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86 Modern Analytical Chemistry

4
EXAMPLE .16
Before determining the amount of Na 2 CO 3 in an unknown sample, a student
decides to check her procedure by analyzing a sample known to contain
98.76% w/w Na 2 CO 3 . Five replicate determinations of the %w/w Na 2 CO 3 in the
standard were made with the following results
98.71% 98.59% 98.62% 98.44% 98.58%
Is the mean for these five trials significantly different from the accepted value at
the 95% confidence level (a = 0.05)?

SOLUTION
The mean and standard deviation for the five trials are
–
X = 98.59 s = 0.0973
–
Since there is no reason to believe that X must be either larger or smaller than
m, the use of a two-tailed significance test is appropriate. The null and
alternative hypotheses are
– –
H 0 : X = m H A : X ≠ m
The test statistic is
m - X ´ n 98 .76 -98 .59 ´ 5
t exp = = = . 391
s . 0 0973
The critical value for t(0.05,4), as found in Appendix 1B, is 2.78. Since t exp is
greater than t(0.05, 4), we must reject the null hypothesis and accept the
alternative hypothesis. At the 95% confidence level the difference between
–
X and m is significant and cannot be explained by indeterminate sources of
error. There is evidence, therefore, that the results are affected by a determinate
source of error.

If evidence for a determinate error is found, as in Example 4.16, its source
should be identified and corrected before analyzing additional samples. Failing to
reject the null hypothesis, however, does not imply that the method is accurate, but
only indicates that there is insufficient evidence to prove the method inaccurate at
the stated confidence level.
–
The utility of the t-test for X and mis improved by optimizing the conditions
–
used in determining X. Examining equation 4.15 shows that increasing the num-
ber of replicate determinations, n, or improving the precision of the analysis en-
hances the utility of this significance test. A t-test can only give useful results,
however, if the standard deviation for the analysis is reasonable. If the standard
deviation is substantially larger than the expected standard deviation, s, the con-
–
fidence interval around X will be so large that a significant difference between
–
X and mmay be difficult to prove. On the other hand, if the standard deviation is
–
significantly smaller than expected, the confidence interval around X will be too
–
small, and a significant difference between X and mmay be found when none ex-
ists. A significance test that can be used to evaluate the standard deviation is the
subject of the next section.