# Introduction to Statistics for Research: Proportions, chi-squared tests and odds ratios

## Qualitative data

Qualitative data are also called nominal or categorical, and happen when we classify subjects into two or more categories. For example, we might classify a patient's condition as 'poor', 'fair', 'good' or 'excellent', or give as options for a question 'yes', 'no', or 'don't know'. This is different from quantitative data, where we have numbers which represent the magnitude of something, such as blood pressure. Even though these may actually be recorded as numerical codes 1, 2, 3, or 4, the number does have any numerical meaning. We could code 'yes' as 1 and 'no' as 2, or 'yes' as 2 and 'no' as 1 and it would not make any difference to the analysis. Categorical variables with only two categories, such as 'alive' or 'dead', or 'female' or 'male' are called dichotomous, attribute, quantal, or binary.

Many statistical methods have been developed to analyse such data. In this lecture I cover the chi-squared test for association, Fisher's exact test, the chi-squared test for trend, risk ratio, relative risk, or rate ratio, odds ratio, and the number needed to treat.

## Contingency tables

A contingency table is a cross-tabulation of two categorical variables. For example, Table 1 shows data from a study of the acceptance of the HIV antibody test in antenatal clinics (Meadows et al., 1994).

Table 1. Acceptance of HIV test grouped by marital status (Meadows et al., 1994)
Marital status Acceptance of HIV test Total
Accepted Rejected
Married 71 415 486
Living with partner 41 181 222
Single 15 35 50
Divorced/widowed/separated 7 23 30
Total 134 654 788

The data are arranged in rows and columns. The part of a table defined by a particular row and column is called a cell of the table. The numbers in a contingency table are frequencies. The number in the first cell of Table 1, which is 71, tells us that in this study there were 71 women who both were married and accepted the HIV test. I have also added the totals for each row and column and for the whole table. The row and column totals are also called marginal totals, the total number of all the observations in the table is called the grand total. This kind of cross tabulation of frequencies is also called a cross classification table. We often refer to tables using the size of the table. Table 1 might be called a '4 by 2' or '4 × 2' table, because it has four rows and two columns. You sometimes hear the general term 'r × c table', where r would denote the number of rows and c the number of columns.

## The Chi-squared test

We often want to test the null hypothesis that there is no relationship between the two variables. We use the term 'association' for a relationship between two categorical variables. If the sample is large, we can do this by a chi-squared test. If the sample is small, we must use a different test, Fisher's exact test, described below.

Our null hypothesis is that there is no association between the two variables. The alternative hypothesis is that there is an association of some type.

The chi-squared test works by calculating the frequencies we would expect to see in the cells if there were absolutely no association. It works like this. For the HIV test data, the proportion who accepted the test is 134/788. Out of 486 married women, we would expect 486 × 134/788 = 82.6 to accept the test if the null hypothesis of not association were true. Similarly, the proportion who refused the test is = 654/788. Out of 899 486 married women, we would expect 486 × 654/788 = 403.4 to accept the test if the null hypothesis were true. Note that 82.6 + 403.4 = 788. The expected frequencies sum to the same total as do the observed frequencies.

In the same way, out of 222 cohabiters we would expect 222 × 134/788 = 37.8 to accept the HIV test if the null hypothesis were true. We would also expect 222 × 654/788 = 184.2 to be refusers if the null hypothesis were true. Note that 37.8 + 184.2 = 222, the second row total. We continue in this way until we have expected frequencies for all the cells of the table (Table 2).

Table 2. Acceptance of HIV test grouped by marital status, with expected frequencies
Marital status Acceptance of HIV test Total
Accepted Rejected
observed expected observed expected
Married 71 82.6 415 403.4 486
Living with partner 41 37.8 181 184.2 222
Single 15 8.5 35 41.5 50
Divorced/widowed/separated 7 5.1 23 24.9 30
Total 134 654 788

Note that 82.6 + 37.8 + 8.5 + 5.1 = 134.0 and 403.4 + 184.2 + 41.5 + 24.9 = 654.0. The observed and expected frequencies have the same row and column totals.

We can see that for each cell of the table, we calculated the expected frequency by

row total × column total / grand total

These would be the expected frequencies if the null hypothesis were true. Here the word 'expected' is used to mean 'the average number of observations we would expect in the cell if we repeated this study over and over again'. It does not mean that we expect to see 82.6 married women accepting HIV testing. Statisticians often torture the language in this way, we regret to say. To be really pedantic, is the expected frequency if the null hypothesis were true and the row and column totals remained the same.

