Download Sampling Distribution Sampling Distribution of Means Sampling

Sampling Distribution and Confidence Interval
6-1
Sampling Distribution of
Means
Sampling Distribution
Theoretical Probability Distribution of the
Sample Statistic.
If a random sample is taken from a normally
distributed population that has a mean µ
and a standard deviation σ , the sampling
distribution of the sample means, x, is
normal with
µx = µ
What is the Shape of this distribution?
What are the values of the parameters
such as mean and standard deviation?
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Central Tendency
An Application
Population Distribution
σσ = 10
µx = µ
Dispersion
σ
σ =
n
x
µµ = 50
X
Sampling Distribution
n=4
σσ X = 5
Waiting times at a certain type of
clinic are normally distribution
with µ = 8 min. & σ = 2 min. If
you select random samples of 25
cases, what is the sampling
distribution of the mean?
n =16
σσ X = 2.5
µ X- = 50
X
Sampling Distribution
Solution*
If you select random samples of 25 cases
from a normal distribution (or population)
with µ = 8 min. & σ = 2 min.
The sampling distribution of the mean is
normal distribution with mean = 8 min.,
and standard deviation = 2/5 = .4 min.
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σ
n
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Sampling from
Normal Populations
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σx =
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Continue the Application
Waiting times at a certain type of
clinic are normally distribution
with µ = 8 min. & σ = 2 min. If
you select random samples of 25
cases, what is the probability that
the sample mean of a random
sample of 25 cases is between
σ
σX = .4
7.8 & 8.2 minutes?
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7.8 8 8.2 X

Sampling Distribution and Confidence Interval
6-2
Standardizing Sampling
Distribution of Mean
Z ==
Sampling
Sampling
Distribution
Distribution
X −− µµ x
X −− µµ
==
σ
σ
σ
σx
n
σ X

Sampling Distribution
Solution*
Z ==
Standardized
Standardized
Normal
Normal Distribution
Distribution
σ=1
Sampling
Sampling
Distribution
Distribution of
of
sample
sample mean
mean
Z ==
X −− µµ 7.8 −− 8
==
== −−.50
σσ n 2 25
X −− µµ 8.2 −− 8
==
== .50
Standardized
Standardized
σσ n 2 25
Normal
Normal Distribution
Distribution
σ
σ =1
σ
σX = .4
.383
µ X

µ =0
X
-.50 0 .50
7.8 8 8.2 X

Z
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Z
The probability that the sample mean would be
between 7.8 & 8.2 minutes is .383.
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Central Limit Theorem
Central Limit Theorem
In the previous problem the data was from
normally distributed population.
What if the population is not normally
distributed?
If a relative large random sample is taken
from a population that has a mean µ and a
standard deviation σ, regardless of the
distribution of the population, the
distribution of the sample means is
approximately normal with
µx = µ
σx =
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Sampling from
Non-Normal Populations
Central Limit Theorem
As
sample
size gets
large
enough
(n ≥ 30) ...
σx =
Central Tendency
σ
n sampling
distribution
becomes
almost
normal.
X
X
µx = µ
Population Distribution
σσ = 10
µx = µ
Dispersion
σx =
n
n
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σ
n
σ
n
Sampling
Sampling with
with
replacement
replacement
µµ = 50
n = 30
σσ X = 1.8
n=4
σσ X = 5
µµX- = 50
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X
Sampling Distribution
X
Sampling Distribution and Confidence Interval
6-3
Central Limit Theorem
Example: Consider the distribution of serum
cholesterol levels for all 20- to 74-year-old
males living in United States has a mean of
211 mg/100 ml, and the standard deviation
of 46 mg/100 ml. If a random sample of
100 individuals from the population, what is
the probability that the average serum
cholesterol level of these 100 individuals is
higher than 225?
