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Sampling Distribution
WELCOME to INFERENTIAL STATISTICS
A Sampling Distribution


We are moving from descriptive statistics
to inferential statistics.
Inferential statistics allow the researcher
to come to conclusions about a
population on the basis of descriptive
statistics about a sample.
For example:
A Sampling Distribution



Your sample says that a candidate gets support from
47%.
Inferential statistics allow you to say that the
candidate gets support from 47% of the population
with a margin of error of +/- 4%.
This means that the support in the population is likely
somewhere between 43% and 51%.
A Sampling Distribution

Margin of error is taken directly from a
sampling distribution.
95% of Possible Sample Means
It looks like this:
43%
47%
51%
Your Sample Mean
A Sampling Distribution


Let’s create a sampling distribution of means…
Take a sample of size 1,500 from the US. Record the mean
income. Our census said the mean is $30K.
$30K
A Sampling Distribution


Let’s create a sampling distribution of means…
Take another sample of size 1,500 from the US. Record the
mean income. Our census said the mean is $30K.
$30K
A Sampling Distribution


Let’s create a sampling distribution of means…
Take another sample of size 1,500 from the US. Record the
mean income. Our census said the mean is $30K.
$30K
A Sampling Distribution


Let’s create a sampling distribution of means…
Take another sample of size 1,500 from the US. Record the
mean income. Our census said the mean is $30K.
$30K
A Sampling Distribution


Let’s create a sampling distribution of means…
Take another sample of size 1,500 from the US. Record the
mean income. Our census said the mean is $30K.
$30K
A Sampling Distribution


Let’s create a sampling distribution of means…
Take another sample of size 1,500 from the US. Record the
mean income. Our census said the mean is $30K.
$30K
A Sampling Distribution


Let’s create a sampling distribution of means…
Let’s repeat sampling of sizes 1,500 from the US. Record the
mean incomes. Our census said the mean is $30K.
$30K
A Sampling Distribution


Let’s create a sampling distribution of means…
Let’s repeat sampling of sizes 1,500 from the US. Record the
mean incomes. Our census said the mean is $30K.
$30K
A Sampling Distribution


Let’s create a sampling distribution of means…
Let’s repeat sampling of sizes 1,500 from the US. Record the
mean incomes. Our census said the mean is $30K.
$30K
A Sampling Distribution


Let’s create a sampling distribution of means…
Let’s repeat sampling of sizes 1,500 from the US. Record the
mean incomes. Our census said the mean is $30K.
The sample means would stack up in
a normal curve. A normal sampling
distribution.
$30K
A Sampling Distribution


Say that the standard deviation of this distribution is $10K.
Think back to the empirical rule. What are the odds you would
get a sample mean that is more than $20K off.
The sample means would stack up in
a normal curve. A normal sampling
distribution.
$30K
-3z
-2z
-1z
0z
1z
2z
3z
A Sampling Distribution


Say that the standard deviation of this distribution is $10K.
Think back to the empirical rule. What are the odds you would
get a sample mean that is more than $20K off.
The sample means would stack up in
a normal curve. A normal sampling
distribution.
2.5%
2.5%
$30K
-3z
-2z
-1z
0z
1z
2z
3z
Central Limit Theorem
(CLT)
Central Limit Theorem: As sample size
increases, the sampling distribution of sample
means approaches that of a normal
distribution with a mean the same as the
population and a standard deviation equal to
the standard deviation of the population
divided by the square root of n (the sample
size).
N(ℳ , σ/√n) with mean ℳ and sd σ/√n
Variability in Sampling Distribution
For example, if the variability is low,
, we can
trust our number more than if the variability is high,
.
•
An Example:
•
A population’s car values are  = $12K with  = $4K.
•
Which sampling distribution is for sample size 625 and
which is for 2500? What are their s.e.’s (standard error)?
95% of M’s
-3
-2
?
$12K
-1
0
?
1
2
95% of M’s
3
? $12K ?
-3-2-1 0 1 2 3
•
An Example:
•
A population’s car values are  = $12K with  = $4K.
•
Which sampling distribution is for sample size 625 and which is for
2500? What are their s.e.’s?
•
(2500 = 50)
(625 = 25)
s.e. = $4K/50 = $80
s.e. = $4K/25 = $160
95% of M’s
-3
-2
?
$12K
-1
0
?
1
2
95% of M’s
3
? $12K ?
-3-2-1 0 1 2 3
Which sample will be more precise? If you get a
particularly bad sample, which sample size will
help you be sure that you are closer to the true
mean?
95% of M’s
-3
-2
?
$12K
-1
0
?
1
2
95% of M’s
3
? $12K ?
-3-2-1 0 1 2 3
So we know in advance of ever collecting a sample, that if
sample size is sufficiently large:

Repeated samples would pile up in a normal distribution

The sample means will center on the true population mean

The standard error will be a function of the population
variability and sample size

The larger the sample size, the more precise, or efficient, a
particular sample is

95% of all sample means will fall between +/- 2 s.e. from the
population mean
What proportion of US teens know that 1492 was
the year in which Columbus “discovered”
America? A Gallup Poll fund that 210 out of a
random sample of 501 American teens aged 1317 knew this historically important date. The
sample proportion:
p
= 210/501 = 0.42
0.42 is the statistic that we use to gain information
about the unknown population parameter p. We
may say that 42% of US teens know that
Columbus discovered America in 1492.
Sampling distribution of
sample proportion
p
=
Count of success in sample
Size of the sample
=
X
n
The mean of the sampling distributionp is exactly p
The standard deviation of the sampling distribution
p
is
p(1-p)
n
√
Applying to college
p
Normal calculation involving
A polling organization asks an SRS (simple random sample) of
1500 1st year college students whether they applied for
admission to any other college. In fact 35% of all the 1st year
students applied to colleges besides the one they are attending.
What is the probability that the random sample of 1500 students
will give a result within 2 percentage point of this true value?
n=1500
p=0.35
ℳ p =0.35
σ= √
=
p(1-p)
n
√
0.35(1-0.35)
1500
=
0.0123
Sampling Distribution
Jeremy, out of boredom, decided to find
the probability of a male student being
72 inches tall in BHS. Mr. Delton told
him that the average height of 857 male
students in BHS is 67 inches with a
standard deviation of 3.5 inches. Show
a statistical procedure on how to help
Jeremy on his quest of getting rid of his
boredom.
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