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6.6 Choosing Sample Sizes for Inferences about 1 2
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two populations were shifted by amounts 0, .4s, and .8s, where s denotes the standard deviation of the distribution. (When the population distribution is Cauchy, s
denotes a scale parameter.)
From Table 6.18, we can make the following observations. The level of the
paired t test remains nearly equal to .05 for uniform and double exponential distributions, but is much less than .05 for the very heavy-tailed Cauchy distribution.
The Wilcoxon signed-rank test’s level is nearly .05 for all four distributions, as
expected, because the level of the Wilcoxon test only requires that the population
distribution be symmetric. When the distribution is normal, the t test has only
slightly greater power values than the Wilcoxon signed-rank test. When the population distribution is short-tailed and uniform, the paired t test has slightly greater
power than the signed-rank test. Note also that the power values for the t test are
slightly less than the t power values when the population distribution is normal.
For the double exponential, the Wilcoxon test has slightly greater power than the t
test. For the Cauchy distribution, the level of the t test deviates significantly from
.05 and its power is much lower than the Wilcoxon test. From other studies, if the
distribution of differences is grossly skewed, the nominal t probabilities may be
misleading. The skewness has less of an effect on the level of the Wilcoxon test.
Even with this discussion, you might still be confused as to which statistical test
or confidence interval to apply in a given situation. First, plot the data and attempt to
determine whether the population distribution is very heavy-tailed or very skewed.
In such cases, use a Wilcoxon rank-based test. When the plots are not definitive in
their detection of nonnormality, perform both tests. If the results from the different
tests yield different conclusions, carefully examine the data to identify any peculiarities to understand why the results differ. If the conclusions agree and there are no
blatant violations of the required conditions, you should be very confident in your
conclusions. This particular “hedging” strategy is appropriate not only for paired
data but also for many situations in which there are several alternative analyses.
6.6
Choosing Sample Sizes for Inferences about 1 2
Sections 5.3 and 5.5 were devoted to sample-size calculations to obtain a confidence
interval about m with a fixed width and specified degree of confidence or to conduct
a statistical test concerning m with predefined levels for a and b. Similar calculations
can be made for inferences about m1 m2 with either independent samples or with
paired data. Determining the sample size for a 100(1 a)% confidence interval
about m1 m2 of width 2E based on independent samples is possible by solving the
following expression for n:
1
1
za2s
E
An
n
Note that, in this formula, s is the common population standard deviation and that
we have assumed equal sample sizes.
Sample Sizes for a
100(1 A)% Confidence
Interval for 1 2 of the
Form y1 y 2 E,
Independent Samples
2z2a2 s2
E2
(Note: If s is unknown, substitute an estimated value to get an approximate
sample size.)
n
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Chapter 6 Inferences Comparing Two Population Central Values
The sample sizes obtained using this formula are usually approximate because
we have to substitute an estimated value of s, the common population standard
deviation. This estimate will probably be based on an educated guess from information on a previous study or on the range of population values.
Corresponding sample sizes for one- and two-sided tests of m1 m2 based on
specified values of a and b, where we desire a level a test having the probability of
a Type II error b(m1 m2) b whenever |m1 m2| , are shown here.
Sample Sizes for Testing
1 2 , Independent
Samples
One-sided test: n 2s2
Two-sided test:
n 2s2
(za zb)2
∆2
(za2 zb)2
∆2
where n1 n2 n and the probability of a Type II error is to be b when
the true difference |m1 m2| ∆. (Note: If s is unknown, substitute an
estimated value to obtain an approximate sample size.)
EXAMPLE 6.10
One of the crucial factors in the construction of large buildings is the amount of
time it takes for poured concrete to reach a solid state, called the “set-up” time.
Researchers are attempting to develop additives that will accelerate the set-up
time without diminishing any of the strength properties of the concrete. A study is
being designed to compare the most promising of these additives to concrete without the additive. The research hypothesis is that the concrete with the additive
will have a smaller mean set-up time than the concrete without the additive. The
researchers have decided to have the same number of test samples for both the
concrete with and without the additive. For an a .05 test, determine the appropriate number of test samples needed if we want the probability of a Type II error
to be less than or equal to .10 whenever the concrete with the additive has a mean
set-up time of 1.5 hours less than the concrete without the additive. From previous
experiments, the standard deviation in set-up time is 2.4 hours.
Solution Let m1 be the mean set-up time for concrete without the additive and m2
be the mean set-up time for concrete with the additive. From the description of the
problem, we have
●
●
●
●
●
One-sided research hypothesis: m1 m2
s 2.4
a .05
b .10 whenever m1 m2 1.5 n1 n2 n
From Table 1 in the Appendix, za z.05 1.645 and zb z.10 1.28. Substituting
into the formula, we have
n 2s2(za zb)2
2(2.4)2(1.645 1.28)2
43.8,
∆2
(1.5)2
or 44
Thus, we need 44 test samples of concrete with the additive and 44 test samples of
concrete without the additive.
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