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Central Limit Theorem
Solution:
The sampling distribution of the mean is normal
with
mean µµ x == µµ = 211 and
standard error
.001
P(X > 225) = P(Z > [225 - 211]/4.6)
= P(Z > 3.04)
= 0.001
1. State What Is Estimated
2. Distinguish Point & Interval Estimates
3. Explain Interval Estimates
4. Compute Confidence Interval Estimates
for Population Mean (& Proportion)
5. Compute Sample Size
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Population
Mean, µ,
µ is
unknown
J J
Sample
J
J
J J
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3.04
Thinking Challenge
Suppose you’re interested in the
average body temperature of
healthy adults in Northeastern
Ohio (the population). How would
you find out?
How can we estimate this average
with a measure of reliability?
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Estimation Process
J
0
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Learning Objectives
J
J
σ
σ
= 4.6 (= 46/10).
n
σ
σ x ==
Random Sample
JMean J
X
 = 98
I am 95% confident
that µµ is between
97 & 99, or 98±±1
Unknown Population
Parameters Are Estimated
Estimate Population
Parameter...
Mean
µ
Proportion
π
Variance
σ
Differences
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2
µ1 - µ 2
with Sample
Statistic
x
p
s2
x1 − x2
z
Sampling Distribution and Confidence Interval
6-4
Estimation Methods
Point Estimation
1. Provides Single Value
n
Estimation
Point
Estimation
Interval
Estimation
Based on Observations from 1 Sample
2. Gives No Information about How Close
Value Is to the Unknown Population
Parameter
3. Example: Sample Mean
MeanX = 3 Is Point
Estimate of Unknown Population Mean
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Key Elements of
Interval Estimation
Interval Estimation
1. Provides Range of Values
n
Based on Observations from 1 Sample
A probability that the population parameter
falls somewhere within the interval.
2. Gives Information about Closeness to
Unknown Population Parameter
n
Confidence
interval
Sample statistic
(point estimate)
Stated in terms of Probability
l Knowing Exact Closeness Requires Knowing
Unknown Population Parameter
3. Example: Unknown Population Mean Lies
Between 50 & 70 with 95% Confidence
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Confidence
limit (lower)
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Sampling Error
Confidence
limit (upper)
x ± Margin of Error
Sampling Distribution of
the Mean
σx_
Confidence
interval
Sample statistic
(point estimate)
µµ - ? σσx
x
µ
x
Sampling Error = | µ– x |
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.025
.025
µ
X
µµ + ?σσx
x
Within how many standard deviations
of the mean will have 95% of the
sampling distribution?
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Sampling Distribution and Confidence Interval
6-5
A Special Notation
Size of Interval
zα = the z score that the proportion of
the standard normal distribution to the
right of it is α.
Z .05
.06
.07
z.025 = 1.96
1.8 .9678 .9686 .9693
z.010 = ?
σx_
µµ - 2.58σ
2.58σσ x
µµ - 1.65σσx
µµ-1.96σ
µ-1.96
σσ x
1.9 .9744 .9750 .9756
µ
X
µµ + 1.65σσ x µµ + 2.58σσx
µµ+1.96σσ x
µ+1.96
90% Samples
z .025
2.0 .9798 .9803 .9808
95% Samples
2.1 .9842 .9846 .9850
99% Samples
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Confidence Interval
Mean ( σ Known)
The Confidence Interval
σx_
Confidence Level
α/2
1.96 = z.025
α/2 = .025
1- α = .95
µ
µµ - 1.96σ
1.96σσx
(1-α
(1α)· 100% Confidence Interval Estimate
X
x
( X − Zα / 2 ⋅
µµ + 1.96σσx
x
or
95% Samples
Confidence Interval => x - 1.96σ
1.96σσx
X ± Zα / 2 ⋅ σ
n
x + 1.96σσx
x
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σ
σ
, X + Zα / 2 ⋅
)
n
n
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Confidence Interval
of Mean (σ
(σ Known)
The Confidence Interval
σx_
Assumptions
n
n
n
Population Standard Deviation Is Known
Population Is Normally Distributed
If Not Normal, Can Be Approximated by
Normal Distribution (n
(n ≥ 30)
95% Samples
µµ - 1.96σ
1.96σσx
µ
X
x
µµ + 1.96σσx
x
95% Confidence
Interval
Confidence Interval => x - 1.96σ
1.96σσx
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x - 1.96σ
1.96σσx
Sampling Distribution and Confidence Interval
6-6
Factors Affecting
Interval Width
The Confidence Interval
σx_
2.5%
1. Data Dispersion
2.5%
95% Samples
µ
n
X
95 % of
intervals
contain µ.
