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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.

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 S12/S22

- Where, S12 gt S22
- Values of F ( a statistical factor ) at

different confidence levels which can be obtained

from statistical F tables. When Fcalculated lt

Ftabulated this is an indication of no

statistical difference between precision or

variances of the two methods.

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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

- F S12/S22
- F 8.8/4.8 1.73 (the subscript is because the

answer is less than the key number) - Since Fcalculated lt Ftabulated 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).

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 tcalculated lt ttabulated . - There are three situations where the t test is

applied

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a. When an Accepted Value is Known

- The tcalc is calculated from the relation below

and compared to ttab - m x ts/N1/2 or more conveniently,
- t (x - m) N1/2/s

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? ttab

2.776

- Substitution into equation below, we get
- t (x - m) N1/2/s
- t (10.8-11.7) 51/2/0.7
- t 2.9
- the tcalc is larger than the ttab. 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.

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

- Where, x1 and xs are means of measurements

using the new and standard methods. N1 and Ns

are number of replicates done using the new and

standard methods, respectively. Sp 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.

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?

- 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 Fcalculated

lt Ftabulated , then there is no significant

statistical difference between the precision of

the two procedures. Therefore, we continue with

calculation of t test.

- t 1.23
- The tabulated t value is 2.262. Since

tcalculated lt ttabulated 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.