The chi-squared test for a contingency table uses the differences between the observed and expected frequencies. The bigger these differences are, the more evidence we will have that the two variables are associated. We cannot just add these differences, because they always sum to zero. Instead we do what we did when calculating standard deviation, we square them. Another problem is that the bigger the frequencies are, the greater is the possible size of the difference between observed and expected. We might expect that big samples would produce bigger differences than small samples, just by chance. It turns out that we can allow for this by dividing the squared difference by the expected frequency, to give:

(oberved - expected)2 / expected

The precise reasons for this choice are rather abstract and mathematical, so I won't go into them here, but Bland (2000) explains it. We work this (observed - expected)2/expected out for each cell of the table and add them together. I don't usually give formulae in this course, but we shall come across this one quite often and it is pretty simple, so I have included it to make it easier to see what is going on. For Table 2, this sum is 9.15. This will be our test statistic (Week 3). Following the usual formulation of a significance test, this should follow some known distribution if the null hypothesis is true. It follows one called the Chi-squared distribution. (See Bland, 2000, for more about this and why it is true.) 'Chi-squared' is often written , where is the Greek letter 'chi', pronounced 'ki' as in 'kite'. The sum of (observed - expected)2/expected is called the chi-squared statistic, sometimes written as X2. I shall give the Chi-squared distribution a capital 'C' to distinguish it from chi-squared statistics.

The Chi-squared distribution is very like the t distribution, to which it is closely related. It is a family of distributions, and the particular member of the family is defined by one parameter, called the degrees of freedom. Figure 1 shows a few members of the Chi-squared family.

Figure 1. Some members of the Chi-squared distribution family. d

When the degrees of freedom is small, the Chi-squared distribution is highly skewed to the right, and as the degrees of freedom increases it becomes more symmetrical. Eventually it becomes like a Normal distribution. We would expect this to happen, because it is obtained by adding lots of things together and this tends to generate a Normal distribution as the number of things added increases. Like the t distribution and the Normal distribution, there is no simple formula for the area under the chi-squared curve and hence for the probability of exceeding any given value. We can use a table of probabilities laboriously calculated by a (very accurate) mathematical approximation. Table 3 shows some percentage points for the Chi-squared distribution with different degrees of freedom.

Table 3. Percentage points of the Chi-squared distribution
Degrees of
freedom
Probability that the tabulated value is exceeded
10% 5% 1% 0.1%
1 2.71 3.84 6.63 10.83
2 4.61 5.99 9.21 13.82
3 6.25 7.81 11.34 16.27
4 7.78 9.49 13.28 18.47
5 9.24 11.07 15.09 20.52
6 10.64 12.59 16.81 22.46
7 12.02 14.07 18.48 24.32
8 13.36 15.51 20.09 26.13
9 14.68 16.92 21.67 27.88
10 15.99 18.31 23.21 29.59
11 17.28 19.68 24.73 31.26
12 18.55 21.03 26.22 32.91
13 19.81 22.36 27.69 34.53
14 21.06 23.68 29.14 36.12
15 22.31 25.00 30.58 37.70
16 23.54 26.30 32.00 39.25
17 24.77 27.59 33.41 40.79
18 25.99 28.87 34.81 42.31
19 27.20 30.14 36.19 43.82
20 28.41 31.41 37.57 45.32
The table shows the upper 5% or 0.05 point, as shown in Figure 2.

Figure 2. Upper 5% point of the Chi-squared distribution with 4 degrees of freedom, as shown in Table 3 d

Of course, computers are very good for laborious calculations and in practice we just let the computer program do the work and calculate the probability for us every time. One useful feature of the Chi-squared distribution is that its mean is equal to its degrees of freedom. So if the observed value of a chi-squared statistic is similar to or less than its degrees of freedom, the data will be consistent with the null hypothesis being tested.

We decide which member of the Chi-squared family applies to our table by calculating the degrees of freedom. For a contingency table, the degrees of freedom are given by:

(number of rows - 1) × (number of columns - 1).

