z Tests and Confidence Intervals for a Difference Between Two Population Means
Basic Assumptions
1. X1, X2,….Xm is a random sample from a distribution with
mean 1 and variance .
2. Y1, Y2,…..Yn is a random sample from a distribution with
mean 2 and variance .
3. The X and Y samples are independent of one another.
1
z Tests and Confidence Intervals for a Difference Between Two Population Means
The use of m for the number of observations in the first
sample and n for the number of observations in the second
sample allows for the two sample sizes to be different.
Sometimes this is because it is more difficult or expensive
to sample one population than another.
In other situations, equal sample sizes may initially be
specified, but for reasons beyond the scope of the
experiment, the actual sample sizes may differ.
2
z Tests and Confidence Intervals for a Difference Between Two Population Means
Proposition
The expected value of
is 1 – 2, so
is an
unbiased estimator of 1 – 2. The standard deviation of
is
3
Test Procedures for Normal Populations with Known Variances
Standardizing
gives the standard normal variable
(9.1)
In a hypothesis-testing problem, the null hypothesis will
state that 1 – 2 has a specified value.
4
Test Procedures for Normal Populations with Known Variances
Rejection regions for Ha: 1 – 2 < 0 and Ha: 1 – 2 ≠ 0
that yield tests with desired significance level  are lowertailed and two-tailed, respectively.
Null hypothesis:H0 : 1 – 2 = 0
Test statistic value: z =
5
Test Procedures for Normal Populations with Known Variances
Alternative Hypothesis
Rejection Region for Level  Test
Ha: 1 – 2 > 0
z  z (upper-tailed)
Ha: 1 – 2 < 0
z  – z (lower-tailed)
Ha: 1 – 2 ≠ 0
either z  z/2 or z  – z/2(twotailed)
Because these are z tests, a P-value is computed as it was
for the z tests [e.g., P-value = 1 – (z) for an upper-tailed
test].
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Large-Sample Tests
The assumptions of normal population distributions and
known values of 1 and 2 are fortunately unnecessary
when both sample sizes are sufficiently large. In this case,
the Central Limit Theorem guarantees that
has
approximately a normal distribution regardless of the
underlying population distributions.
Furthermore, using
and in place of
and
in
Expression (9.1) gives a variable whose distribution is
approximately standard normal:
7
Large-Sample Tests
Use of the test statistic value
along with the previously stated upper-, lower-, and twotailed rejection regions based on z critical values gives
large-sample tests whose significance levels are
approximately .
These tests are usually appropriate if both m > 40 and n >
40. A P-value is computed exactly as it was for our earlier z
tests.
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Confidence Intervals for 1 – 2
Manipulation of the inequalities inside the parentheses to
isolate 1 – 2 yields the equivalent probability statement
This implies that a 100(1 – )% CI for 1 – 2 has lower
limit
upper limit
where
is
the square-root expression. This interval is a special case
of the general formula
9
Confidence Intervals for 1 – 2
Provided that m and n are both large, a CI for 1 – 2 with a
confidence level of approximately 100(1 – )% is
where – gives the lower limit and the upper limit of the
interval. An upper or a lower confidence bound can also be
calculated by retaining the appropriate sign (+ or –) and
replacing z/2 by z.
Our standard rule of thumb for characterizing sample sizes
as large is m > 40 and n > 40.
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The Two-Sample t Test and Confidence Interval
We could, for example, assume that both population
distributions are members of the Weibull family or that they
are both Poisson distributions. It shouldn’t surprise you to
learn that normality is typically the most reasonable
assumption.
Assumptions
Both population distributions are normal, so that
X1, X2,…, Xm is a random sample from a normal distribution
and so is Y1,…,Yn (with the X’s and Y’s independent of one
another).
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The Two-Sample t Test and Confidence Interval
Theorem
When the population distribution are both normal, the
standardized variable
(9.2)
has approximately a t distribution with df v estimated
from the data by
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The Two-Sample t Test and Confidence Interval
where
(round v down to the nearest integer).
Manipulating T in a probability statement to isolate 1 – 2
gives a CI, whereas a test statistic results from replacing
1 – 2 by the null value 0.
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The Two-Sample t Test and Confidence Interval
The two-sample t confidence interval for 1 – 2 with
confidence level 100(1 –  ) % is then
A one-sided confidence bound can be calculated as
described earlier.
The two-sample t test for testing H0: 1 – 2 = 0 is as
follows:
Test statistic value: t =
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The Two-Sample t Test and Confidence Interval
Alternative Hypothesis Rejection Region for
Approximate Level  Test
Ha: 1 – 2 > 0
t  t,v (upper-tailed)
Ha: 1 – 2 < 0
t  – t,v (lower-tailed)
Ha: 1 – 2  0
either t  t/2,v or t  –t/2,v
(two-tailed)
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Pooled t Procedures
Alternatives to the two-sample t procedures just described
result from assuming not only that the two population
distributions are normal but also that they have equal
variances
.
That is, the two population distribution curves are assumed
normal with equal spreads, the only possible difference
between them being where they are centered.
