A Critical Comparison of Two Procedures for Antiperspirant Evaluation
A Critical Comparison of Two Procedures for Antiperspirant Evaluation
A Critical Comparison of Two Procedures for Antiperspirant Evaluation
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266 JOURNAL OF THE SOCIETY OF COSMETIC CHEMISTS<br />
(b) Compute the difference between the antiperspirant and control means<br />
(in terms <strong>of</strong> logs)<br />
Tx ---- 5.6656: T2 '- 6.2820<br />
D •- 5.6656 - 6.2820 = --0.6164<br />
note that the difference used is antiperspirant minus control, not the re-<br />
verse.<br />
(c) Compute the 95% confidence interval abouthe population difference<br />
CL0.95 =• +- S• t0.0.•<br />
where i5 represents the above difference; S• represents the standard error <strong>of</strong><br />
the difference; and t0.05 represents Student's t at ot = 0.05.<br />
In the example<br />
so that<br />
or<br />
__<br />
D = -0.6164<br />
S•= 0.0504<br />
t.05 = 2.228*<br />
CL0.9, = -0.6164 -+ (0.0594)(2.228)<br />
= -0.6164 - (0.1323)<br />
-0.4841 to -0.7487<br />
(d) A difference between the logarithms <strong>of</strong> two numbers is the ratio <strong>of</strong> the<br />
two numbers when antilogs are taken. Thus<br />
antilog • - •<br />
T2<br />
where T• and T,are in milligrams. If we take antilogs <strong>of</strong> the two confidence<br />
limits, we obtain a maximum and minimum ratio <strong>of</strong> antiperspirant<br />
to control milligrams, representing 95% confidence limits abouthe popula-<br />
tion mean ratio:<br />
antilog -0.4841 = 0.6163<br />
antilog -0.7487 = 0.4730<br />
PR• = (1-0.6163) 100 = 38.37<br />
PR (from step 3) = 46.01<br />
PR2 = (1-0.4730) 100 = 52.70<br />
*At 10 df (the error df in fl•e ANOVA).