Ординатура / Офтальмология / Английские материалы / Study Design and Statistical Analysis a practical guide for clinicians_Katz _2006

Use Spearman’s rank coefficient to test for linear relationships with non-normally distributed interval variables.

When you have an interval variable that is non-normally distributed you cannot use Pearson’s correlation coefficient to test for a linear association with another interval variable. Instead, use the Spearman’s rank correlation coefficient. This test is the same as the Pearson’s correlation coefficient except that the correlation coefficient is based on the rankings of the subjects instead of the actual value of the values of the subjects. In other words, to calculate the Spearman’s rank correlation coefficient you first rank the observations separately for each group as I did in Table 5.14 (not together, as I did for calculating the Mann-Whitney test in Table 5.17). Next you use the rankings of the two groups on the variable to calculate the Pearson’s correlation coefficient.

For example, in Section 5.6.B, I reviewed a study examining the relationship between coronary inflammation and unstable angina. The investigators measured inflammation using the neutrophil myeloperoxidase index. They found low levels of the index in the coronary vascular circulation, indicating inflammation, in patients with unstable angina. The index does not follow a normal distribution. Therefore, to assess the correlation between the neutrophil myeloperoxidase index and the C-reactive protein level in the blood, a measure of systemic inflammation, they used the Spearman’s rank correlation coefficient. The results are shown in Figure 5.6.

From Figure 5.6 there is no way to tell that the correlation is based on the rankings of the subjects rather than their actual scores on these variables. As with Pearson’s correlation coefficient, Spearman’s rank correlation coefficient does not detect non-linear relationships.

Analyse ordinal variables using nonparametric statistics that are based on rankings.

You will remember that ordinal variables are categorical variables with multiple categories that can be ordered, but for which there is not a fixed interval between the categories, such as stage of cancer (Section 2.10).

Since there is not a fixed interval between the categories, ordinal variables should be analysed by using non-parametric statistics that are based on rankings. As you can see from Table 5.1, many of the tests for ordinal variables are similar to those for non-normally distributed interval variables. Therefore, I will not repeat the explanations in this section. One test for ordinal variables that is important to review is the chi-square for trend. It is not used with interval variables but can be used when you are looking at the association between an ordinal variable and a dichotomous variable (Section 5.8.A).

As the availability of non-parametric statistics is limited, especially if multivariable analysis is needed, investigators will sometimes transform an ordinal variable into a dichotomous variable. This can be done using a natural cut-off, a median split, or comparing extreme categories, just as you would do for a nonnormally distributed interval variable (Section 5.4). For example, in a study of health literacy and glycemic (sugar) control among diabetes, health literacy, was

0

Number at risk

Revascularization No revascularization

No coronary artery revascularization

Coronary artery revascularization

Use the log-rank test to compare the time to outcome of different groups.

Tip

Do not rely exclusively on statistical tests to compare survival curves: visually compare them as well.

revascularization prior to elective vascular surgery.90 As you can see the two Kaplan–Meier curves are essentially superimposed on one another, indicating that coronary artery revascularization does not make a difference.

For Figures 5.7 and 5.8 you do not really need any statistical test to draw the appropriate conclusions. However, most curves are not this obvious, and even when they are, you will still want to present the statistical comparison of the curves.

The most commonly used test to assess the difference between two Kaplan– Meier curves is the log-rank test. We seek to disprove the null hypothesis: that there is no difference in the survival experience of the two groups.

For each time interval, the log-rank test compares the observed number of outcomes in each group to what would have been expected if the two groups had the same survival experience. A time interval is defined by an outcome occurring in either of the groups. The differences between the observed and the expected outcomes for each time interval are then summed. If the difference is large relative to the size of the standard error then the log-rank will be large and the P-value will be small. This is the case with Figure 5.7. The P-value associated with the log-rank test is 0.001. We can therefore reject the null hypothesis and conclude that there is a difference in mortality between persons who received a transfusion and those who did not.

In contrast, if the difference between the groups is small relative to the size of the standard error then the log-rank is small and the P-value is large. This is the case with Figure 5.8. The P-value for the log-rank test is equal to 0.92. We therefore would not reject the null hypothesis that there is no difference in the survival of persons who received coronary artery revascularization compared to those who did not.

