Download The standard error of the sample mean and confidence

The standard error of the sample
mean and confidence intervals
How far is the average sample mean
from the population mean?
In what interval around mu can we
expect to find 95% or 99% or sample
means
An introduction to random samples
• When we speak about samples in statistics, we are
talking about random samples.
• Random samples are samples that are obtained in
line with very specific rules.
• If those rules are followed, the sample will be
representative of the population from which it is
drawn.
• One way that it will be representative of the
population is that the sample mean will be close to
the population mean.
• Specifically, on the average, sample means are
closer to mu than are individual scores.
Random samples: Some principles
• In a random sample, each and every score must have
an equal chance of being chosen each time you add a
score to the sample.
• Thus, the same score can be selected more than once,
simply by chance. (This is called sampling with
replacement.)
• The number of scores in a sample is called “n.”
• Sample statistics based on random samples provide
least squared, unbiased estimates of their population
parameters.
The variance and the standard deviation
are the basis for the rest of this chapter.
• In Chapter 1 you learned to compute the
average squared distance of individual
scores from mu. We called it the variance.
• Taking a square root, you got the standard
deviation.
• Now we are going to ask a slightly different
question and transform the variance and
standard deviation in another way.
As you add scores to a random
sample
• Each randomly selected score tends to
correct the sample mean back toward mu
• If we have several samples, as we add
scores the sample means get closer to each
other and closer to mu
• The larger the samples, the closer they will
be to mu, on the average.
Let’s see how that happens
Population is 1320 students taking a test.
 is 72.00,  = 12
Let’s randomly sample one student at a time and see what
happens.We’ll create a random sample with 8 students’ scores in the
sample.
Test Scores
F
r
e
q
u
e
n
c
y
Standard
deviations
Scores
Mean
score
3
2
1
0
1
2
3
36
48
60
72
84
96
108
Sample scores:
102
Means:
72
87
66
80
76
79
66
76.4
78
69
76.7 75.6
63
74.0
How much closer to mu does the sample mean get
when you increase n, the size of the sample? (1)
• The average squared distance of individual
scores is called the variance. You learned to
compute it in Chapter 1.
• The symbol for the mean of a sample is the
letter X with a bar over it.We will write that
as X-bar.
How much closer to mu does the sample mean get
when you increase n, the size of the sample? (2)
• The average squared distance of sample
means from mu is the average squared
distance of individual scores from mu
divided by n, the size of the sample.
• Let’s put that in a formula
• sigma2X-bar = sigma2/n
Let’s take that one step further
• As you know, the square root of the variance is
called the standard deviation. It is the average
unsquared distance of individual scores from mu.
• The average unsquared distance of sample means
from mu is the square root of
sigma2X-bar = sigmaX-bar.
sigmaX-bar is called the standard error of the sample
mean or, more briefly, the standard error of the
mean. Let’s look at the formulae:
sigma2X-bar = sigma2/n
sigmaX-bar = sigma/ n
The standard error of the mean
• Let’s translate the formula into English, just to be
sure you understand it. Here is the formula again:
sigmaX-bar = sigma/ n
• In English: The standard error of the sample mean
equals the ordinary standard deviation divided by
the square root of the sample size.
• Another way to say that: The average unsquared
distance of the means of random samples from mu
equals the average unsquared distance of
individual scores from the population mean
divided by the square root of the sample size.
The standard error of the mean is
the standard deviation of the
sample means around mu.
• We could compute the average unsquared distance of
sample means from mu by 1. subtracting mu from
each sample mean. 2. squaring the differences, 3.
getting a sum of squares 4. dividing by the number
of sample means and 5. taking the square root.
• We would need to do that for all possible samples of
a particular size from a population. That’s a lot of
calculations. (A real lot.)
Example: Start with a tiny population
N=5
• The scores in this population
form a perfectly
•
•
•
•
rectangular distribution.
Mu = 5.00
Sigma = 2.83
We are going to list all the possible samples of
size 2 (n=2)
First see the population, then the list of
samples
If we did compute a standard deviation
of sample means from mu, it should give
the same result as the formula
• Let’s see if it does.
• We can only do all the computations if we
have a very small population and an even
tinier sample.
• Let’s use an example with N=5 and n, the
size of each sample = 2.
Figur e 4.5: Scores of the 5 research pa rticipants in this population.
