*
`X = 3.7
S = 3.8987
n = 6, df = n - 1 = 6 - 1 = 5
S / n = 3.8987 / 6 = 1.592
t.05,5 = 2.0150
3.7 - (2.015)(1.592) 3.7 + (2.015)(1.592)
0.492  6.908
5 - 44
Estimation of Mean
for Finite Populations
5 - 45
Confidence Interval
Estimates
CCoonnfid
fideennccee
In
ls
rvaals
terv
Inte
M
Meeaann
ssK
Knnoowwnn
5 - 46
PPro
rtionn
roppoortio
ss UUnnkknnoowwnn
VVaaria
riannccee
FFin
ite
inite
PPooppuula
tionn
latio
Estimation for
Finite Populations
n
Assumptions
l
Sample is large relative to population
s n / N > .05
n
Use finite population correction factor
n
Confidence interval (mean, s unknown)
X  t aa //22,, nn 11 
5 - 47
S
n

N  n
N 1
   X  t aa //22,, nn 11 
S
n

N  n
N 1
Confidence Interval Estimate
of Proportion
5 - 48
Confidence Interval
Estimates
CCoonnfid
fideennccee
In
ls
rvaals
terv
Inte
M
Meeaann
ssK
Knnoowwnn
5 - 49
PPro
rtionn
roppoortio
ss UUnnkknnoowwnn
VVaaria
riannccee
FFin
ite
inite
PPooppuula
tionn
latio
n
Assumptions
l
l
l
n
Confidence Interval
Proportion
Two categorical outcomes
Population follows binomial distribution
Normal approximation can be used
s n·p  5 & n·(1 - p)  5
Confidence interval estimate
p ss  Z 
5 - 50
p ss  (1  p ss )
n
 p  p ss  Z 
p ss  (1  p ss )
n
Estimation Example
Proportion
A random sample of 400 graduates
showed 32 went to grad school. Set
up a 95% confidence interval estimate
for p.
5 - 51
Estimation Example
Proportion
A random sample of 400 graduates
showed 32 went to grad school. Set
up a 95% confidence interval estimate
for p.
p ss  Z aa //22 
.0 8  1.9 6 
p ss  (1  p ss )
n
.0 8  ( 1  .0 8 )
400
 p  p ss  Z aa //22 
 p  .0 8  1.9 6 
.0 5 3  p  .1 0 7
5 - 52
p ss  (1  p ss )
n
.0 8  ( 1  .0 8 )
400
Thinking Challenge
You’re a production
manager for a newspaper.
You want to find the %
defective. Of 200
defects. What is the
90% confidence interval
estimate of the population
proportion defective?
5 - 53
Alone
Group Class
Confidence Interval
Solution*

n·p  5
n·(1 - p)  5
p ss  Z aa //22 
.1 7 5  1.6 4 5 
p ss  (1  p ss )
n
.1 7 5  (.8 2 5 )
200
 p  p ss  Z aa //22 
 p  .1 7 5  1.6 4 5 
.1 3 0 8  p  .2 1 9 2
5 - 54
p ss  (1  p ss )
n
.1 7 5  (.8 2 5 )
200
This Class...