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**Application:
**Consider a 2×2 cross-over design contains two
sequences (treatment orderings) and two time periods (occasions). One sequence
receives treatment A followed by treatment B. The other sequence receives B and
then A. This procedure is used to test the following hypotheses:

**Procedure**:

- Enter

a)
value
of ** α**,
the probability of type I error

b)
value
of ** β**,
the probability of type II error

c)**
**value of

d) value of *population
variance*

e)
value of ** δ**>0, the equivalence limit.

- Click the button “Calculate” to obtain result sample size of
each group
*n*.

**Formula:
**_{}** (*)**

**Notations:**

**α****:** The
probability of type I error (significance level) is the *probability of rejecting the true null
hypothesis. *

**β****:** The
probability of type II error (1 – power of the test) is the *probability of
failing to reject the false null hypothesis.*

** δ**: The largest

** μ_{2} – μ_{1}**: The value of

**Examples**

**Example 1:
**Suppose a difference of 5% (i.e., δ=5%) in percent
change of low density lipidproteins (LDLs) is considered of clinically
meaningful difference. By using (*), assuming that the standard deviation is
=0.2 (i.e., ** population variance is 0.04**), the required sample size of each
group to achieve an 80% power (

**Reference**: Chow, Shao and Wang, *Sample
Size Calculations In Clinical Research*,