Frequency
3
2
1
x
1.00
x
2.00
3.00
x
4.00
5.00
x
6.00
7.00
x
8.00
9.00
Computing sigma
• SS=(1-5)2+(3-5)2+(5-5)2+ (7-5)2+ (9-5)2=40
• sigma2=SS/N=40/5=8.00
• sigma = 2.83
Figure 4.6: Means of all possible 25 samples (n=2) from this population
Frequency
5
4
3
2
1
Score
X
1.00
X
X
2.00
X
X
X
3.00
X
X
X
X
4.00
X
X
X
X
X
X
X
X
X
5.00
X
X
X
6.00
X
X
7.00
X
8.00
9.00
Table 4.10: List of all 25 possible samples (n=2 ) of sco res from the tiny population o f
five scores shown in Table 4.9 and Figure 4.5 and direct computation of the standa rd
error of thes e means. In this case, the standa rd error is computed from these means
(n=2) just as we would the stand ard dev iation o f an ordinary set of scores (n=1 ).
Sample Scores
AA
AB
AC
AD
AE
BA
BB
BC
BD
BE
CA
CB
CC
CD
CE
1,1
1,3
1,5
1,7
1,9
3,1
3,3
3,5
3,7
3,9
5,1
5,3
5,5
5,7
5,9
X
1.00
2.00
3.00
4.00
5.00
2.00
3.00
4.00
5.00
6.00
3.00
4.00
5.00
6.00
7.00
Sample Scores
DA
DB
DC
DD
DE
EA
EB
EC
ED
EE
7,1
7,3
7,5
7,7
7,9
9,1
9,3
9,5
9,7
9,9
X
4.00
5.00
6.00
7.00
8.00
5.00
6.00
7.00
8.00
9.00
Summary statisti cs
(all samples, n=2 )
 X = 125.00
N = 25
mu = 5.00
SS X = 100.00
sigma X 2 = 4.00
sigma X = 2.00
The standard error = the standard
deviation divided by the square
root of n, the sample size
• In the example you just saw, sigma = 2.83.
Divide that by the square root of n (1.414)
and you get the standard error of the mean
(2.00).
• The formula works. And it works every
time.
Let’s see what sigmaX-bar can tell us
• We know that the mean of SAT/GRE scores = 500 and sigma
= 100
• So 68.26% of individuals will score between 400 and 600 and
95.44% will score between 300 and 700
• But if we take random samples of SAT scores, with 4 people
in each sample, the standard error of the mean is sigma
divided by the square root of the sample size = 100/2=50.
• 68.26% of the sample means (n=4) will be within 1.00
standard error of the mean from mu and 95.44% will be within
2.00 standard errors of the mean from mu
• So, 68.26% of the sample means (n=4) will be between 450
and 550 and 95.44% will fall between 400 and 600
• NOTE THAT SAMPLE MEANS FALL CLOSER TO MU,
ON THE AVERAGE, THAN DO INDIVIDUAL SCORES.
Let’s make the samples larger
• Take random samples of SAT scores, with 400 people in each sample, the
standard error of the mean is sigma divided by the square root of 400 =
100/20=5.00
• 68.26% of the sample means will be within 1.00 standard error of the mean
from mu and 95.44% will be within 2.00 standard errors of the mean from
mu.
• So, 68.26% of the sample means (n=400) will be between 495 and 505 and
95.44% will fall between 490 and 510.
• Take random samples of SAT scores, with 2500 people in each sample, the
standard error of the mean is sigma divided by the square root of 2500 =
100/50=2.00.
• 68.26% of the sample means will be within 1.00 standard error of the mean
from mu and 95.44% will be within 2.00 standard errors of the mean from
mu.
• 68.26% of the sample means (n=2500) will be between 498 and 502 and
95.44% will fall between 496 and 504
What happens as n increases?
• The sample means get closer to each other and to mu.
• Their average squared distance from mu equals the
variance divided by the size of the sample.
• Therefore, their average unsquared distance from mu
equals the standard deviation divided by the square
root of the size of the sample.
• The sample means fall into a more and more perfect
normal curve.
• These facts are called “The Central Limit
Theorem” and can be proven mathematically.
CONFIDENCE INTERVALS
We want to define two intervals
around mu:
One interval into which 95% of
the sample means will fall.
Another interval into which 99%
of the sample means will fall.
95% of sample means will fall in a symmetrical
interval around mu that goes from 1.960 standard
errors below mu to 1.960 standard errors above mu
• A way to write that fact in statistical language is:
CI.95: mu + 1.960 sigmaX-bar
or
CI.95: mu - 1.960 sigmaX-bar < X-bar < mu + 1.960 sigmaX-bar
As I said, 95% of sample means will fall in a
symmetrical interval around mu that goes from 1.960
standard errors below mu to 1.960 standard errors
above mu
• Take samples of SAT/GRE scores (n=400)
• Standard error of the mean is sigma divided by the square root
of n=100/ 400 = 100/20.00=5.00
• 1.960 standard errors of the mean with such samples = 1.960
(5.00)= 9.80
• So 95% of the sample means can be expected to fall in the
interval 500+9.80
• 500-9.80 = 490.20 and 500+9.80 =509.80
CI.95: mu + 1.960 sigmaX-bar = 500+9.80
CI.95: 490.20 < X-bar < 509.20
or
99% of sample means will fall within 2.576
standard errors from mu
• Take the same samples of SAT/GRE scores (n=400)
• The standard error of the mean is sigma divided by the square
root of n=100/20.00=5.00
• 2.576 standard errors of the mean with such samples =
2.576 (5.00)= 12.88
• So 99% of the sample means can be expected to fall in the
interval 500+12.88
• 500-12.88 = 487.12 and 500+12.88 =512.88
CI.99: mu + 2.576 sigmaX-bar = 500+12.88
or
CI.99: 487.12 < the sample mean < 512.88
Let’s do another one.
