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

Table 4.9. SGOT levels and diagnosis data for 255 patients transplanted for malignancy: a initial tabulation; b revised tabulation

aInitial tabulation

bRevised tabulation

1 The values in parentheses are expected numbers if the row and column classifications are independent.

sented in table 4.9a was prepared. If SGOT levels and diagnosis group are independent, the expected values which are given in parentheses involve too many small values to carry out a reliable test of the null hypothesis of independence. However, combining the diagnosis groups D4 and D5, and also the SGOT categories 2–3 !, 3–4 ! and 14 ! Normal results in a 3 ! 4 table with larger expected numbers in many of the cells (see table 4.9b). Notice, too, that in general, the observed and expected numbers agree fairly closely in the cells which disappear through the combining of rows and columns. The observed value of the test statistic, T, was calculated for the revised table and its value was determined to be to = 4.52. Since the revised table has three rows (a = 3) and four columns (b = 4), the approximate 2 distribution of T has 2 ! 3 = 6 degrees of freedom. Therefore, the significance level of the data with respect

Fig. 4.2. Chi-squared probability curves. a 24, 27 and 210. b The location of selected critical values for 24.

table for every degree of freedom, it would appear that 2 tables must be very unwieldy.

Although sets of 2 tables do exist which follow the pattern we have described above, a much more compact version has been devised, and these are usually quite adequate. Since the significance level of the 2 test is approximate, it is frequently satisfactory to know a narrow range within which the approximate significance level of a test lies. It follows that, for any member of the 2 family of probability distributions, we only require a dozen or so special reference points, called critical values. These are values of the random variable which correspond to specified significance levels or, equivalently, values which cut off predetermined amounts of probability (area) in the right-hand tail of the probability curve. For example, we might wish to know the critical values for 24 which correspond to the right-hand tail probabilities 0.5, 0.25, 0.10, 0.05, 0.025, 0.01 and 0.005. The actual numbers which correspond to these probability levels for 24 are 3.357, 5.385, 7.779, 9.488, 11.14, 13.28 and 14.86. Figure 4.2b shows a sketch of the 24 probability curve indicating the location of these critical values and the corresponding right-hand tail probabilities. Such a table requires only one line of critical values for each degree of freedom. Thus, an entire set of tables for the 2 family of probability distributions can be presented on a single page. This is the method of presentation which most 2 tables adopt, and an example of 2 statistical tables which follow this format is reproduced in table 4.10. The rows of the table correspond to different values of k, the degrees of freedom parameter. The columns identify predetermined righthand tail probabilities, i.e., 0.25, 0.05, 0.01, etc., and the entries in the table specify the corresponding critical values for the 2 probability distribution identified by its unique degrees of freedom.

The actual use of 2 tables is very straightforward. In order to calculate Pr( 2k 6 to), locate the row for k degrees of freedom. In that row, find the two numbers, say tL and tU, which are closest to to so that tL ^ to ^ tU. For example, if we want to evaluate Pr( 24 6 10.4), these numbers are tL = 9.488 and tU = 11.14 since 9.488 ! 10.4 ! 11.14. Since tL = 9.488 corresponds to a right-hand tail probability of 0.05 and tU = 11.14 corresponds to a right-hand tail probability of 0.025, it follows that Pr( 24 6 10.4) is smaller than 0.05 and larger than 0.025, i.e., 0.05 1 Pr( 24 6 10.4) 1 0.025. And if Pr( 24 6 10.4) is the approximate significance level of a test, then we know that the p-value of the test is between 0.025 and 0.05.

Table 5.1. A 2 ! 2 table summarizing binary data collected from two groups

in a group, and that the outcome of the experiment for any subject may not influence the outcome for any member of either group.

Unfortunately, the first of these assumptions, namely that the probability of success is the same for each member of a group, is often not true; there may be factors other than group membership which influence the probability of success. If the effect of such factors is ignored and the observed data are summarized in a single 2 ! 2 table, the test of significance that is used to analyze the data could mask important effects or generate spurious ones. The following example, which is similar to one described in Bishop [6], illustrates this phenomenon.

A study of the effect of prenatal care on fetal mortality was undertaken in two different clinics; the results of the study are summarized, by clinic, in figure 5.1a. Clearly, there is no evidence in the data from either clinic to suggest that the probability or rate of fetal mortality varies with the amount of prenatal care delivered. However, if we ignore the fact that this rate is roughly three times higher in Clinic 2 and combine the data in the single 2 ! 2 table shown in figure 5.1b, the combined data support the opposite conclusion. Why? Notice that in Clinic 1 there are fewer deaths overall and much more intensive care, while in Clinic 2 there are more deaths and much less intensive care. Therefore, the significant difference in the two rates of fetal mortality observed in the combined table is due to the very large number of more intensive care patients contributed by Clinic 1 with its low death rate.

The preceding example illustrates only one of the possible hazards of combining 2 ! 2 tables. If there are other factors, in addition to the characteristic of major interest, which might affect the probability of success, then it is important to adjust for these other factors, which are sometimes called confounding factors, in analyzing the data. In the preceding example, the primary focus of the study is the relationship between the amount of prenatal care and the rate of fetal mortality. However, we observed that the amount of prenatal care received depended on the additional factor representing clinic location (Clinic 1 or Clinic 2) which also influenced the rate of fetal mortality. Thus, clinic

Table 5.2. A 2 ! 2 table summarizing the binary data for level i of a confounding factor

els’) for a possible confounding factor, e.g., k different clinics participating in the fetal mortality and prenatal care study. Stratifying the study data on the basis of this confounding factor will give us k distinct 2 ! 2 tables like the one shown in table 5.2, where the possible values of i, representing the distinct levels of the confounding factor, could be taken to be the numbers 1 through k.

As we saw in chapter 4, the 2 test for a single 2 ! 2 table evaluates the discrepancy between the observed and expected values for each cell in the table, assuming that the row and column totals are fixed and the probability of success in the two groups is the same. The test of significance which correctly combines the results in all k stratified 2 ! 2 tables calculates, for each distinct 2 ! 2 table, an expected value, ei, corresponding to ai; this expected value, ei, is based on the usual assumptions that the row and column totals Ai, Bi, ri and Ni – ri are fixed and that the probability of success in the two groups is the same. Notice, however, that the probability of success may now vary from one stratum (2 ! 2 table) to another; the test no longer requires that the probability of success must be the same for each level of the confounding factor. Instead, the null hypothesis for this test of significance specifies that, for each distinct level of the confounding factor, the probabilities of success for Groups 1 and 2 must be the same; however, this hypothesis does allow the common probabilities of success to differ from one level of the confounding factor to another. In terms of the fetal mortality example, the null hypothesis requires a constant fetal mortality rate in Clinic 1, regardless of prenatal care received, and a constant, but possibly different, fetal mortality rate in Clinic 2, regardless of prenatal care received.

Of course, evaluating the expected numbers e1, e2, …, ek automatically determines the corresponding expected values for the remaining three cells in each of the k distinct 2 ! 2 tables. As we might anticipate, the test statistic evaluates the discrepancy between the sum of the k observed values, O = a1 +

k

a2 + … + ak = ai, and the sum of the corresponding k expected values, E =

i = 1