The standard normal distribution is a normal distribution with
a mean of 0 and a standard deviation of 1. Normal
distributions can be transformed to standard normal
distributions by the formula:
Z=X−μσ
Z
X
μ
σ
(1)
where
XX is a score from the
original normal distribution,
μμ is the mean of the original
normal distribution, and
σσ
is the standard deviation of original normal distribution.
Given a population with a mean of
mm and a standard deviation of
σσ, the sampling
distribution of the mean has a mean of
mm and a standard deviation of
σN
σ
N
, where NN is the sample
size. The standard deviation of the sampling distribution of
the mean is called the standard error of the
mean.It is designated by the symbol:
σ
M
σ
M
.
The central limit theorem states that given a distribution
with a mean mm and variance
s2
s
2
, the sampling distribution of the mean
approaches a normal distribution with a mean
(mm) and a variance
s2N
s
2
N
as NN, the sample size,
increases.
This section applies only when the means are computed from
independent samples. This distribution can be understood by
thinking of the following sampling plan: Sample n scores
from the population of people and compute the mean. This
mean will be designated as
M
1
M
1
. Then, sample nn scores from
a second population of people not taking the drug and
compute the mean. This mean will be designated as
M
2
M
2
. Finally compute the difference between
M
1
M
1
and
M
2
M
2
. This difference will be called
M
d
M
d
where the "d" stands for "difference." The mean and the
variance of the sampling distribution of
M
d
M
d
are:
μ
M
d
=
μ
1
−
μ
2
μ
M
d
μ
1
μ
2
(2)
and
μ
M
d
2=
σ
1
2
n
1
+
σ
2
2
n
2
μ
M
d
2
σ
1
2
n
1
σ
2
2
n
2
(3)
Finally the standard error of
M
d
M
d
is simply the square root of the variance of the sampling
distribution of
M
d
M
d
.
Assuming the means from which
LL is computed are independent,
the mean and standard deviation of the sampling distribution
of LL are:
m
L
=
a
1
m
1
+
a
2
m
2
+…+
a
k
m
k
m
L
a
1
m
1
a
2
m
2
…
a
k
m
k
and
σ
L
=∑
a
i
2nσ22
σ
L
a
i
2
n
σ
2
2
where
m
i
m
i
is the mean for population ii,
σ2
σ
2
is the variance of each population, and
nn is the number of elements
sampled from each population.
If NN pairs of scores were
sampled over and over again the resulting Pearson
rr's would form a
distribution. When the absolute value of the correlation in
the population is low (say less than about 0.4) then the
sampling distribution of Pearson's
rr is approximately
normal. However, with high values of correlation, the
distribution has a negative skew. Fisher's
z
′
z
′
transformation converts Pearson's
rr to a value that is normally
distributed and with a standard error of
σ
z
′
=1N−32
σ
z
′
1
N
3
2
(4)
Since
NN is in the denominator
of the formula, the larger the sample size, the smaller the
standard error. The number of standard deviations from the
mean can be calculated with the formula:
z=
z
′
−
μ
z
′
σ
z
′
z
z
′
μ
z
′
σ
z
′
(5)
where:
zz is the number of
standard deviations above the
z
′
z
′
associated with the population correlation,
z
′
z
′
is the value of Fisher's
z
′
z
′
for the sample correlation,
mm
is the value of
z
′
z
′
for the population correlation and is the mean of the
sampling distribution of
z
′
z
′
.
σ
z
′
σ
z
′
is the standard error of Fisher's
z
′
z
′
.
The standard error of the standard deviation is:
σ
s
=0.71σN
σ
s
0.71
σ
N
(7)
The distribution of the standard deviation is positively
skewed for small N but is approximately normal if
NN is 25 or greater. Thus,
procedures for calculating the area under the normal curve
work for the sampling distribution of the standard deviation
as long as
NN is at least 25
and the distribution is approximately normal.
The sampling distribution of a proportion is equal to the
binomial distribution. The mean and standard deviation of
the binomial distribution are
m=p
m
p
and
σ
p
=π(1−π)N
σ
p
π
1
π
N
(8)
The mean of the sampling distribution of the difference
between two independent proportions
p
1
−
p
2
p
1
p
2
is
m
p
1
-
p
2
=
p
1
−
p
2
m
p
1
-
p
2
p
1
p
2
. The standard error of
p
1
−
p
2
p
1
p
2
is
σ
p
1
-
p
2
=
π
1
(1−
π
1
)
n
1
+
π
2
(1−
π
2
)
n
2
σ
p
1
-
p
2
π
1
1
π
1
n
1
π
2
1
π
2
n
2
(9)
The sampling distribution
of
p
1
−
p
2
p
1
p
2
is approximately normal as long as the proportions
are not too close to 1 or 0 and the sample sizes are not too
small. As a rule of thumb, if
n
1
n
1
and
n
2
n
2
are each at least 10 and neither nor are within 0.10 of 0 or
1 then the approximation is satisfactory for most purposes.