• What are the 95% and 99% confidence intervals
for samples of 25 randomly selected IQ scores
• First compute the standard error for samples of 25
IQ scores
• IQ: mu =100, sigma = 15
• Standard error for samples of size 25 is 15 divided
by the square root of 25 = 15/5.00 =3.00
IQ scores – CI.95 and CI.99 for
samples n=25
• The standard error of the mean is sigma divided by the square root of
n=15/5.00=3.00
• 1.960 standard errors of the mean with such samples =
1.960 (3.00)= 5.88 points
• So, 95% of the sample means (n=25) can be expected to fall in the
interval 100 + 5.88
• 100-5.88 = 94.12 and 100+5.88 =105.88
CI.95: mu + 1.960 sigmaX-bar = 100+5.88
or
CI.95: 94.12 < X-bar < 105.88
99% of the sample means (n=25) can be expected to fall in the interval
100 + (2.576)(3.00) = 100 + 7.73
CI.99: 100+7.73
or
CI.99: 92.27 < X-bar < 107.73
Here is another example. This time we
start with an even smaller population
(N=4) and take all possible samples of
size 3. There are 64 of them. Let’s see
that again the means form a normal
curve around mu and the standard error
equals sigma divided by the square root
of the sample size (3).
Even tinier population (N=4)
Frequency
1
x
Score1.00
x
x
x
2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00
Sample Scores
X
Sample
AAA
AAB
AAC
AAD
ABA
ABB
ABC
ABD
ACA
ACB
ACC
ACD
ADA
ADB
ADC
ADD
BAA
BAB
BAC
BAD
BBA
BBB
BBC
BBD
BCA
BCB
BCC
BCD
BDA
BDB
BDC
BDD
1.00
2.00
3.00
4.00
2.00
3.00
4.00
5.00
3.00
4.00
5.00
6.00
4.00
5.00
6.00
7.00
2.00
3.00
4.00
5.00
3.00
4.00
5.00
6.00
4.00
5.00
6.00
7.00
5.00
6.00
7.00
8.00
CAA
CAB
CAC
CAD
CBA
CBB
CBC
CBD
CCA
CCB
CCC
CCD
CDA
CDB
CDC
CDD
DAA
DAB
DAC
DAD
DBA
DBB
DBC
DBD
DCA
DCB
DCC
DCD
DDA
DDB
DDC
DDD
1,1,1
1,1,4
1,1,7
1,1,10
1,4,1
1,4,4
1,4,7
1,4,10
1,7,1
1,7,4
1,7,7
1,7,10
1,10,1
1,10,4
1,10,7
1,10,10
4,1,1
4,1,4
4,1,7
4,1,10
4,4,1
4,4,4
4,4,7
4,4,10
4,7,1
4,7,4
4,7,7
4,7,10
4,10,1
4,10,4
4,10,7
4,10,10
ScoresX
7,1,1
3.00
7,1,4
4.00
7,1,7
5.00
7,1,10
6.00
7,4,1
4.00
7,4,4
5.00
7,4,7
6.00
7,4,10
7.00
7,7,1
5.00
7,7,4
6.00
7,7,7
7.00
7,7,10
8.00
7,10,1
6.00
7,10,4
7.00
7,10,7
8.00
7,10,10
9.00
10,1,1
4.00
10,1,4
5.00
10,1,7
6.00
10,1,10
7.00
10,4,1
5.00
10,4,4
6.00
10,4,7
7.00
10,4,10
8.00
10,7,1
6.00
10,7,4
7.00
10,7,7
7.00
10,7,10
9.00
10,10,1
7.00
10,10,4
8.00
10,10,7
9.00
10,10,10 10.00
Summary statistics
(all samples, n=3)
X
= 352.00
X
N = 64
X mu
= 5.50
X
SS = 240.00
2
X
Sigma
= 3.75
X
Sigma = 1.94
12
11
10
9
8
7
6
5
4
3
2
1
X
Score 1.00
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00 10.00
X
Standard error of the mean - 2
• The standard deviation of the individual scores
was 3.35
• Sample size was 3
• 3.35 divided by the square root of 3 = 1.94
• Computing the standard error directly from the
sample means shows the standard error = 1.94