Chap. 5: The Normal Distribution & Sampling Distributions

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© 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
© 1997 Prentice-Hall, Inc.
n
n
n
n
‘Bell-shaped’ &
symmetrical
f(X)
Mean, median,
mode are equal
‘Middle spread’ is
1.33 s
Random variable
has infinite range
5-2
X
Mean
Median
Mode
Standardize the
Normal Distribution
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X 
Z
s
Normal
Distribution
Standardized
Normal Distribution
s
s= 1

X
= 0
One table!
5-3
Z
Standardizing Example
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Normal
Distribution
s = 10
= 5 6.2 X
5-4
Standardizing Example
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X   6.2  5
Z

 .12
s
10
Normal
Distribution
s = 10
= 5 6.2 X
5-5
Standardized
Normal Distribution
s=1
= 0 .12 Z
Obtaining
the Probability
© 1997 Prentice-Hall, Inc.
Standardized Normal
Probability Table (Portion)
Z
.00
.01
s=1
.02
0.0 .0000 .0040 .0080
.0478
0.1 .0398 .0438 .0478
0.2 .0793 .0832 .0871
= 0 .12 Z
0.3 .1179 .1217 .1255
Probabilities
5-6
Shaded area
exaggerated
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5-7
Example
P(3.8  X  5)
Example
P(3.8  X  5)
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X   3.8  5
Z

  .12
s
10
Normal
Distribution
Standardized
Normal Distribution
s = 10
s=1
.0478
3.8 = 5
5-8
X
-.12  = 0
Shaded area exaggerated
Z
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5-9
Example
P(2.9  X  7.1)
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Example
P(2.9  X  7.1)
Z
Normal
Distribution
Z
X   2.9  5

  .21
s
10
X   7.1  5

 .21
Standardized
s
10
Normal Distribution
s = 10
s=1
.1664
.0832 .0832
2.9 5 7.1 X
5 - 10
-.21 0 .21
Shaded area exaggerated
Z
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5 - 11
Example
P(X  8)
Example
P(X  8)
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X  85
Z

 .30
s
10
Normal
Distribution
Standardized
Normal Distribution
s = 10
s=1
.5000
.3821
.1179
=5
5 - 12
8 X
=0
Shaded area exaggerated
.30 Z
Central Limit Theorem
© 1997 Prentice-Hall, Inc.
As
sample
size gets
large
enough
( 30) ...
sampling
distribution
becomes
almost
normal.
X
X
5 - 13
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Introduction
to Estimation
5 - 14
Statistical Methods
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Statistical
Statistical
Methods
Methods
Descriptive
Descriptive
Statistics
Statistics
Inferential
Inferential
Statistics
Statistics
Estimation
Estimation
5 - 15
Hypothesis
Hypothesis
Testing
Testing
Estimation Process
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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.
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Population Parameters Are
Estimated
Estimate population
population
parameter...
parameter...
Mean

with sample
sample
statistic
statistic
`x
Proportion
Proportion
p
pss
Variance
s
22
s
11
22
`x11 -`x22
Differences
Differences
5 - 17
22
Point Estimation
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n
Provides single value
l
n
n
Based on observations from 1 sample
Gives no information about how
close value is to the unknown
population parameter
Example: Sample mean`X = 3 is
point estimate of unknown
population mean
5 - 18
Interval Estimation
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n
Provides range of values
l
n
Gives information about closeness to
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
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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
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Parameter =
Statistic ± Error
© 1984-1994
T/Maker Co.
5 - 21
(1)
  X  Error
(2)
Error  X   or X  
X 
(3)
Z
(4)
Error  Zs xx
(5)
  X  Zs x
sx

Error
s xx
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Many Samples Have
Same Interval
`X =  ± Zs`x
sx_
-1.65s`x

