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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).

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