2.5 % do not.
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Measured by σ
Intervals Extend from
x - zα/2
σ X to x + zα/2
σX
αα/2· σ
αα/2 · σ
2. Sample Size
n
σX =
σ
n
3. Level of Confidence
(1 - α)
n
Affects Z
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Confidence Interval
Mean ( σ Known)
(1-α
(1α)· 100% Confidence Interval Estimate
( X − Zα / 2 ⋅
σ
σ
, X + Zα / 2 ⋅
)
n
n
or
X ± Zα / 2 ⋅ σ
n
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Estimation Example
Mean ( σ Known)
The mean of a random sample of n = 25
isX = 50. Set up a 95% confidence
is
interval estimate for µ if σ = 10.
1 − α = .95 ,
α = .05, α
2
= .025, z α 2 = 1.96.
σ
σ
, X + Zα / 2 ⋅
)
n
n
10
10
( 50 − 1 . 96 ⋅
, 50 + 1 .96 ⋅
)
25
25
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( X − Zα / 2 ⋅
Confidence Interval Estimate
We can be 95% confident that the
population mean is in (46.08, 53.92).
We can be 95% confident that the
maximum sampling error using this
interval estimate for estimating mean is
within 3.92.
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You’re a Q/C inspector for Coke. The σ for 2-liter
bottles is .05 liters. A random sample of 100
bottles showed
showed X = 1.99 liters. What is the 90%
confidence interval estimate of the true mean
amount in 2-liter bottles?
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Sampling Distribution and Confidence Interval
6-7
Confidence Interval
Solution*
Confidence Interval
Mean ( σ Unknown)
1 - α = .90, α = .1, α /2 = .05, Z α / 2 = 1.64
( X − Z α /2 ⋅
( 1. 99 − 1. 64 ⋅
σ
σ
, X + Zα / 2 ⋅
)
n
n
1. Assumptions
n
n
.05
.05
, 1.99 + 1.64 ⋅
)
100
100
Population Standard Deviation Is Unknown
Population Must Be Normally Distributed
2. Use Student’s t Distribution
3. Confidence Interval Estimate
( X − tα / 2 ,n −−1 ⋅
( 1.982 , 1. 998 )
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S
S
, X + t α / 2,n −−1 ⋅
)
n
n
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Student’s t Table
Student’s t Distribution
Standard
Normal (Z)
x−µ
t=
s
n
t (df
( df = 13)
Bell-Shaped
Symmetric
t (df
( df = 5)
‘Fatter’ Tails
α/2
df
.90
.95
.975
1 3.078 6.314 12.706
For a 90% C.I.:
n=3
df = n - 1 = 2
α
α = .10
α
α/2
α/2 =.05
2 1.886 2.920 4.303
.05
3 1.638 2.353 3.182
0
Z
t
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0
t values
2.920
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Estimation Example
Mean ( σ Unknown)
A random sample of n = 25 has
hasx = 50 &
s = 8. Set up a 95% confidence interval
estimate for µ.
1 - α = .95, α = .05, α/2 = .025, t α 2 , df =24 = 2. 064
S
S
, X + tα / 2 ,n−1 ⋅
)
n
n
8
8
( 50 − 2. 064 ⋅
, 50 + 2 .064 ⋅
)
25
25
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( 46 . 69 , 53 . 30 )
Thinking Challenge
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.
3.1.
What is the 90% confidence interval estimate
of the population mean task time?
( X − tα / 2,n −1 ⋅
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t
Sampling Distribution and Confidence Interval
6-8
Confidence Interval
Solution*
X = 3.7
One-Sided C. I.
