Estimation
Statistics with Confidence
Estimation
Before we collect our sample, we know:
Repeated sampling sample means would stack up in a normal curve,
Centered on the true population mean,
With a standard error (measure of dispersion) that depends on
1. population standard deviation
2. sample size

-3z
-2z
-1z
0z
1z
2z
3z
What are they doing?

Estimation
But we do not know:
1. True Population Mean
2. Population Standard Deviation
Repeated sampling sample means would stack up in a normal curve,
Centered on the true population mean,
With a standard error (measure of dispersion) that depends on
1. population standard deviation
2. sample size

-3z
-2z
-1z
0z
1z
2z
3z
Estimation
Will our sample be one of these (accurate)?
Or one of these (inaccurate)?

-3z
-2z
-1z
0z
1z
2z
3z
Estimation
Which is more likely?
accurate?
or inaccurate?

-3z
-2z
-1z
0z
68%
95%
1z
2z
3z
Estimation
We’re most likely to get close to the true
population mean…
Our sample’s mean is the best guess of the
population mean, but it is not precise.

-3z
-2z
-1z
0z
68%
95%
1z
2z
3z
Estimation
And if we increase our sample size (n)…

-3z
-2z
-1z
0z
68%
95%
1z
2z
3z
Estimation
And if we increase our sample size our sample mean is an
even better estimate of the
population mean, we are
more precise!
-3
-3z
-2z
-1z
-2
-1

0
0z
68%
95%
1
2 3
1z
2z
3z
Estimation
We know that the standard deviation of this pile of samples
(standard error) equals the population standard deviation
() divided by the square
root of the sample size (n).
-3
-2
-1

0
68%
95%
1
2 3
Estimation
But we do not know the population standard deviation!
What is our best guess
of that?
-3
-2
-1

0
68%
95%
1
2 3
Estimation
Our best guess of the population standard deviation is our
sample’s s.d.! On average, this s.d. gives population .
In fact, when we calculate that,
we use “n – 1” to make our
“estimate” larger to reflect
that dispersion of a sample
 (Yi – Y)2
s=
n-1
is smaller than a population’s.
= Cases in the sample
0 5 10 15 20 25 30 35
Population Dispersion
0 5 10 15 20 25 30 35
Sample Dispersion
Estimation
So now we know that we can use the sample standard deviation to stand in for the
population’s standard deviation.
So we can use the formula
for standard error with that
estimate and get a good estimate
of the dispersion of the
sampling distribution.
-3
-2
s.e =
-1

0
68%
95%
1
2 3
s
n
Estimation
Now we know some limits on how far off our sample
mean is likely to be from the true population mean!
68% of means will
be within +/- 1 s.e.
s
95% of means will
be within +/- 2 s.e.
-3
s.e. =
-2
-1

0
68%
95%
1
2 3
n
Estimation
For example, if we took GPAs from a sample of 625
students and our s was .50…
68% of means would
be within +/- 1*(.02)
.5
95% of means would
be within +/- 2*(.02)
s.e. =
-3
-2
-1

0
1
2 3
[0.02]
68%
95%
 625 = 0.02
Estimation
GPAs from a sample of 625 students with s = .50…
If our sample were
this one,
our estimate of
the mean would
be correct!
.5
s.e. =
-3
-2
-1

0
68%
95%
1
2 3
 625 = 0.02
Estimation
GPAs from a sample of 625 students with s = .50…
But what if it were
this one?
.5
We’d be slightly
wrong, but well within
+/- 2 *(.02)
s.e. =
-3
95% of samples would be!
-2
-1

0
68%
95%
1
2 3
 625 = 0.02
Estimation
A sample’s mean is the
best estimate of the
population mean.
But what if we base
our estimate on this
erroneous sample?
s
s.e. =
-3
-2
-1

0
68%
95%
1
2 3
n
Estimation
Let’s create a “measuring device” with our sampling
distribution and center it over our sample’s mean.
s
Check it Out!
s.e. =
The true mean falls within
the 95% bracket.
-3
-2
-1
0
1

2 3
68%
95%
n
Estimation
What if the sample we collected were this one?
…and we used the measuring device again?
s
Check it Out!
s.e. =
The true mean falls within
the 95% bracket.
-3
-2

-1
0
68%
95%
1
2 3
n
Estimation
The sampling distribution allows us to:
1. Be humble and admit that our sample
statistic may not be the population’s
and 2. Forms a measuring device
with which we can determine a range
where the true population mean
is likely to fall...
this is called a confidence interval.
Estimation
If you calculate your sampling distribution’s standard error,
you can form a device that tells you that
if your sample mean
is wrong, there is a
documented a range in
which the true
s
population mean is likely
s.e. =  n
2Xist.
Sample
Check it Out!
The true mean falls within
the 95% bracket.
-3
-2
-1
0

