Ординатура / Офтальмология / Английские материалы / Using and Understanding Medical Statistics_Matthews, Farewell_2007

Table 4.4. The 2 ! 2 table of expected values corresponding to the graft rejection data summarized in table 4.2

Notice that this is the product of the row total for Group 1 (R1) and the column total for category I (C1) divided by the total number of observations (N). This makes the formula particularly easy to recall and use. The other expected values in the 2 ! 2 table are calculated from similar formulae, viz.

e12 R1 C2 , e21 R2 C1 and e22 R2 C2 . N N N

Of course, since the row and column totals for the 2 ! 2 table of expected numbers are already fixed (R1, R2, C1 and C2), we know from the discussion of Fisher’s test (see § 3.2) that, once we have calculated e11, we can obtain the other entries in the table of expected numbers by subtraction. One possible set of formulae for obtaining e12, e21 and e22 in this way is

e12 = R1 – e11, e21 = C1 – e11 and e22 = R2 – e21.

Although we could list other sets of formulae for calculating e12, e21 and e22, these versions will suffice. Any correct set will always produce a table of expected numbers having row and column totals R1, R2, C1 and C2. Remember that these expected values are based on the assumption that the null hypothesis, H0, is true, i.e., the probability of category I (success) is the same for Groups 1 and 2. Table 4.4 shows the table of expected values for the data concerning bone marrow transplantation, including details of the calculations.

In §2.3 we specified that an appropriate test statistic, T, is needed to carry out a test of significance. In the case of Fisher’s test, T is the number of suc-

Though we certainly recommend use of the concise formula for T, there is an advantage in our having introduced the longer version first. For a given set of row and column totals (R1, R2, C1 and C2), the test statistic T could take on many different values, depending on the observed values O11, O12, O21 and O22. Whatever this set of possible values of T might be, there are some common features which we can deduce by examining the longer formula for T. To begin with, each of the four terms in the long version of T must be positive; therefore, T will always be positive. Secondly, if each observed value, Oij, is very close to its corresponding expected number, eij, then T will be very close to 0. Thus, a small value of T suggests that the observed data are consistent with the null hypothesis, i.e., the probability of category I (success) is the same for Groups 1 and 2. Conversely, if each Oij is very different from the corresponding eij, then T should be rather large. Thus, a large value of T suggests that the observed data are not consistent with the null hypothesis. And so we see that if the observed value of T, for a particular set of data, is the number to (for example, 8.01), then values of T which exceed to correspond to possible 2 ! 2 tables which are less consistent with the null hypothesis, H0, than the 2 ! 2 table that actually was observed (T = to). In order to calculate the significance level of the data, we simply need to sum the probabilities associated with all possible values of T which are greater than or equal to the observed value to, i.e., calculate Pr(T 6 to). In the bone marrow transplant example this is precisely Pr(T 6 8.01).

As we indicated earlier, the heart of the 2 approximation to Fisher’s test lies in the choice of the test statistic, T. We have already learned that 2 ! 2 tables which are no more consistent with the null hypothesis than the observed 2 ! 2 table (T = to) correspond to the set described by the inequality T 6 to. To obtain the probability of this set, we require the probability distribution of T when H0 is true. And herein lies the importance of the choice of T as a test statistic. By means of complex mathematical arguments, statisticians have proved that when H0 is true, i.e., when the probability of category I is the same in Groups 1 and 2, the test statistic T has a distribution which is approximately the same as the probability distribution called 21 (chi-squared with one degree of freedom). Therefore, we can determine the significance level of the data (recall that this is Pr(T 6 to)) by calculating the probability that 21 6 to. The shorthand notation Pr( 21 6 to) represents the probability that a random variable, whose probability distribution is 21, exceeds to. Of course, this means that the significance level we calculate will be approximate; however, a great deal of statistical research has been devoted to showing that the approximation is quite accurate for most cases not covered by exact calculation of the significance level for Fisher’s test.

the total numbers of observations in the rows and columns of the 2 ! 2 table. A conservative rule-of-thumb for ensuring that the approximation is accurate requires that all the expected numbers, eij, exceed five. A more liberal rule allows one expected number to be as low as two. If the values in the 2 ! 2 table of expected numbers are small, then Fisher’s test should be used to evaluate the significance level of the data with respect to H0.

4.3. The 2 Test for Rectangular Contingency Tables

In the preceding section, we considered the problem of analyzing binary data (i.e., Success/Failure, Response/No Response) collected from independent samples drawn from two populations (i.e., Control, Treatment). However, the simplicity of the 2 ! 2 table is also a major disadvantage if initial tabulations of the observed data are more extensive than the binary categories Success and Failure. Clearly, reducing more detailed observations to simple binary categories results in lost information. The rectangular contingency table with, say, a rows and b columns generalizes the simple 2 ! 2 table.