(Error in the Note)
Z C.I.:
σ
( − ∞ , X + Zα ⋅
)
Lower interval
n
σ
Upper interval ( X − Zα ⋅
, ∞)
S = 0.38987
n = 6, df = n - 1 = 6 - 1 = 5
n
S / √n = 0.38987 / √6 = 0.1592
t .05,5 = 2.015
( 3.7 - (2.015)(0.1592) , 3.7 + (2.015)(0.1592) )
( 3.379, 4.021 )
t C.I.:
σ
( − ∞ , X + tα ⋅
)
Lower interval
n
σ
Upper interval ( X − tα ⋅
, ∞)
n
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Size of Interval
σx_
0.025
µµ-1.96σ
µ-1.96
σσx
σx_
0.025
0.05
.95
.95
µ
Lower Interval
X
µ
µµ+1.96σσx
µ+1.96
x
X
µµ+ 1.64σσx
µ+
x
z α = z .05
95% Samples
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95% Samples
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Estimation Example
Mean ( σ Known)
The mean of a random sample of n = 25
isX = 50. Set up a upper 95% confidence
is
interval estimate for µ if σ = 10.
1 − α = .95 ,
α = .05, zα = 1.64.
σ
( X − Zα ⋅
, − ∞)
n
10
( 50 − 1 .64 ⋅
, − ∞)
25
( 46 .72 , − ∞ )
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Finding Sample Sizes
for Estimating µ
C.I. : x ± z 2 ⋅ σ
n
α
Margin of Error = B = Zα 2 ⋅ σ
n
2
2
zα 2 ⋅ σ
2
n=
B
B = Margin of Error or Bound
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I don’t want to
sample too much
or too little!
Sampling Distribution and Confidence Interval
6-9
Sample Size Example
What sample size is needed to be 90%
confident of being correct within ± 5? A
pilot study suggested that the standard
deviation is 45.
Z.05 σ2 (1.645) (45 )
=
= 219.2 ≅ 220
B2
(5)2
2
n=
2
2
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C.I. Example
You plan to survey employees to find their
average auto insurance. You want to be 95%
confident that the sample mean is within ±
$50..
$50
A pilot study showed that σ was about $400
$400..
What sample size do you use?
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Sample Size Solution*
2
n=
Z0. 0252σ
B
Sampling Distribution of
Proportion
The sampling distribution of sample
proportion, p, if nπ ≥ 5 and n(1
(1−π
−π)) ≥ 5,
is approximately normal, with
2
2
2
1.96) (400)
=(
(50)
2
µp = π ,
= 245. 86 ≅ 246
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1. Assumptions
n
n
π (1 − π )
n
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Confidence Interval
Proportion
n
and σ p =
Two Categorical Outcomes
Population Follows Binomial Distribution
Normal Approximation Can Be Used
l
np ± 3 np(1 − p )
Does Not Include 0 or 1
Estimation Example
Proportion
A random sample of 400 graduates
showed 32 went to grad school. Set up a
95% confidence interval estimate for π.
( p − Zα / 2 ⋅
2. Confidence Interval Estimate
p ⋅ (1 − p )
p ⋅ (1 − p )
( p − zα 2 ⋅
, p + zα 2 ⋅
)
n
n
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( .08 − 1.96 ⋅
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p ⋅ (1 − p )
p ⋅ (1 − p)
, p + Zα / 2 ⋅
)
n
n
.08 ⋅ (1 − .08)
. 08 ⋅ (1 − .08)
, .08 + 1.96 ⋅
)
400
400
( . 053 , . 107 )
Sampling Distribution and Confidence Interval
6-10
Confidence Interval
Solution*
Example
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?
( p − zα / 2 ⋅
( .175 − 1 .645 ⋅
p ⋅ (1 − p)
, p + zα / 2 ⋅
n
.175 ⋅ (. 825)
.175 ⋅ (. 825 )
, .175 + 1.645 ⋅
)
200
200
( .1308 , .2192 )
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p ⋅ (1 − p )
)
n