Figure 5.9 represents a more complicated situation: the Kaplan–Meier curves cross. Event-free survival is higher among those persons who receive angioplasty initially, but in the latter part of the study (after about 150 days) eventfree survival is higher among those who were stented.91

Although, the log-rank test is significant (P 0.04) indicating that over the course of the study event-free time is greater with stenting, the log-rank test does not adequately reflect the true complexity of the situation. If you did not review the graphical presentation and went straight to the log-rank test, you would not be able to fully inform your patients that with stenting they are taking on an initially higher risk of an adverse event, but ultimately their chance of avoiding an adverse event is better with a stent. Often when the curves cross, the log-rank

90McFalls, E.O., Ward, H.B., Moritz, T.E., et al. Coronary-artery revascularization before elective major vascular surgery. New Engl. J. Med. 2004; 351: 2795–804.

91Erbel, R., Haude, M., Hopp, H.W., et al. Coronary-artery stenting compared with balloon angioplasty for restensosis after initial balloon angioplasty. New Engl. J. Med. 1998; 339: 1672–8.

Stent group

Angioplasty group

Days after procedure

The log-rank test does not require that the time to event be normally distributed.

will be non-significant because survival advantages of one group in the beginning of the trial are averaged with survival advantages of the other group at the end of the trial.

The log-rank test is a non-parametric test and therefore does not require that the time to event be normally distributed. In fact, in studies of time to outcome, the data are rarely normally distributed because there are usually a small group of subjects who have substantially longer times to outcome than average.

The log-rank test can also be used to compare time to outcome for more than two groups. In such cases you are testing the null hypothesis that the survival experience does not significantly differ among the groups.

As with performing a chi-squared test or ANOVA with more than two groups, a significant log-rank does not tell you where the difference lies. To determine this you can perform pairwise comparisons of the curves. But to avoid capitalizing on chance, set a more stringent standard (i.e., a lower P-value) before concluding that differences are not due to chance. You can do this by using the Bonferroni correction (Section 5.6.A). For example, if you are performing all possible pairwise comparisons of three curves – three – than you would set the threshold for disproving the null hypothesis of 0.017 (0.05/3).

Tip

The Wilcoxon test weighs outcomes that occur early in the study more heavily than outcomes that occur later on.

Data from Laheij, R.J.F., et al. Risk of community-acquired pneumonia and use of gastric acid-suppressive drugs. J. Am. Med. Assoc. 2004; 292: 1955–60.

Some investigators use the Wilcoxon test92 (also known as Geham’s test) to compare the survival experience of different groups of subjects. The Wilcoxon test weighs outcomes that occur early in the study more heavily than outcomes that occur later on. For this reason, if there is a large difference between the groups in the number of early outcomes you may find that the Wilcoxon test is statistically significant while the log-rank test is not. However, generally we weigh early and late outcomes about equally and for this reason you will rarely see the Wilcoxon test used.

Repeated observations of the same subject must be analysed with statistics that take into account that the observations are correlated.

A common clinical research design involves observing the same subjects on multiple occasions (e.g., at 6-month intervals) or under different conditions (e.g., before or after treatment).

Repeated observations of the same subject may occur due to multiple observations over time or due to subjects receiving different types of treatments.

In either case, we have repeated measurements of the same subjects. In the simplest cases we are evaluating differences in the repeated measurements of a single sample of subjects. In more complex models we are comparing changes in the repeated measurements of two or more samples of subjects.

The bivariate statistics that we have reviewed so far assume that the observations are independent of one another. This is not the case with repeated observations of the same subjects. The observations are not independent of one another because the same subject is more likely to respond in a similar way at a repeat examination or under a different condition than a different subject. Repeated observations of the same subject must be analysed with statistics that take into account that the repeated observations are correlated.95

95The analysis of correlated outcome data is a complicated issue. Besides repeated observations of the same individuals, there are other circumstances that lead to correlated outcomes including matched studies (Section 5.11), clustered study designs where subjects have been recruited from multiple settings of related individuals (e.g., families, doctor’s practices, or hospitals) and observations of different body parts of the same person. Analysis of correlated outcome data often requires multivariable modeling. Readers who want to know more about this important area of clinical research should see: Katz, M.H. Multivariable Analysis: A Practical Guide for Clinicians (2nd edition). Cambridge University Press, 2005, pp. 158–78.

Use McNemar’s test to assess paired observations on a

dichotomous outcome.

When you have paired observations (i.e., before and after) of a dichotomous variable use McNemar’s test.

For example, Kuipers and colleagues sought to determine the likelihood that long-term acid suppression would result in gastritis.96 They followed 59 patients with gastroesophageal reflux disease and Heliobacter pylori. All patients were treated with an acid suppresser (omeprazole); the average duration of treatment was 5 years.