Download The standard error of the sample mean and

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.
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.”
(Small n, not capital N.)
• Sample statistics based on random samples provide
least squared, unbiased estimates of their population
parameters.
The first way a random sample is
representative of its population
• One way a random sample will be
representative of the population is that the
sample mean will be a good estimate of the
population mean.
• Sample means are better estimates of mu
than are individual scores.
• Thus, on the average, sample means are
closer to mu than are individual scores.
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 drawn from a
single population, as we add scores to each
sample, each sample mean gets closer to
mu.
• Since they are all getting closer to mu, they
will also be getting closer to each other.
As you add scores to a random
sample – larger vs. smaller
samples
• The larger the random samples, the closer
their means will be to mu, on the average.
• The larger the random samples, the closer
their means will be to each other, 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
Which sample means are likely to
be closer to each other.
1. The means of three random
samples, each with an n of 10
2. The means of three random
samples, each with an n of 5
The larger samples should be
closer, on the average, to mu, and
therefore closer to each other.
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 or X
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 = sigma2/n
X
The standard error of the sample mean
• 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
• 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. Here are 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 sample means
from the population mean (mu)
equals the average unsquared
distance of individual scores from
the population mean divided by the
square root of the sample size.
Still another way to say the same thing.
The standard error of the mean is the
standard deviation of the sample means
around mu.
This sometimes confuses people. If it
confuses you, just remember:
The standard error of the mean is the
averaged unsquared distance of sample
means from mu.
Let’s check and make sure that the
formula is correct. Let’s see that the
standard error equals the ordinary
standard deviation divided by the square
root of n. To do that, let’s start with a
tiny population: N=5
• Here are all the scores in a population: 1,3,5,7,9.
• The scores in this population form a perfectly
rectangular distribution.
• Mu = 5.00
Figure 4.5: Scores of the 5 research participants 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
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 the population of 5 scores we just looked at
(sigma = 2.83). We’ll use samples with n = 2.
• If the formula is right, the average unsquared distance of
sample means, n=2, should be 2.83/ 2= 2.83/1.414 = 2.00.
• Is that right? To find out, let’s compute the standard error
of the mean from the differences between the means of all
possible samples (n=2) from mu (the mean from the
population of scores 1,3,5,7,9).
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
• The formula works. And it works every time.
• By the way the mean of the samples
(125/25=5.00 = mu) is the mean of the population.
And that works every time too.
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
• REMEMBER THAT SAMPLE MEANS FALL
CLOSER TO MU, ON THE AVERAGE, THAN DO
INDIVIDUAL SCORES.
What happens when we take
random samples with n=4?
• The standard error of the mean is sigma divided by the
square root of the sample size
= 100/ 4 =100/2=50.00.
These sample means will form a lovely normal curve around
the population mean (mu=500) with a standard deviation
equal to the standard error of the mean = 50.00
• 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
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/ 400 = 100/20=5.00.
• These sample means will form a lovely normal curve around
the population mean (mu=500) with a standard deviation
equal to the standard error of the mean = 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.
You try it with random samples of
verbal SAT scores, with 2500 people in
each sample.
• In what interval can we expect that 68.26%
of the sample means will fall?
• In what interval can we expect 95.44% of
the sample means to fall?
• With n= 2500, the standard error of the mean is sigma
divided by the square root of 2500 = 100/50=2.00.
• So on the verbal part of the SAT, the average (unsquared)
distance of samples (n=2500) is 2.00 points
• 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
A slightly tougher question
• Using SAT scores, with n=2500:
• Into what interval should 95% of the sample
means fall?
A slightly tougher question
• 95% of the sample means should fall within
1.960 standard errors of the mean from mu.
• Given that sigmaX-bar =2.00, you multiply
1.960 * sigmaX-bar = 1.960 x 2.00 = 3.92
Thus:
• 95% of the sample means should fall in an
interval that goes 3.92 points in both
directions around mu
• 500 – 3.92 = 496.08
• 500 + 3.92 = 503.92
• So 95% of sample means (n=2500) should
fall between 496.08 and 503.92
The Central Limit Theorem
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 with n=400 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
You do one.
• What is the 95% confidence interval for
samples of 25 randomly selected IQ scores?
• To do the problem, first compute the
standard error for samples of 25 IQ scores
• IQ: mu =100, sigma = 15
Compute sigmaX-bar
• 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 – Interval into which 95%
of samples n=25 should fall,
sigmaX = 3.00
• 95% of the sample means should fall within
1.960 standard errors of mu = 1.960 (3.00) =
5.88 points
• 100-5.88 = 94.12 and 100+5.88 =105.88
So, 95% of sample means (n=25) for IQ scores
should fall into the range from 94.12 to
105.58.
How do we express that as a 95%
confidence interval?
• 95% of the sample means should fall within 1.960
standard errors of mu = 1.960 (3.00) = 5.88 points
• 100-5.88 = 94.12 and 100+5.88 =105.88
• To express that as a confidence interval, we can
write it in either of two ways
CI.95: mu + 1.960 sigmaX-bar = 100+5.88
or
CI.95: 94.12 < X-bar < 105.88