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Understandable Statistics
Seventh Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Chapter Seven
Introduction to Sampling
Distributions
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1
Review of Statistical Terms
•
•
•
•
Population
Sample
Parameter
Statistic
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2
Population
the set of all measurements (either
existing or conceptual) under
consideration
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3
Sample
a subset of measurements from a
population
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4
Parameter
a numerical descriptive measure of
a population
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5
Statistic
a numerical descriptive measure of
a sample
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6
We use a statistic to make
inferences about a population
parameter.
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7
Principal types of inferences
• Estimate the value of a population
parameter
• Formulate a decision about the value of a
population parameter
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8
Sampling Distribution
a probability distribution for the
sample statistic we are using
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9
Example of a Sampling
Distribution
Select samples with two elements
each (in sequence with
replacement) from the set
{1, 2, 3, 4, 5, 6}.
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10
Constructing a Sampling
Distribution of the Mean for
Samples of Size n = 2
List all samples and compute the mean of each
sample.
sample:
mean:
sample:
mean
{1,1}
{1,2}
{1,3}
{1,4}
{1,5}
1.0
1.5
2.0
2.5
3.0
{1,6}
{2,1}
{2,2}
…
3.5
1.5
4
...
There are 36 different samples.
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11
Sampling Distribution of the Mean
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x
p
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
12
Sampling Distribution
Histogram
6
36
1
36
|
1
|
|
|
|
|
|
|
|
|
|
|
1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
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13
Let x be a random variable
with a normal distribution with
mean  and standard deviation
. Let x be the sample mean
corresponding to random
samples of size n taken from
the distribution .
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14
Facts about sampling
distribution of the mean:
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15
Facts about sampling
distribution of the mean:
• The
x
distribution is a normal distribution.
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16
Facts about sampling
distribution of the mean:
• The x distribution is a normal distribution.
• The mean of the x distribution is  (the
same mean as the original distribution).
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17
Facts about sampling
distribution of the mean:
• The x distribution is a normal distribution.
• The mean of the x distribution is  (the
same mean as the original distribution).
• The standard deviation of the x
distribution is  n (the standard deviation
of the original distribution, divided by the
square root of the sample size).
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18
We can use this theorem to
draw conclusions about means
of samples taken from normal
distributions.
If the original distribution is
normal, then the sampling
distribution will be normal.
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19
The Mean of the Sampling
Distribution
x
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20
The mean of the sampling
distribution is equal to the
mean of the original
distribution.
x  
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21
The Standard Deviation of the
Sampling Distribution
x
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22
The standard deviation of the
sampling distribution is equal to the
standard deviation of the original
distribution divided by the square
root of the sample size.

x 
n
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23
The time it takes to drive between
cities A and B is normally
distributed with a mean of 14
minutes and a standard deviation
of 2.2 minutes.
• Find the probability that a trip between
the cities takes more than 15 minutes.
• Find the probability that mean time of
nine trips between the cities is more than
15 minutes.
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24
Mean = 14 minutes, standard
deviation = 2.2 minutes
• Find the probability that a trip between
the cities takes more than 15 minutes.
15  14
z
 0.45
14 15
2.2
P( z  0.45)  1.00  0.6736  0.3264
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Find this
area
25
Mean = 14 minutes, standard
deviation = 2.2 minutes
• Find the probability that mean time of
nine trips between the cities is more than
15 minutes.
 x    14
 2.2
x 

 0.73
n
9
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26
Mean = 14 minutes, standard
deviation = 2.2 minutes
• Find the probability that mean time of
nine trips between the cities is more than
15 minutes.
Find this
area
15  14
z
 1.37
0.73
14 15
P( z  1.37 )  0.5  0.4147  0.0853
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27
What if the Original
Distribution Is Not Normal?
Use the Central Limit Theorem.
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28
Central Limit Theorem
If x has any distribution with mean  and
standard deviation , then the sample
mean based on a random sample of size n
will have a distribution that approaches
the normal distribution (with mean  and
standard deviation  divided by the
square root of n) as n increases without
bound.
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29
How large should the sample
size be to permit the
application of the Central
Limit Theorem?
In most cases a sample size of
n = 30 or more assures that the
distribution will be approximately
normal and the theorem will apply.
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30
Central Limit Theorem
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31
Central Limit Theorem
• For most x distributions, if we use a
sample size of 30 or larger, the x
distribution will be approximately
normal.
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32
Central Limit Theorem
• The mean of the sampling distribution is
the same as the mean of the original
distribution.
• The standard deviation of the sampling
distribution is equal to the standard
deviation of the original distribution
divided by the square root of the sample
size.
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33
Central Limit Theorem
Formula
x  
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34
Central Limit Theorem
Formula

x 
n
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35
Central Limit Theorem
Formula
z 
x  