Again, I won't go into the reasons for this, see Bland (2000) if you are curious. But we can see something of the logic by looking at how many cells of the table can be filled before the remaining cells can be calculated from the row and column totals. We can start filling the cells of the first row, until we reach the last cell. This must be fixed, because all the cell frequencies must sum to the row total. So the number of free choices in the first row is the number of columns minus one. We can fill in cells in the second row in the same way, getting another 'number of columns minus one' free choices. We go on doing this until we reach the last row. Now these frequencies will al be fixed, because the frequencies in each column must sum to the column totals. For all the rows but one, we have 'number of columns minus one' free choices. Hence the total number of free choices is 'number of columns minus one' multiplied by 'number of rows minus one'. This gives us the degrees of freedom for the table.

For Table 2, we have (4 - 1) × (2 - 1) = 3 degrees of freedom. If we look in Table 3 at the 3 d.f. row, we see that the 5% point is 7.81 and the 1% point is 11.34. Our observed chi-squared statistic, 9.15, is between these two, so our probability is between 5% and 1%. We would write this as P<5% or P<0.05. Using a computer to do the calculation, we get the more accurate P = 0.027, which we could round to one significant figure and report as P = 0.03.

Like most significance tests, there are some assumptions we have to make about the data or conditions that the data must fulfil for the test to be valid. These are that the sample is large enough and that the observations are independent. The conventional criterion for the sample size is this: the chi-squared test is valid if at least 80% of the expected frequencies exceed 5 and all the expected frequencies exceed 1. This is a large sample test. The smaller the expected values become, the more dubious will be the test. For Table 2, they all exceed 5.0, so there is no problem. As there are 8 expected frequencies, we could accept 8 × 0.2 = 1.6 expected frequencies less than five. We should round this down to 1.0 and say that one expected frequency between one and five would not be a problem. If we have a 2 by 2 table, 20% of the cells is 4 × 0.20= 0.80, which is less than one, so no cell should have an expected frequency less than 5. Note that the observed frequencies could be zero without affecting the validity of the test. As long as the expected frequencies are greater than 5.0, the test is valid.

What should we do if this condition is not met? We can make the expected frequencies bigger by combining rows or columns. In Table 2, for example, we might combined the single and divorced/widowed/separated categories. Another approach is to drop a category altogether, if there is no obvious combination including it. It is easy to see how combining categories will affect expected frequencies, because when we do it the expected frequencies will simply be added together. The alternative approach is to use Fisher's exact test, or, for 2 by 2 table, Yates' correction, both described below.

The chi-squared test for association in a contingency table is also known as the Pearson chi-squared test.

The chi-squared statistic is not an index of the strength of the association. If we double the frequencies, this will double chi-squared, but the strength of the association is unchanged. There are indices of strength for use in special circumstances, but they are not seen much.

## Fisher's exact test

When the chi-squared test is not valid because the expected frequencies are too small, we can use a different test, Fisher's exact test, also called the Fisher-Irwin exact test. This works for any sample size, though it used to be used only for small samples in 2 by 2 tables, because of computing problems. Now that we have powerful computers with efficient statistical programs it can be done for any table.

For Fisher's exact test, we calculate the probability of every possible table with the given row and column totals. We then sum the probabilities for all the tables as or less probable than the observed. You can do this by hand for a 2 by 2 table with small frequencies (Bland 2000 gives a formula) but you would not want to. For Table 1, Fisher's exact test gives P = 0.029. Compare this to the chi-squared test P = 0.027. They are very similar. This is not always the case. Table 4 shows the results of a clinical trial.

Table 4. Leg ulcer wound healing by type of bandage in a randomised trial, with row percentages (Callam et al., 1992)
Bandage Healed Did not heal Total
n % n % n %
Elastic 35 53.8 30 46.2 65 100.0
Inelastic 19 28.4 48 71.6 67 100.0
Total 54 40.9 78 59.1 132 100.0

Fisher's exact test for this table gives P = 0.004. The chi-squared test gives chi-squared = 8.87, 1 d.f., P = 0.0029. These are rather different, though both would lead to the same conclusion. Which should we choose? I think that Fisher's exact test is always to be preferred, because it is exact. So why have two tests? Fisher's can only be done for tables with more than two rows or columns or with a large number of subjects if you have a modern computer. Researchers' behaviour changes slowly, textbooks are always out of date, and it takes a long time for practice to change. You will see lots of chi-squared tests. There are many other situations where chi-squared statistics are used and the principles will be much the same. I consider one below under the chi-squared test for trend.