1
68%
95%
2 3
Estimation
For example, if we took GPAs from a sample of 625 students
and our mean was 2.5 and s.d. was .50…
We make a confidence
interval (C.I.)by…
.5
s.e. =
Calculating the s.e. (.02)
and
-3 -2
Going +/- 2 * s.e.
from the mean.
-1
0
68%
 625 = 0.02
=2.52
1
2 3
95%
95% C.I. = 2.5 +/- 2(.02) = 2.46 to 2.54
We are 95% confident that the true mean is in
this range!
Estimation
Guys… This is power!
Knowing that the spread of
95% of normally distributed
sample means has outer
limits…
We know that if we put these
limits around our sample
mean…
We have defined the range
where the population mean
has a 95% probability of
being!
Estimation
Our sample statistics provide enough
information to give us a great
estimation (highly educated guess)
about population statistics.
We do this without needing to know the
population mean—without needing to
have a census.
Estimation
Another Example:
Sample of 2,500 with an average income of $28,000
with a standard deviation of $8,000.
Provide a 95% C.I. = M +/- 2 * (s.e.)
s
1.
s.e. = $8,000/2,500 = $160
s.e. =  n
2.
2 * $160 = $320
3.
C.I. = $28,000 +/- $320
C.I. >>> $27,680 to $28,320
Estimation
Another Example:
Sample of 2,500 with an average income of $28,000
with a standard deviation of $8,000.
Provide a 95% C.I. = M +/- 2 * (s.e.)
1.
s.e. = $8,000/2,500 = $160
2.
2 * $160 = $320
3.
C.I. = $28,000 +/- $320
We are 95% confident
C.I. >>> $27,680 to $28,320
that the true mean falls
from $27,680 up to
$28,320
Estimation
NO WAIT! We’re wrong!
Technically speaking, on a normal curve,
95% of cases fall between +/- 1.96
standard deviations rather than 2.
(Check your book’s table.)
Empirical Rule
vs.
Actuality
68%
1z
0.99z
95%
2z
1.96z
99.9973%
3z
3z
Estimation
Another Example:
Sample of 2,500 with an average income of $28,000
with a standard deviation of $8,000.
Provide a 95% C.I. = M +/- 1.96 * (s.e.)
1.
s.e. = $8,000/2,500 = $160
2.
1.96 * $160 = $313.6
3.
C.I. = $28,000 +/- $313.6
We are 95% confident
C.I. >>> $27,686.4 to $28,313.6
that the population mean
falls between $27,686.4
and $28,313.6
Estimation
Another Example:
Sample of 2,500 with an average income of
$28,000 with a standard deviation of
$8,000.
What if we want a 99% confidence interval,
What z do we use?
Check the table in your book!
Estimation
Another Example:
Sample of 2,500 with an average income of
$28,000 with a standard deviation of
$8,000.
What if we want a 99% confidence interval?
99% fall between +/- 2.58 z’s
Estimation
Another Example:
Sample of 2,500 with an average income of $28,000 with a standard deviation of $8,000.
What if we want a 99% confidence interval?
1.
2.
3.
s.e. = $8,000/2,500 = $160
2.58 * $160 = $412.8
C.I. = $28,000 +/- $412.8
CI >>> $27,587.2
to
$28,412.8
We are 99% confident that the population mean falls between these values.
Why did the interval get wider than 95% CI’s which was $27,686.4 to $28,313.6???
Estimation
99% CI >>> $27,587.2to
$28,412.8
Why did the interval get wider than 95% CI’s which was
$27,686.4 to $28,313.6???
-3
-2
-1
M
0
68%
95%
1
2 3
99%
…
Estimation
Let’s recap: We can say that 95% of the sample means in repeated sampling
will always be in the range marked by -1.96 over to +1.96 standard errors.
Self-esteem
15
20
25
30
35
40
1.96
Z-3 -2 -1 0 1 2 3
-1.96
z -3
-2
95% Range
-1
0
1
2
3
Estimation
And remember: If we
don’t know the true
population mean,
95% of the time a
95% confidence
interval would
contain the true
population mean!
Self-esteem
15
20
25
30
35
40
95% Ranges for different samples.
Estimation
If we want that range to
contain the true
population mean 99% of
the time (99%
confidence interval) we
just construct a wider
interval, corresponding
with 2.58 z’s.
Self-esteem
15
20
25
30
35
40
99% Ranges for different samples, overlaying 95% intervals.
Estimation
1.96z
The sampling distribution’s
standard error is a
measuring stick that we can
use to indicate the range of
a specified middle
percentage of sample means
in repeated sampling.
95%
1z
68%
3z
-3 -1.96 -1
25
0
99.99%
1 1.96 3
68%
95%
99.99%
Estimation
Another Confidence Interval Example:
I collected a sample of 2,500 with an average self-esteem score of 28 with a standard
deviation of 8.
What if we want a 99% confidence interval? CI = Mean +/- z * s.e.
1.
2.
3.
Find the standard error of the sampling distribution:
s.d. / n = 8/50 = 0.16
Build the width of the Interval. 99% corresponds with a z of 2.58.
2.58 * 0.16 = 0.41
Insert the mean to build the interval:
99% C.I. = 28 +/- 0.41
The interval: 27.59 to 28.41
We are 99% confident that the population mean falls between these values.
Estimation
And if we wanted a 95% Confidence Interval instead?
I collected a sample of 2,500 with an average self-esteem score of 28 with a standard deviation of 8.
X
95%
What if we want a 99% confidence interval? CI = Mean +/- z * s.e.
Find the standard error of the sampling distribution:
s.d. / n = 8/50 = 0.16
1.
95%
X
1.96
X
Build the width of the Interval. 99% corresponds with a z of 2.58.
2.58 * 0.16 = 0.41
2.
X
1.96
3.
95%
X
0.31
Insert the mean to build the interval:
99% C.I. = 28 +/- 0.41
X
The interval:
X
X 0.31
27.59 to 28.41
X
X
27.69 to 28.31
We are 99% confident that the population mean falls between these values.
95%
Estimation
By centering my sampling distribution’s +/1.96z range around my sample’s mean...


I can identify a range that, if my sample is
one of the middle 95%, would contain the
population’s mean.
Or
I have a 95% chance that the population’s
mean is somewhere in that range.
Estimation
By centering my sampling distribution’s +/1.96z
X range around my sample’s mean...
2.58z


I can identify a range that, if my sample is
one of the middle 95%,
X would contain the
population’s mean. 99%
Or
99%
I have a 95%
X chance that the population’s
mean is somewhere in that range.