**Help Aids** Top

**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* allowable difference*

d)
value of *population variance*

- 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.*

** μ_{2} – μ_{1}**: The value of

**Examples**

**Example 1:
**Suppose a low density lipidproteins
(LDLs) is considered of clinically meaningful
difference. By using (*), assuming that the standard deviation is 10% (i.e.,
population variance is 0.01), the required sample size of each group to achieve
an 80% power (*β*=0.2) at *α*=0.05 for correctly detecting such
difference of *μ _{2} – μ_{1}=*0.5 change obtained by normal
approximation as

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