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Chapter 18 The Diversity of Samples
from the Same Populations
population – set of all possible
measurements
sample – a subset of the population
parameter – a numerical measure of a
population
statistic – a numerical measure of a
sample
Since the population is often not
available, we use statistics to estimate
parameters
In statistical application, we take a
random sample from the population and
compute a statistic, say x
The value of the statistic x depends on
which items are selected for the sample
Therefore, x is a random variable …
different samples yield different values
of x
The probability of a statistic over all
possible samples is known as its
sampling distribution
Sampling Distribution of the Sample
Mean
The statistic x estimates the population
mean 
We hope x is very close to 
We want the sampling distribution to be
centered at the value of the parameter
and to have little variation.
Facts about sampling distribution of x
x  
The average value of x across all
possible samples in  , the population
mean
x 

n
The standard deviation of the
sampling distribution of x is the
population standard deviation 
divided by n
Notice that as n increases the sample to
sample variability in x decreases
If our sample comes from a normal
distribution with mean  and standard
x
deviation  then Z 
has a

n
standard normal distribution
Central Limit Theorem
If we sample from a population with
mean  and standard deviation  then
x
is approximately standard
Z

n
normal for large n .
If n  30 or larger, the central limit
theorem will apply in almost all cases
Example
A population of soft drink cans has
amounts of liquid following a normal
distribution with   12 and   0.2 oz.
What is the probability that a single can
is between 11.9 and 12.1 oz.
P(11.9  X  12.1)
 P(0.5  Z  0.5)
 .69  .31
 .38
What is the probability that x is between
11.9 and 12.1 for n = 16 cans
P (11.9  x  12.1)
 P (2  Z  2)
 .975  .025
 .95
Example
A population of trees have heights with a
mean of 110 feet and a standard
deviation of 20 feet.
A sample of 100 trees is selected
Find  x
 x    110
Find  x

20
x 

2
n
100
Find P( x  108 feet)
P ( x  108)
 P ( Z  1)
 1  .16
 .84
Notice this probability for x is
approximately correct even though the
population is not normally distributed
(because of central limit theorem)
What about P( X  108) ?
This cannot be done for x since we do
not know that the population is normally
distributed.
Sampling Distribution of the Sample
Proportion
Population Proportion
# in population with characteristic
p
# in population
Sample Proportion
# in sample with characteristic
pˆ 
n
p̂ is a point estimate of p
 pˆ  p
 pˆ 
p1  p 
n
If we sample from a population with a
proportion of p, then Z 
pˆ  p
is
p1  p 
n
approximately standard normal for large
n.
Example
Suppose the president’s approval rating
is 56% and we look at samples of size
100. Find the following.
Find  p̂
 pˆ  p  .56
Find  p̂
p1  p 
.561  .56

n
100
.2464

 .002464  .0496
100
 pˆ 
Example
A survey of 120 registered voters yields
54 who plan to vote for the republican
candidate.
p = proportion of all voters who plan to
vote for the republican candidate
pˆ 
54
 0.45  45%
120
Do you think there is much of a chance
that the republican candidate will get at
least 50% of the vote?
Calculate the margin of error
1
120
 .0913
Think about the variance of the sampling
distribution
pˆ 1  pˆ 
.451  .45
.2475


 .0454
n
120
120
The empirical rule says that 95% of data
should be within 2 standard deviations
2.0454   .0908
Do you see where the margin of error
comes from?