Page 106 - Modern Analytical Chemistry

1400-CH04 9/8/99 3:55 PM Page 89

Chapter 4 Evaluating Analytical Data 89

ts A
m A = X A ± 4.17
n A

ts B
m B = X B ± 4.18
B n
where n A and n B are the number of replicate trials conducted on samples A and B. A
–
–
comparison of the mean values is based on the null hypothesis that X A and X B are
identical, and an alternative hypothesis that the means are significantly different.
A test statistic is derived by letting m A equal m B , and combining equations 4.17
and 4.18 to give
ts A ts B
X A ± = X B ±
n A n B
– –
Solving for ÷X A – X B ÷and using a propagation of uncertainty, gives
s 2 A s 2 B
X A - X B = t ´ +
n A n B
Finally, solving for t, which we replace with t exp , leaves us with
X - X B
A
t exp = 4.19
2 2
n ) +
s (/ A ( s n )
/ B
A
B
The value of t exp is compared with a critical value, t(a, n), as determined by the cho-
sen significance level, a, the degrees of freedom for the sample, n, and whether the
significance test is one-tailed or two-tailed.
It is unclear, however, how many degrees of freedom are associated with t(a, n)
2 2
since there are two sets of independent measurements. If the variances s A and s B esti-
2
mate the same s , then the two standard deviations can be factored out of equation
4.19 and replaced by a pooled standard deviation, s pool , which provides a better esti-
mate for the precision of the analysis. Thus, equation 4.19 becomes
X - X B
A
t exp = 4.20
1 n ) + (1 )
s pool (/ A n / B
with the pooled standard deviation given as
1
( n A - ) s 2 A +( n B -) s 2 B
1
s pool = 4.21
n A + n B -2
As indicated by the denominator of equation 4.21, the degrees of freedom for the
pooled standard deviation is n A + n B –2.
If s A and s B are significantly different, however, then t exp must be calculated
using equation 4.19. In this case, the degrees of freedom is calculated using the fol-
lowing imposing equation.

s 2 ) s 2 )] 2
[( /n A +( /n B
B
A
n= – 2 4.22
1
1
s 2 2 + )] s 2 2 +)]
A
[( /n A ) /(n A +[( /n B ) /(n B
B
Since the degrees of freedom must be an integer, the value of nobtained using
equation 4.22 is rounded to the nearest integer.
Regardless of whether equation 4.19 or 4.20 is used to calculate t exp , the null hy-
pothesis is rejected if t exp is greater than t(a, n), and retained if t exp is less than or
equal to t(a, n).