1. 1 Tests of Significance In this section we deal with two tests used for comparing two analytical methods, one is a new or proposed method and the other is a standard method. The two methods are compared in terms of whether they provide comparable precision ( the F test ), based on their standard deviations or variances. The other test ( t test ) tells whether there is a statistical difference between results obtained by the two methods.

2. 2 The F Test The precision of two methods could be compared based on their standard deviations using the F test which can be defined as the ratio between the variances ( the variance is the standard deviation squared ) of the two methods. The ratio should always be larger than unity. That is, the larger variance of either method is placed in the nominator. F = S 1 2 /S 2 2

3. 3 Where, S 1 2 > S 2 2 Values of F ( a statistical factor ) at different confidence levels which can be obtained from statistical F tables. When F calculated < F tabulated this is an indication of no statistical difference between precision or variances of the two methods.

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5. 5 Example In the analysis of glucose using a new developed procedure and a standard procedure, the variances of the two procedures were 4.8 and 8.3. If the tabulated F value at 95% confidence level at the number of degrees of freedom used was 4.95. Determine whether the variance of the new procedure differs significantly from that of the standard method

6. 6 F = S 1 2 /S 2 2 F = 8.8/4.8 = 1.7 3 (the subscript is because the answer is less than the key number) Since F calculated < F tabulated there is no significant statistical difference between the variances of the two methods (i.e. there is no significant statistical difference between the precision of the two methods).

7. 7 The Student t Test To check whether there is a significant statistical difference between the results of a new or proposed procedure and a standard one, the t test is used. As we did above, we calculate t and compare it to the tabulated value at the required confidence level and at the used degrees of freedom. There is no significant statistical difference between the results of the two methods when t calculated < t tabulated. There are three situations where the t test is applied:

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9. 9 a. When an Accepted Value is Known The t calc is calculated from the relation below and compared to t tab  = x + ts/N 1/2 or more conveniently, + t = (x -  ) N 1/2 /s

10. 10 Example A new procedure for determining copper was used for the determination of copper in a sample. The procedure was repeated 5 times giving an average of 10.8 ppm and a standard deviation of +0.7 ppm. If the true value for this analysis was 11.7 ppm, does the new procedure give a statistically correct value at the 95% confidence level? t tab = 2.776

11. 11 Substitution into equation below, we get: + t = (x -  ) N 1/2 /s + t = (10.8-11.7) 5 1/2 /0.7 + t = 2. 9 the t calc is larger than the t tab. Therefore, there is a significant statistical difference between the two results which also means that it is NOT acceptable to use the new procedure for copper determination.

12. 12 b. Comparison between two means When an accepted value is not known and the sample is analyzed using the new procedure and a standard procedure. Here, we have two sets of data, a standard deviation for each set of data and a number of data points or results in each set. Under these conditions, we use the pooled standard deviation for the two sets. The same equation in a is used but with some modifications. The t value is calculated from the relation (N 1 N s ) 1/2 (N 1 + N s ) 1/2 ( x 1 – x s ) S p + t = *

13. 13 Where, x 1 and x s are means of measurements using the new and standard methods. N 1 and N s are number of replicates done using the new and standard methods, respectively. S p is the pooled standard deviation. In such calculations it is wise to apply the F test first, and if it passes the t test is then applied.

14. 14 Example Nickel in a sample was determined using a new procedure where six replicate samples resulted in a mean of 19.65% and a variance of 0.4524. Five replicate analyses where conducted using a standard procedure resulting in a mean of 19.24% and a variance of 0.105. If the pooled standard deviation was +0.546, is there a significant difference between the two methods?

15. 15 First, let us find whether there is a significant difference in precision between the two procedures, by applying the F test F = 0.4524/0.105 = 4.31 The tabulated F value is 6.26. Since F calculated < F tabulated, then there is no significant statistical difference between the precision of the two procedures. Therefore, we continue with calculation of t test.

16. 16 + t = 1.2 3 The tabulated t value is 2.262. Since t calculated < t tabulated for nine degrees of freedom at 95% confidence level, we conclude that there is no significant statistical difference between the results of the two methods. (N 1 N s ) 1/2 (N 1 + N s ) 1/2 ( x 1 – x s ) S p + t = * (6*5) 1/2 (6+ 5) 1/2 ( 19.65 – 19.24) 0.546 + t = *