Anal Bioanal Chem (2007) 389:11–12 DOI 10.1007/s00216-007-1466-x

ANALYTICAL CHALLENGE

Solution to quality assurance challenge 5 Manfred Reichenbächer & Jürgen W. Einax

Published online: 28 July 2007 # Springer-Verlag 2007

The t-test is based on the assumptions:

Solution Two independent sample means, x1 and x2 , are compared by means of the t-test [1]. The test value bt is given by: jx1  x2 j bt ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi s2p 1=n1 þ 1=n2

ð1Þ

in which n1 and n2 are the number of values in samples 1 and 2, respectively, and s2p is the pooled variance, which is calculated by use of Eq. (2): s2p ¼

ðn1  1Þs21 þ ðn2  1Þs22 ð n1 þ n2  2Þ

ð2Þ

The calculated t-value is compared with the critical t-value, tcrit, at the chosen significance level, P, and dftotal ¼ n1 þ n2  2 degrees of freedom in both laboratories. (The significance level for all tests was given by P=95%.) For a two-sided test the hypothesis that there is no difference between the mean values of sample 1, x1 , and sample 2, x2 is accepted if bt < tcrit .

M. Reichenbächer : J. W. Einax (*) Institute of Inorganic and Analytical Chemistry, Friedrich Schiller University Jena, Lessingstraße 8, 07743 Jena, Germany e-mail: [email protected]

1. The samples with means x1 and x2 are drawn from normally distributed populations, which can be tested by use of the David test [2]. 2. The variances s21 and s22 are statistically equal. This is tested by use of the F-test [3]. 3. There must be no outliers in the data sets [3]. The test values for normal distribution (David test) and for outliers (Grubbs test), TV and brm , respectively, are listed in Table 1. For Laboratory 1 and Laboratory 2 the test values, TV, lie within the boundaries of the David table, in our example 2.28 and 3.16 for a sample size of n=6 and at a confidence level of P=95%. The analytical results may be regarded as normally distributed. The critical rm value for n=6 at P=95% is 1.938. The test value of the lowest value, xmin, of Laboratory 2 is larger than the critical value. As a consequence, xmin is regarded as an outlier at the 0.95 level of significance. It must be removed from the data set of Laboratory 2. No outlier could be detected in the data set of Laboratory 1. The test values of the homogeneity of the variance by means of the F-test with the outlier-free data set of Laboratory 2 are listed in Table 2. The calculated F-value b ¼ 2:796) is smaller than the critical value F (P=95%; (F df1 =5; df2 =4) = 6.256. (Note, for the F-test s21 must always be higher than s22 .) The null hypothesis that both variances are equal is accepted. The t-test may be carried out. With the pooled standard deviation sp =0.41209 the test value bt is calculated to be 2.739 and the critical t-value (from Excel with the function TINV(0.05; dftotal =9)) is 2.262. Note, the test value bt is higher than the critical t-value.

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Anal Bioanal Chem (2007) 389:11–12

Table 1 Results from tests for normal distribution, by the David test, and for outliers, by the Grubbs test

Laboratory 1 Laboratory 2

x

s

xmin

xmax

TV

brm;min

brm;max

6 6

94.92 95.27

0.4875 0.8571

94.1 93.6

95.6 95.9

3.077 2.683

1.675 1.945

1.402 0.739

n

x

s

df

laboratory. With the data set listed in Table 2 the following t-test values are calculated:

6 5

94.92 95.60

0.48751 0.29155

5 4

laboratory 1: bt1 = 0.419, laboratory 2: bt2 = 4.602.

Answer to question 1: There is a significant difference between the results from the two laboratories. The laboratory-to-laboratory transfer is not successful. Note, if you do not remove the outlier, the t-test value calculated is 0.869. bt is smaller than tcrit, which means the mean values in both laboratories are equal, but this result is wrong. The t-test can be used as a test for accuracy. For each laboratory the test value bt is calculated by use of Eq. (3): bt ¼

jx  mj pffiffiffi  n s

Grubbs test

n

Table 2 Test values for the F-test

Laboratory 1 Laboratory 2

David test

ð3Þ

x, s, and n are the mean value, standard deviation of the analytical values, and the number of (outlier-free) data in each laboratory, respectively and μ is the “true” value, in our example the known assay of the chosen batch is μ= 95.0% m/m. The analytical result is regarded as accurate if the bt-value calculated is smaller than the critical t-value at the significance level P (in our example 95%) and the number of degrees of freedom df of the data set in each

The critical t-value for laboratory 1 at P=95% and df=5 is 2.571. This value is greater than the test value, bt1 . The analytical result of the reference laboratory (laboratory 1) is accurate. With df=4 for laboratory 2 the critical value is 2.776. This value is smaller than bt2 . The analytical result is not accurate. Answer to question 2: The answer is no. The analytical result of laboratory 1 is accurate but the result of laboratory 2 is wrong. Note that with the outlier in laboratory 2 the test t-value calculated (0.762) is smaller than tcrit and the analytical result should be accurate. But this is not correct!

References 1. Massart DL, Vandeginste BGM, Buydens LMC, De Jong S, Lewi PJ, Smeyers-Verbeke J (1997) Data handling in science and technology, Handbook of chemometrics and qualimetrics: part A, vol 20. Elsevier, Amsterdam 2. Reichenbächer M, Einax JW (2007) Anal Bioanal Chem 387:1585– 1586 3. Reichenbächer M, Einax JW (2006) Anal Bioanal Chem 384:14–18