Chap. 5: The Normal Distribution & Sampling Distributions
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Transcript Chap. 5: The Normal Distribution & Sampling Distributions
© 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
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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
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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 85
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
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As
sample
size gets
large
enough
( 30) ...
sampling
distribution
becomes
almost
normal.
X
X
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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
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Hypothesis
Hypothesis
Testing
Testing
Estimation Process
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Population
J
J
Mean, , is
unknown
J
J J
Sample
J
J
J J
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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
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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
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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
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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)
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Sample statistic
(point estimate)
Confidence
limit (upper)
Confidence Limits
for Population Mean
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Parameter =
Statistic ± Error
© 1984-1994
T/Maker Co.
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(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
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`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
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`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
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`X
Level of Confidence
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n
n
Probability that the unknown
population parameter falls within
interval
Denoted (1 - a)%
l
n
ais probability that parameter is not
within interval
Typical values are 99%, 95%, 90%
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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
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_
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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.
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Confidence Interval
Estimates
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Confidence
Confidence
Intervals
Intervals
Mean
Mean
sKnown
sKnown
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Proportion
Proportion
ss Unknown
Unknown
Variance
Variance
Finite
Finite
Population
Population
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Confidence Interval Estimate
Mean
(s Known)
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Confidence Interval
Estimates
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Confidence
Confidence
Intervals
Intervals
Mean
Mean
ss Known
Known
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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)
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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
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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.
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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
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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
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.05
100
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Confidence Interval Estimate
Mean
(s Unknown)
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Confidence Interval
Estimates
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Confidence
Confidence
Intervals
Intervals
Mean
Mean
ss Known
Known
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Proportion
Proportion
sUnknown
sUnknown
Variance
Variance
Finite
Finite
Population
Population
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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
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Student’s t Distribution
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Standard
Bellnormal
shaped
Symmetric
t (df = 13)
‘Fatter’ tails
t (df = 5)
0
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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
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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
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2.920
t
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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
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S
n
8
25
Thinking Challenge
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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?
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Alone
Group Class
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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
sKnown
sKnown
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Proportion
Proportion
ss Unknown
Unknown
Variance
Variance
Finite
Finite
Population
Population
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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
Nn
X t aa //22,, nn11
X t aa //22,, nn11
n
N 1
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S
Nn
n
N 1
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Confidence Interval Estimate
of Proportion
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Confidence Interval
Estimates
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Confidence
Confidence
Intervals
Intervals
Mean
Mean
sKnown
sKnown
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Proportion
Proportion
ss Unknown
Unknown
Variance
Variance
Finite
Finite
Population
Population
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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.
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© 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
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Thinking Challenge
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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
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This Class...
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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