Two samples: (in)dependent and
(non)parametric.
Two independent samples, normal model
Comparing the means or variances of two
(sub)populations. We assume we have:
 Two independent random samples (sizes m
and n) from normally distributed population:
 X1,..., Xm are independent and Xi ~N(µX, ),
Y1,..., Yn are independent and Yj ~N(µY,
).
 The sample means and sample variances are
denoted by:
,
resp.
The observed values as

.
are unbiased estimators of
:

,
=
)=
.
are unbiased estimators of
µX and µY , so
µX - µY
1

are independent so
=
+

So
~ N(
+
) or:
Based on this variable and its distribution we
can derive a confidence interval for
and a test statistic for a specific value of
. But this is only usable if
are known. Usually this is not the case.
If we use
as estimators of
and
, then we apply the Student’s t-distribution.
We distinguish three cases (assumptions):
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1.
and
are unknown, but equal:
is the best
estimator of the common variance
(
called the pooled sample variance):
is
Similar to the one-sample-problem we derive:
(1- α)100%-CI(
where
and c from
the t(m+n-2)-table such that P(Tm+n-2 ≥ c) =
.
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The test statistic for the test on
H1 can be one-sided:
is
or
or two-sided:
2.
and
are unknown and different:
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Test statistic:
Where df = number the degrees of freedom:
 df = between min(m-1, n-1) and m + n – 2
 Computer software, like SPSS, computes df
 without software we use: df = min(m-1, n-1)
(1- α)100%-CI(
=
Where df = min(m-1, n-1) and
c such that P(Tdf ≥ c) = α/2
Note: df = min(m-1, n-1) is a “safe” estimate
of the real number. That is why this is called
“the conservative method”.
3. Large m and n (m > 30 and n > 30):
Use the N(0, 1)-distribution as an
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approximation, even if the populations are
not normally distributed (robustness!)
Testing the equality of expectations for
m > 30 and n > 30: if
And (also for large m and n):
(1- α)100%-CI(
=
where c is such that P( Z ≥ c) = α/2
Testing the equality of variances
(Levene’s test for the proportion
We test H0:
H1 can be
=
>
,

H0:
<
or
/
):
=1

.
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Test statistic is the proportion of the sample
variances which has a so called Fisher
distribution with m -1 degrees of freedom in
the numerator and n -1 degrees of freedom
in the denominator.
Notation:
Note 1: If H0 is true T will have values near 1.
Note 2: The Fisher table contains values c,
such that
=α
 c in this equation is the critical value for the
right sided alternative H1:
>
 The critical value c for the left sided
alternative H1:
<
can be found using:
α
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 Two sided test for H0:

, H1:

If T ≤ c1 or T ≥ c2 , then reject H0 , where:
and
When testing the equality of expectations
using SPSS always first check the p-value of
Levene’s test (α = 0.05) to choose between
equal variances assumed and not assumed.
Two pairwise dependent random samples:
the t-test for the differences.
Whenever we have two observation per object
(individual in the sample), we can compute
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the differences and apply the t-procedure for
the differences. Usually this is the case when
we have observations per object i before (xi)
and after treatment (yi).
Statistical assumptions for the (dependent)
random samples x1,...,xn and y1,...,yn:
The differences Zi = Yi - Xi (i =1,...,n) are
independent and Zi ~ N(µ, σ2), where expected
difference µ and variance σ2 are unknown.
t-test on H0: µ = 0 (expected difference is 0):
Confidence interval for the expected
difference (c from the t(n-1)-table):
The relation between two-sided tests and
confidence intervals:
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Suppose θ is the population parameter and we
have a (1-α)100%-CI(θ), based on a sample,
then H0: θ = θ0 will be rejected in favour of
H1: θ  θ0 at significance level α if θ0 is in the
interval.
e.g. A one sample t-test:
95%-CI(µ) = (22.4, 28.1) => reject H0: µ = 30
versus H1: µ  30 at 5%-level
(because 30 is not in the interval).
A binomial test:
90%-CI(p) = (0.42, 0.51) => Do not reject
H0: p = ½ versus H1: p  ½ at 10%-level.
A two sample t-test:
99%-CI(
) = (-6.8, -1.2), then:
reject
in favour of
at 1%-level,
because the difference 0 is not in the interval.
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Non-parametric or distribution-free tests
are used in cases where the assumption of a
distribution (having unknown parameters)
does not apply.
These tests offer for example an alternative
method if the assumption of a normal
distribution is evidently incorrect and the
number of observations is small: this can be
the case if the histogram is skewed to the right
or the left or when there are outliers. SPSS
also provides tests on normality.
1. The Sign Test as an alternative for the ttest on the differences for two pairwise
dependent random samples.
Statistical assumptions:
the differences Zi have an unknown (notnormal) distribution. p = the probability of a
positive difference. We test H0: p = ½
versus H1: p > ½ (or p < ½ or p  ½)
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Note: for symmetric distributions of the
differences p = ½ is equivalent to µ = 0, but
for skewed distributions p = ½ is equivalent to
Median = 0. That’s why this test is also called
the sign test on the median.
The test statistic is X = the number of positive
differences: X ~ B(n, ½) if H0: p = ½ is true.
 n = the number of observed non-zero
differences (just cancel the zero differences)
 The rejection region is:
H1
Rejection region
p>½
{c, c +1, ....., n}
p<½
{0,...., c}
p  ½ {0,...., c1} {c2, c2+1, ....., n}
 Determine the critical value(s), using:
- For n ≤ 25 the binomial table if available.
- Normal approximation is valid for n > 10:
X is approximately N(½n, ¼n)-distributed.
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2. The Wilcoxon Rank Sum Test, as a nonparametric alternative for the two
(independent) samples t-test, on the
equality of the expectations
.
The only statistical assumption we need is that
we have two independent random samples of
numerical variables X1,..., Xm and Y1,..., Yn.
We will test H0: the population distributions
are the same
versus
H1: the population distribution of Y is shifted
compared to the distribution of X.
Example: shift to the left  H1 :
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The test statistic, Wilcoxon’s W, is defined as:
W = sum of the ranks of the X-observations
if we order all X- and Y-observations.
One sided alternatives:
 Y is shifted to the left  H1:
If the alternative is true, the X-values will be
large and W is large: W ≥ c => reject H0.
 Y is shifted to the right  H1:
If the alternative is true, the X-values will be
small and W is small: W ≤ c => reject H0.
Two sided alternative:
Y is shifted to either side  H1:
If the alternative is true, the X-values will be
large or small and W is small or large:
if W ≤ c1 or W ≥ c2, then reject H0.
The distribution of W
We will use a normal approximation of the
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(discrete) distribution of W and apply
continuity correction if m > 5 and n > 5.
This is an approximate distribution (N=m+n).
The Wilcoxon rank sum test using
1. the critical value c (right sided test):
If W ≥ c , then reject . And c such that:
P(W ≥ c)= P(W ≥ c+½ ) ≈ P(Z ≥
2. the p-value p-value ≤ α => reject
P(W ≥ w)= P(W ≥ w+½ ) ≈ P(Z ≥
)=α
:
)
Note: for m ≤ 5 and/or n ≤ 5 the exact
distribution of W is computed, using
combinatorics (not part of this course).
Ties: if two or more observations have the
same value, then we call this group of
observations a tie. All observations of a tie are
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assigned the same rank: the mean rank of
these observations. We can use the
distribution above, if there are just a few ties.
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