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Pooled t Procedures
The following weighted average of the two sample
variances, called the pooled (i.e., combined) estimator of
 2,adjusts for any difference between the two sample
sizes:
The first sample contributes m – 1 degrees of freedom to
the estimate of  2, and the second sample contributes
n – 1 df, for a total of m + n – 2 df.
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Pooled t Procedures
Statistical theory says that if
replaces  2 in the
expression for Z, the resulting standardized variable has a t
distribution based on m + n – 2 df.
In the same way that earlier standardized variables were
used as a basis for deriving confidence intervals and test
procedures, this t variable immediately leads to the pooled t
CI for estimating 1 – 2 and the pooled t test for testing
hypotheses about a difference between means.
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Analysis of Paired Data
Assumptions
The data consists of n independently selected pairs (X1,Y1),
(X2, Y2),…(Xn, Yn), with E(Xi) = 1 and E(Yi) = 2. Let
D1 = X1 – Y1, D2 = X2 – Y2,…, Dn = Xn – Yn so the Di’s are
the differences within pairs.
Then the Di’s are assumed to be normally distributed with
mean value D and variance
(this is usually a
consequence of the Xi’s and Yi’s themselves being normally
distributed).
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The Paired t Test
Because different pairs are independent, the Di’s are
independent of one another. Let D = X – Y, where X and Y
are the first and second observations, respectively,
within an arbitrary pair.
Then the expected difference is
D = E(X – Y) = E(X) – E(Y) = 1 – 2
(the rule of expected values used here is valid even when
X and Y are dependent). Thus any hypothesis about
1 – 2 can be phrased as a hypothesis about the mean
difference D.
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The Paired t Test
But since the Di’s constitute a normal random sample (of
differences) with mean D, hypotheses about D can be
tested using a one-sample t test.
That is, to test hypotheses about 1 – 2 when data is
paired, form the differences D1, D2,…, Dn and carry out a
one-sample t test (based on n – 1 df) on these differences.
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The Paired t Test
The Paired t Test
Null hypothesis: H0: D = 0
(where D = X – Y is the difference between the first and
second observations within a pair, and D = 1 – 2)
Test statistic value:
(where and sD are the sample mean and standard
deviation, respectively, of the di’s)
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The Paired t Test
Alternative Hypothesis
Ha: D > 0
Rejection Region for Level 
Test
t  t,n –1
Ha: D < 0
t  – t,n – 1
Ha: D ≠ 0
either t  t/2,n–1 or t  – t/2,n–1
A P-value can be calculated as was done for earlier t tests.
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The Paired t Confidence Interval
In the same way that the t CI for a single population mean
 is based on the t variable T =
at
confidence interval for D (= 1 – 2) is based on the fact
that
has a t distribution with n – 1 df. Manipulation of this t
variable, as in previous derivations of CIs, yields the
following 100(1 – )% CI: The paired t CI for D is
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Inferences Concerning a Difference Between Population Proportions
Proposition
Let
and
where X ~ Bin(m, p1) and
Y ~ Bin(n, p2 ) with X and Y independent variables. Then
So
is an unbiased estimator of p1 – p2, and
(where qi = 1 – pi)
(9.3)
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A Large-Sample Test Procedure
The most general null hypothesis an investigator might
consider would be of the form H0: p1 – p2 =
Although for population means the case
no difficulties, for population proportions
must be considered separately.
 0 presented
= 0 and  0
Since the vast majority of actual problems of this sort
involve = 0 (i.e., the null hypothesis p1 = p2). we’ll
concentrate on this case.
When H0: p1 – p2 = 0 is true, let p denote the common
value of p1 and p2 (and similarly for q).
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A Large-Sample Test Procedure
Then the standardized variable
(9.4)
has approximately a standard normal distribution when H0
is true.
However, this Z cannot serve as a test statistic because the
value of p is unknown—H0 asserts only that there is a
common value of p, but does not say what that value is.
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A Large-Sample Test Procedure
A test statistic results from replacing p and q in (9.4) by
appropriate estimators.
Assuming that p1 = p2 = p, instead of separate samples of
size m and n from two different populations (two different
binomial distributions), we really have a single sample of
size m + n from one population with proportion p.
The total number of individuals in this combined sample
having the characteristic of interest is X + Y.
The natural estimator of p is then
(9.5)
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A Large-Sample Test Procedure
The second expression for shows that it is actually a
weighted average of estimators and obtained from the
two samples.
Using and = 1 – in place of p and q in (9.4) gives a
test statistic having approximately a standard normal
distribution when H0 is true.
Null hypothesis: H0: p1 – p2 = 0
Test statistic value (large samples):
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A Large-Sample Test Procedure
Alternative Hypothesis
Rejection Region for
Approximate Level  Test
Ha: p1 – p2 > 0
z  za
Ha: p1 – p2 < 0
z  –za
Ha: p1 – p2  0
either z  za/2 or z  –za/2
A P-value is calculated in the same way as for previous z
tests.
The test can safely be used as long as
are all at least 10.
and
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We skip Sec 9.5
31