The basic model for the rectangular contingency table assumes that the outcome for each of N experimental units (patients, animals, lab tests, etc.) may be classified according to two schemes, A and B, say. If the classification categories for scheme A are labelled A1, A2, …, Aa, then the observation for each member of a sample belongs to exactly one of these categories. For example, in a clinical trial design to compare the effects of PAS and streptomycin in the treatment of tuberculosis, each participant in the trial would be treated with one of PAS or streptomycin, or perhaps a combination of both drugs. In this case, A1 might represent the PAS-only group, A2 the streptomycin-only group and A3 the combined drugs group. Similarly, if the classification categories under scheme B are labelled B1, B2, …, Bb, then the outcome for each member of the sample belongs to exactly one of these categories as well. In the pulmonary tuberculosis clinical trial mentioned above, the response of each patient to treatment, as measured by the analysis of a sputum sample, would be one of positive smear, negative smear and positive culture or negative smear and negative culture. In this case, we might label the patients having a positive smear B1, those having a negative smear but positive culture B2, and those with a negative smear and negative culture B3.

The results of classifying each sample unit according to both schemes, A and B, may be concisely summarized in the rectangular contingency table shown in table 4.5. In this a ! b contingency table with a rows and b columns, each cell corresponds to a unique row and column combination. Therefore, each cell identifies the number of sample members which were classified as

hypothesis of independence means that a patient’s condition at the conclusion of the trial did not depend on the treatment received.

However, there is an equivalent null hypothesis which is tested in exactly the same way, and which may occasionally be more appropriate. If one of the classification schemes, say A, corresponds to sampling from a different groups, then H0 can also be interpreted as specifying that the probability distribution of the B classification scheme is identical in all of the a groups. This is simply a generalization of the null hypothesis which we discussed in §4.2 in connection with the 2 test in 2 ! 2 contingency tables. For example, suppose that the categories of scheme A represent three strains of mice, and the categories of B describe three types of tumors which are observed in a sample of 75 mice, 25 of each strain, all of which are tumor-bearing. Then this second interpretation of the null hypothesis specifies that the proportions of the three tumor types arising in each strain are identical.

The procedure for determining the significance level of the data in an a ! b contingency table with respect to the null hypothesis, H0, is a simple generalization of the 2 test for a 2 ! 2 table. The steps in the procedure are the following:

(1) Compute a corresponding a ! b table of expected numbers based on the row totals R1, R2, ..., Ra and the column totals C1, C2, …, Cb in the observed table. The formula for the expected number, eij, in the cell specifying the joint category Ai and Bj is

Notice that this number is obtained by first multiplying the unique row and column totals corresponding to the joint category Ai and Bj. The resulting product is divided by N, the overall sample size. As we indicated above, this formula for calculating eij is a simple extension of the rule which we discussed for 2 ! 2 tables in §4.2. Recall, also, that in the 2 ! 2 case the table of expected values could be completed by subtraction from row and column totals after calculating e11. Likewise, in the a ! b case, the last entry in each row and column of the table of expected numbers could be obtained by subtracting the other calculated values in the row and column from the corresponding row or column total. This means that a total of (a – 1) ! (b – 1) values of eij could be calculated using the formula given above, and the remaining (a + b – 1) values can then be determined by subtraction. Table 4.7 shows the table of expected values corresponding to the a ! b contingency table displayed in table 4.5. Detailed calculations for the PAS, streptomycin clinical trial example, showing the 3 ! 3 table of expected numbers, are given in table 4.8.

to = 0.62 + 1.54 + 5.51 + 0.24 + 0.26 + 0.04 + 1.70 + 0.65 + 7.06 = 17.62.

Fig. 4.1. Calculating the quantities (Oij – eij)2/eij and to, the observed value of the test statistic, for the sputum analysis data.

(3) Using the two a ! b tables of observed and expected numbers, calculate for each of the a ! b joint categories the quantity

Figure 4.1 shows a 3 ! 3 table displaying these values for the PAS, streptomycin clinical trial.

(4) The test statistic, T, is the sum of the individual values of (Oij – eij)2/eij for all a ! b joint categories. If we call the resulting sum to, then to is the observed value of the test statistic. A mathematical expression which describes all this calculation is the simple formula

similar to the one specified in §4.2 for the 2 ! 2 table. However, the formula has been generalized to accommodate a rows and b columns. Notice, also, that with the more general a ! b table a continuity correction is not used.