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

P 0.007 by McNemar’s test.

Data from Kuipers, E.J., et al. Atrophic gastritis and Helicobacter pylori in patients with reflux esophagitis treated with omeprazole or fundoplication. New Engl. J. Med. 1996; 334: 1018–22.

At baseline 41% of patients had normal mucosa and 59% had gastritis (Table 5.23, last column) while at follow-up only 19% had normal mucosa and 81% had gastritis (bottom row).

For calculating McNemar’s test the only two cells that matter are the two cells that indicate change (i.e., going from normal at baseline to having gastritis at follow-up or going from gastritis at baseline to having a normal examination at follow-up). I have bolded these two cells in Table 5.23. Adding these two cells together, we find that 21 patients had a change in status. If there were no tendency toward gastritis in patients with gastroesophageal reflux we would expect that about half of these 21 patients (10.5 patients) would go in each direction (half from normal to gastritis and half from gastritis to normal).

However, our distribution (4 and 17) is clearly different from 10.5. Is it possible that the difference is due to chance? We use McNemar’s test to determine the probability of obtaining these results if the null hypothesis (no difference) were true. In the case of this example, the probability is equal to 0.007. We can therefore reject the null hypothesis and consider alternative hypotheses, such as long-term acid suppression leads to the development of gastritis.

To appreciate the importance of using an analytic tool that incorporates the pairing of observations, let us assume that you analysed the data shown in Table 5.23 as if the observations were unpaired. Feeling smart from having read Section 5.2 you bypass chi-squared because you recognize that the expected number of subjects in one of the cells (normal at baseline and normal at follow-up) is 5 (0.41 11 4.5). The P-value associated with a two-tailed Fisher’s exact test is 0.10. Therefore, you would wrongly assume that the null hypothesis is correct: that there is no significant difference between the baseline and the follow-up examination.

Values are represented as n (%).

P-value of Wilcoxon signed rank test 0.006.

Data from Herrstedt, J., et al. Ondansetron plus metopimazine compared with ondansetron alone in patients receiving moderately emetogenic chemotherapy. New Engl. J. Med.

1993; 328: 1076–80.

under each condition (or time point) and compares the observed rank sums for each condition to what would be expected by chance.

What you would expect by chance is that the observed rank sums for the different conditions would be about the same. However, if the rankings come out substantially higher (or lower) for one of the conditions, then the Friedman statistic will yield a small P-value and you can reject the null hypothesis that there are no differences in the observations at the different conditions/time points.

For example, Darzins and colleagues assessed differences in the type of care subjects preferred depending on whether the care was for an unfamiliar person, a family member, or for himself or herself.104 The type of care was an interval scale range from palliative care to intensive care. The sample included doctors, nurses, health professionals, high school students and members of the general public. Respondents were given a clinical vignette about an 82-year-old-man with dementia who arrives in the emergency department with life-threatening gastrointestinal bleeding. No guidance on the type of care that the patient would want is available from the patient or his family. The respondent is asked to say how the patient should be treated under three different conditions: that the vignette patient is unfamiliar to the subject, that the patient is a family member of the subject, or that the subject is the patient, himself or herself.

There were dramatic differences in the subjects’ choice of care depending on whether they were choosing it for an unfamiliar person, a family member,

104Darzins, R., Molloy, D.W., Harrison, C. Treatment for life-threatening illness. New Engl. J. Med. 1993; 329: 736.

For example, Kujala and colleagues compared the mortality of twin pairs due to smoking.106 There were 84 twin pairs that were discordant on both the risk factor (one twin smoked and one twin did not smoke) and the outcome (one twin died and one did not). If smoking were unrelated to mortality then we would expect that for half of the pairs it would be the smoker who died and for the other half it would be the nonsmoking twin who died. In fact, in 67 of the pairs it was the smoker in the pair who died, and in only 17 of the pairs was it the nonsmoker who died. McNemar’s test was significant at P 0.001.

Besides McNemar’s test, the matched odds ratio is often used to report the results for matched studies. It is easy to compute. It is:

number of pairs where the one with the

risk factor experiences the outcome

matched odds ratio

number of pairs where the one without the risk factor experiences the outcome

In the case of Kujala and colleagues study of smoking and mortality among twins, the matched odds ratio is:

67 3.9

17

106Kujala, U.M., Kaprio, J., Koskenvuo, M. Modifiable risk factors as predictors of all-cause mortality: the roles of genetics and childhood environment. Am. J. Epidemiol. 2002; 156: 985–93.

Table 5.29. Comparison of sleep characteristics of 1st-degree relatives with sleep apnea/hypopnea compared to controls

*Cases are 1st-degree relatives of patients with sleep apnea.

†P-value is based on paired t-test.

Data from Mathur, R. and Douglas, N.J. Family studies in patients with the sleep apnea–hyponea syndrome. Ann. Intern. Med. 1995; 122: 174–8.

5.11.B Matched measurements of a normally distributed interval variable

When you have matched data on a normally distributed interval variable use a paired t-test (just as you would with paired measures of the same person on a normally distributed interval variable, Section 5.10.C).

For example, Mathur and colleagues conducted a case–control study of whether familial factors were associated with development of sleep apnea/ hypopnea.107 The investigators matched 51 1st-degree relatives of patients with sleep apnea (cases) to 51 controls of similar age, sex, height, and weight. They used paired t-tests to compare cases to their matched controls on several normally distributed interval variables.

The paired t-test for comparing matched results is the same as the paired t-test for comparing two measurements of the same person. The only difference is that with the former the “pair” is the case and the control and with the latter the pair is the two measurements.

As you can see, 1st-degree relatives had significantly less slow wave (deep) sleep, more light sleep, and more 2% and 3% oxyhemoglobin desaturations per hour than controls (Table 5.29).

5.11.C Matched measurements of a non-normally distributed interval or ordinal variable

When you have matched data on a non-normally distributed interval or ordinal variable use the Wilcoxon signed rank test (just as you would with

107Mathur, R., Douglas, N.J. Family studies in patients with the sleep apnea–hypopnea syndrome. Ann. Intern. Med. 1995; 122: 174–8.