x
x
x  

 / n
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36
Application of the Central
Limit Theorem
Records indicate that the packages shipped by a
certain trucking company have a mean weight of
510 pounds and a standard deviation of 90
pounds. One hundred packages are being shipped
today. What is the probability that their mean
weight will be:
a.
b.
c.
more than 530 pounds?
less than 500 pounds?
between 495 and 515 pounds?
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37
Are we authorized to use the
Normal Distribution?
Yes, we are attempting to
draw conclusions about
means of large samples.
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38
Applying the Central Limit
Theorem
What is the probability that their mean weight will
be more than 530 pounds?
Consider the distribution of sample means:
 x  510 ,  x  90 / 100  9
P( x > 530): z = 530 – 510 = 20 = 2.22
9
9
.0132
P(z > 2.22) = _______
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39
Applying the Central Limit
Theorem
What is the probability that their mean weight
will be less than 500 pounds?
P( x < 500): z = 500 – 510 = –10 = – 1.11
9
9
.1335
P(z < – 1.11) = _______
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40
Applying the Central Limit
Theorem
What is the probability that their mean weight
will be between 495 and 515 pounds?
P(495 < x < 515) :
for 495: z = 495 – 510 =  15 =  1.67
9
9
for 515: z = 515 – 510 = 5 = 0.56
9
9
.6648
P(  1.67 < z < 0.56) = _______
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41
Sampling Distributions for
Proportions
Allow us to work with the
proportion of successes rather than
the actual number of successes in
binomial experiments.
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42
Sampling Distribution of the
r
Proportion pˆ  n
•
•
•
•
n= number of binomial trials
r = number of successes
p = probability of success on each trial
q = 1 - p = probability of failure on each
trial
r
ˆ 
 p
is read " p - hat"
n
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43
Sampling Distribution of the
r
ˆ
Proportion p  n
If np > 5 and nq > 5 then p-hat = r/n can be
approximated by a normal random variable (x)
with:
 pˆ  p and  p̂ 
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pq
n
44
The Standard Error for
p̂
The standard deviation of
the p̂ sampling distributi on 
 p̂ 
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pq
n
45
Continuity Correction
• When using the normal distribution
(which is continuous) to approximate phat, a discrete distribution, always use
the continuity correction.
• Add or subtract 0.5/n to the endpoints of
a (discrete) p-hat interval to convert it to
a (continuous) normal interval.
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46
Continuity Correction
If n = 20, convert a phat interval from 5/8 • Since n = 20,
.5/n = 0.025
to 6/8 to a normal
interval.
Note: 5/8 = 0.625
6/8 = 0.75
So p-hat interval is
0.625 to 0.75.
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• 5/8 - 0.025 = 0.6
• 6/8 + 0.025 = 0.775
• Required x interval is
0.6 to 0.775
47
Suppose 12% of the population
is in favor of a new park.
• Two hundred citizen are surveyed.
• What is the probability that between
10 % and 15% of them will be in favor of
the new park?
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48
Is it appropriate to the normal
distribution?
• 12% of the population is in favor of a
new park.
p = 0.12, q= 0.88
• Two hundred citizen are surveyed.
n = 200
• Both np and nq are greater than five.
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49
Find the mean and the
standard deviation
 pˆ  p  0.12
 pˆ 
pq
.12(.88)

 0.023
n
200
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50
What is the probability that
between 10 % and 15%of them
will be in favor of the new park?
• Use the continuity correction
• Since n = 200, .5/n = .0025
• The interval for p-hat (0.10 to 0.15)
converts to 0.0975 to 0.1525.
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51
Calculate z-score for x = 0.0975
0.0975  0.12
z
 0.98
0.023
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52
Calculate z-score for x = 0.1525
0.1525  0.12
z
 1.41
0.023
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53
P(-0.98 < z < 1.41)
0.9207 -- 0.1635 = 0.7572
There is about a 75.7% chance that
between 10% and 15% of the
citizens surveyed will be in favor of
the park.
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54
Control Chart for Proportions
P-Chart
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55
Constructing a P-Chart
• Select samples of fixed size n at regular
intervals.
• Count the number of successes r from the
n trials.
• Use the normal approximation for r/n to
plot control limits.
• Interpret results.
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56
Determining Control Limits for
a P-Chart
• Suppose employee absences are to be
plotted.
• In a daily sample of 50 employees, the
number of employees absent is recorded.
• p/n for each day = number absent/50.For
the random variable p-hat = p/n, we can
find the mean and the standard
deviation.
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57
Finding the mean and the
standard deviation
Suppose  pˆ  p  0.12
then  pˆ 
pq
.12(.88)

 0.046
n
50
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58
Is it appropriate to use the
normal distribution?
•
•
•
•
•
The mean of p-hat = p = 0.12
The value of n = 50.
The value of q = 1 - p = 0.88.
Both np and nq are greater than five.
The normal distribution will be a good
approximation of the p-hat distribution.
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59
Control Limits
Control limits are placed at two and three
standard deviations above and below the
mean.
pq
0.12(0.88)
p2
 0.12  2
 0.12  0.092
n
50
pq
0.12(0.88)
p3
 0.12  3
 0.12  0.138
n
50
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60
Control Limits
The center line is at 0.12.
Control limits are placed at -0.018, 0.028,
0.212, and 0.258.
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61
Control Chart for Proportions
Employee Absences
0.3
+3s = 0.258
0.2
+2s = 0.212
0.1
mean = 0.12
0.0
-2s = 0.028
-0.1
-3s = -0.018
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62
Daily absences can now be
plotted and evaluated.
Employee Absences
0.3
+3s = 0.258
0.2
+2s = 0.212
0.1
mean = 0.12
0.0
-2s = 0.028
-0.1
-3s = -0.018
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63