+1.65s`x
90% Samples
5 - 22
`X
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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
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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
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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
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Intervals &
Level of Confidence
Sampling
Distribution a/2
of Mean
sx__
1 -a
a/2
`x = 
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
_
© 1997 Prentice-Hall, Inc.
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
© 1984-1994 T/Maker Co.
5 - 27
Confidence Interval
Estimates
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Confidence
Confidence
Intervals
Intervals
Mean
Mean
sKnown
sKnown
5 - 28
Proportion
Proportion
ss Unknown
Unknown
Variance
Variance
Finite
Finite
Population
Population
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Confidence Interval Estimate
Mean
(s Known)
5 - 29
Confidence Interval
Estimates
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Confidence
Confidence
Intervals
Intervals
Mean
Mean
ss Known
Known
5 - 30
Proportion
Proportion
ss Unknown
Unknown
Variance
Variance
Finite
Finite
Population
Population
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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
© 1997 Prentice-Hall, Inc.
n
Assumptions
l
l
l
n
Confidence Interval
Mean (s Known)
Population standard deviation is known
Population is normally distributed
If not normal, can be approximated by
normal distribution (n  30)
Confidence interval estimate
s
X  Za / 2 
   X  Za / 2 
n
5 - 32
s
n
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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
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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.
s
s
X  Za / 2 
   X  Za / 2 
n
n
10
10
50  1.96 
   50  1.96 
25
25
46.08    53.92
5 - 34
Confidence Interval
Solution*
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X  Za / 2 
s
n
   X  Za / 2 
s
n
.05
   1.99  1.645 
1.99  1.645 
100
1.982    1.998
5 - 35
.05
100
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Confidence Interval Estimate
Mean
(s Unknown)
5 - 36
Confidence Interval
Estimates
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Confidence
Confidence
Intervals
Intervals
Mean
Mean
ss Known
Known
5 - 37
Proportion
Proportion
sUnknown
sUnknown
Variance
Variance
Finite
Finite
Population
Population
© 1997 Prentice-Hall, Inc.
n
Assumptions
l
l
n
n
Confidence Interval
Mean (s Unknown)
Population standard deviation is
unknown
Population must be normally distributed
Use Student’s t distribution
Confidence interval estimate
S
S
X  t a / 2, n 1 
   X  t a / 2, n 1 
n
n
5 - 38
Student’s t Distribution
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Standard
Bellnormal
shaped
Symmetric
t (df = 13)
‘Fatter’ tails
t (df = 5)
0
5 - 39
Z
t
Student’s t Table
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Upper Tail Area
df
.25
.10
Assume:
n=3
df
=n-1
=2
a = .10
a/2 =.05
a/2
.05
1 1.000 3.078 6.314
2 0.817 1.886 2.920
.05
3 0.765 1.638 2.353
0
t values
5 - 40
t
Student’s t Table
© 1997 Prentice-Hall, Inc.
Upper Tail Area
df
.25
.10
Assume:
n=3
df
=n-1
=2
a = .10
a/2 =.05
a/2
.05
1 1.000 3.078 6.314
2 0.817 1.886 2.920
.05
3 0.765 1.638 2.353
0
t values
5 - 41
2.920
t
© 1997 Prentice-Hall, Inc.
Estimation Example
Mean (s Unknown)
A random sample of n = 25 has`X = 50
& S = 8. Set up a 95% confidence
interval estimate for .
S
X  t a / 2, n 1 
   X  t a / 2, n 1 
n
8
50  2.0639 
   50  2.0639 
25
46.69    53.30
5 - 42
S
n
8
25
Thinking Challenge
© 1997 Prentice-Hall, Inc.
You’re a time study
analyst in manufacturing.
You’ve recorded the
following task times (min.):
3.6, 4.2, 4.0, 3.5, 3.8, 3.1.
What is the 90%
confidence interval
estimate of the population
mean task time?
5 - 43
Alone
Group Class
© 1997 Prentice-Hall, Inc.
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
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Estimation of Mean
for Finite Populations
5 - 45
Confidence Interval
Estimates
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Confidence
Confidence
Intervals
Intervals
Mean
Mean
sKnown
sKnown
5 - 46
Proportion
Proportion
ss Unknown
Unknown
Variance
Variance
Finite
Finite
Population
Population
© 1997 Prentice-Hall, Inc.
n
Estimation for
Finite Populations
Assumptions
l
Sample is large relative to population
s n / N > .05
n
Use finite population correction factor
n
Confidence interval (mean, s unknown)
S
Nn
X  t aa //22,, nn11 

   X  t aa //22,, nn11 
n
N 1
5 - 47
S
Nn

n
N 1
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Confidence Interval Estimate
of Proportion
5 - 48
Confidence Interval
Estimates
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Confidence
Confidence
Intervals
Intervals
Mean
Mean
sKnown
sKnown
5 - 49
Proportion
Proportion
ss Unknown
Unknown
Variance
Variance
Finite
Finite
Population
Population
© 1997 Prentice-Hall, Inc.
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
pss  (1  pss )
pss  (1  pss )
pss  Z 
 p  pss  Z 
n
n
5 - 50
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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
© 1997 Prentice-Hall, Inc.
Estimation Example
Proportion
A random sample of 400 graduates
showed 32 went to grad school. Set
up a 95% confidence interval estimate
for p.
pss  (1  pss )
pss  (1  pss )
pss  Z aa//22 
 p  pss  Z aa//22 
n
n
.08  (1 .08)
.08  (1 .08)
.08  1.96 
 p  .08  1.96 
400
400
.053  p  .107
5 - 52
Thinking Challenge
© 1997 Prentice-Hall, Inc.
You’re a production
manager for a newspaper.
You want to find the %
defective. Of 200
newspapers, 35 had
defects. What is the
90% confidence interval
estimate of the population
proportion defective?
5 - 53
Alone
Group Class
© 1997 Prentice-Hall, Inc.
Confidence Interval
Solution*

n·p  5
n·(1 - p)  5
pss  (1  pss )
pss  (1  pss )
pss  Z aa//22 
 p  pss  Z aa//22 
n
n
.175  (.825)
.175  (.825)
.175  1.645 
 p  .175  1.645 
200
200
.1308  p  .2192
5 - 54
This Class...
© 1997 Prentice-Hall, Inc.
Please take a moment to answer the
following questions in writing:
n
What was the most important thing
you learned in class today?
n
What do you still have questions
about?
n
How can today’s class be improved?
5 - 55
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