Major topics covered in this chapter
Exercises 35 balance with a single internal reference weight.) Carefully considered procedures
3.3 Comparison of two experimental means
Another way in which the results of a new analytical method may be tested is by comparing them with those obtained by using a second (perhaps a reference) method. In this case the two methods give two sample means, and The null hypothesis is that the two methods give the same result, i.e. or , so we need to test whether ( ) differs significantly from zero. As we saw in the previous section, the t-test can be used to compare an experimental result, ( ) in this case, with a standard value, obviously zero here. However, we must allow for the fact that the results from the two methods might have different sample sizes, n1and n2, and that we also have two different standard deviations,
x1 - x2
x1- x2 m1 - m2 = 0
H0: m1 = m2, x2. x1
40 3: Significance tests
s1and s2. If these standard deviations are not significantly different (see Section 3.5 for a method of testing this assumption), a pooled estimate, s, of the standard devi-ation can first be calculated using the equdevi-ation:
The next example shows another application of this test, where it is used to decide whether a change in the conditions of an experiment affects the result.
(3.3.1) s = (n1 - 1)s21 + (n2- 1)s22
(n1 + n2 - 2)
(3.3.2)
where t has n1 + n2- 2degrees of freedom.
t = x1 - x2 sA1
n1
+ 1 n2
To decide whether the difference between the two means, and is significant, i.e. to test the null hypothesis, H0: m1 = m2,the statistic t is then calculated from:
x2, x1
Example 3.3.1
In a comparison of two methods for the determination of chromium in rye grass, the following results were obtained:
Method 1: mean 1.48; standard deviation 0.28 Method 2: mean 2.33; standard deviation 0.31 For each method five determinations were made.
(Sahuquillo, A., Rubio, R. and Rauret, G., 1999, Analyst, 124: 1)
Do these two methods give results having means which differ significantly?
From Eq. (3.3.1), the pooled value of the standard deviation is given by:
so
From Eq. (3.3.2):
There are eight degrees of freedom, so (Table A.2) the critical value is Since the experimental value of is greater than this, the difference between the two results is significant at the 5% level and the null hypothesis is rejected. In fact since the critical value of t8for P 0.01 is 3.36, the difference is significant at the 1% level. In other words, if the null hypoth-esis is true the probability of such a large difference arising by chance is less than 1 in 100.
ƒt ƒ t8 = 2.31 (P = 0.05).
t = 2.33 - 1.48 0.295A1
5 + 1 5
= 4.56 s = 0.295.
s2 = ([4 * 0.282] + [4 * 0.312])>(5 + 5 - 2) = 0.0872 (mg kg-1 Cr)
Comparison of two experimental means 41
Example 3.3.2
In a series of experiments on the determination of tin in foodstuffs, samples were boiled with hydrochloric acid under reflux for different times. Some of the results are shown below:
Refluxing time (min) Tin found (mg kg-1)
30 55, 57, 59, 56, 56, 59
75 57, 55, 58, 59, 59, 59
(Analytical Methods Committee, 1983, Analyst, 108: 109).
Does the mean amount of tin found differ significantly for the two boiling times?
The mean and variance (square of the standard deviation) for the two times are:
30 min 57.00 2.80 75 min 57.83 2.57
The null hypothesis is that the refluxing time has no effect on the amount of tin found. From Eq. (3.3.1), the pooled value for the variance is given by:
From Eq. (3.3.2):
There are 10 degrees of freedom so the critical value is
The observed value of is less than the critical value so the null hypothesis is retained: there is no evidence that the refluxing time affects the amount of tin found.
The table below shows the result of performing this calculation using Excel®: ƒt ƒ ( = 0.88)
t10 = 2.23 (P = 0.05).
= - 0.88
t = 57.00 - 57.83 1.64A1
6 + 1 6 s = 1.64
s2 = ([5 * 2.80] + [5 * 2.57])>10 = 2.685 s22
x2
s12
x1
t-Test: Two-sample assuming equal variances
Variable 1 Variable 2
Mean 57 57.833
Variance 2.8 2.567
Observations 6 6
Pooled variance 2.683
Hypothesized mean difference 0
df 10
t Stat -0.881
P(T<=t) one-tail 0.199
t Critical one-tail 1.812
P(T<=t) two-tail 0.399
t Critical two-tail 2.228
42 3: Significance tests
The distinction between ‘one-tail’ and ‘two-tail’ will be covered in Section 3.5.
For the present, it is sufficient to consider only the two-tail values. These show that Since this probability is much greater than 0.05, the result is not significant at the 5% level.
P ( ƒ t ƒ 7 0.88) = 0.399.
If the population standard deviations are unlikely to be equal then it is no longer appropriate to pool sample standard deviations in order to give an overall estimate of standard deviation. An approximate method in these circumstances is given below:
In order to test when it cannot be assumed that the two samples come from populations with equal standard deviations, the statistic t is calcu-lated where
(3.3.3)
(3.3.4)
the value obtained being truncated to an integer.
with the number of degrees of freedom =
as21 n1
+ s22 n2b2 s41
n211n1 - 12 + s42
n221n2 - 12 t = 1x1 - x22
B s21 n1 +
s22 n2 H0: m1 = m2
Several different equations have been suggested for the number of degrees of freedom for t when s1and s2differ, reflecting the fact that the method is an approximate one.
Equation (3.3.4) is used by both Minitab®and Excel®, but Minitab®, erring on the side of caution in giving a significant result, rounds the value down, while Excel® rounds it to the nearest integer. For example, if the equation gave a value of 4.7, Minitab®would take four degrees of freedom and Excel®would take five.
Example 3.3.3
The data below give the concentration of thiol (mM) in the blood lysate of the blood of two groups of volunteers, the first group being ‘normal’ and the second having rheumatoid arthritis:
Normal: 1.84, 1.92, 1.94, 1.92, 1.85, 1.91, 2.07 Rheumatoid: 2.81, 4.06, 3.62, 3.27, 3.27, 3.76
(Banford, J.C., Brown, D.H., McConnell, A.A., McNeil, C.J., Smith, W.E., Hazelton, R.A. and Sturrock, R.D., 1983, Analyst, 107: 195)
The null hypothesis adopted is that the mean concentration of thiol is the same for the two groups.
Paired t-test 43
Two sample T for Normal vs Rheumatoid
N Mean StDev SE Mean
Normal 7 1.9214 0.0756 0.029 Rheumato 6 3.465 0.440 0.18 The reader can check that:
n1 7 1.921 0.076 n2 6 3.465 0.440
Substitution in Eq. (3.3.3) gives and substitution in Eq. (3.3.4) gives 5.3, which is truncated to 5. The critical value is so the null hypothesis is rejected: there is evidence that the mean concentration of thiol differs between the groups.
The result of performing this calculation using Minitab® (where the non-pooled test is the default option) is shown below.
Two sample t-test and confidence interval
t5 = 4.03 (P = 0.01) t = - 8.48
s2 x2
s1 x1
95% CI for mu Normal - mu Rheumato: (-2.012, -1.08) T-Test mu Normal = mu Rheumato (vs not =):T= -8.48 P = 0.0004 DF = 5
This confirms the values above and also gives the information that This probability is extremely low: the result is in fact significant at P 0.001.
P