Not all statisticians would agree that Fisher's exact test is right and this is a matter of occasional lively debate among statisticians (yes, it really happens). I might be wrong. You will hear Fisher's exact test described as 'conservative' because it gives larger P values than does the chi-squared test. I think that the opposite is true, that the chi-squared tests gives P values which are too small and is 'anti-conservative'. This only matters when the frequencies are small,

There are two different ways of doing Fisher's exact test. The test above was done using the statistical program Stata. I also tried it using my own program Clinstat. This gave 0.0049. This is not because I am a poor programmer, but because there are two different ways to calculate the Fisher probability. When the test was first devised, it was strictly for 2 by 2 tables. You could do a one-tailed test, by summing the probability for your observed table and for all the more extreme ones in the same direction. For Table 4, this meant making the number in the first cell (35) progressively bigger (36, 37, etc., up to 54) and calculating the probability for each of them. These probabilities would be added. This gave a one-sided test. To make a two-sided test it was not clear what tables you would use, nothing is symmetrical. What was done originally was to double the one-tailed P value. This is what Clinstat does. It could give probabilities greater than 1.0, but these would be set equal to 1.0. When you move to a table with more than two rows or columns, the direction of the effect doesn't mean anything and all tests will be two-sided. The way Stata does it is the only way for larger tables and so I now recommend it as the better way. Programs have been written which do both and let you make up your own mind. I would just do whatever your software does, but you should always state which program you use so that anybody who really wants to know can find out which method was used.

Even with the computers available at the time of writing, very large tables can defeat them. There are so many probabilities to compute that they cannot store them all and cannot calculate them before the summer holidays. Some programs, such as SPSS, offer a Monte Carlo option. This means that tables are generated at random and a sample of all the possible tables is produced and their probabilities calculated. This is used to estimate the P value. If you do not have enough simulations, you might get a slightly different answer if you do it again, so make sure you ask it for a few thousand.

## Yates' correction

For tables with small expected frequencies, the chi-squared test gives smaller probabilities than Fisher's exact test. If you believe the exact probability, then the chi-squared test is wrong. Yates introduced a modified chi-squared test for a 2 by 2 table which approximates the Fisher's exact probability very closely. It works by making the difference between observed and expected frequency closer to zero by 0.5 before squaring. It works extremely well. For Table 4, Fisher's exact test gives P = 0.0049 and the ordinary chi-squared test gives the chi-squared statistic = 8.87, d.f. = 1, P = 0.0029. The chi-squared test with Yates' correction gives a smaller chi-squared statistic = 7.84, d.f. = 1, P = 0.0051. This is very close to the 0.0049 given by the exact test. It is a remarkably good modification to the chi-squared test and much easier to calculate without a computer than is the exact test.

Yates' correction is also called the continuity correction for the chi-squared test. This is because it makes allowance for the fact that for a table with few observations there are only a small number of possible values for the chi-squared statistic. We approximate the discrete distribution of the test statistic with the continuous Chi-squared distribution and Yates' correction improves this approximation.

Even Yates' correction makes the calculation without a computer more time-consuming than the simple chi-squared test and it used to be recommended only when there was at least one expected frequency less than five. This convention continued when people began routinely to do these calculations using computers. However, if Yates' correction is better for small tables, it must be better for all tables.

Yates' correction is now obsolete as we can always do the exact test and we will start to see it less often.

## The chi-squared test for trend

Table 5 shows data from a study of maternity care in Scotland.

Table 5. Number of antenatal visits by type of unit in a survey of women in Scotland (Hundley et al., 2002)
Type of maternity unit Number of antenatal visits that women received Total
0-4 5-9 10-14 15 or more
Traditional model n 10 82 167 72 331
% 37.0 30.8 40.7 46.2 38.5
New model n 17 184 243 84 528
% 63.0 63.0 69.2 59.3 61.5
Total n 27 266 410 156 859
% 100.0 100.0 100.0 100.0 100.0

We might carry out a chi-squared test for association on this table, which would give chi-squared = 11.36, d.f. = 3, P = 0.01. Many researchers would do exactly this. Hundley et al. (2002) did not, however, they reported chi-squared = 9.33, d.f. = 1, P = 0.002. They used a different test called the chi-squared test for trend.

In Table 5, the categories '0-4', '5-9', '10-14', and '15 or more' have a clear order, unlike the marital status categories in Table 1. The chi-squared test for association does not take the ordering of the categories into account. We would get the same chi-squared statistic however we shuffled the rows of Table 1. This is as it should be, we would not want to treat 'married', 'living with partner', 'single', and 'divorced/widowed/separated' as a meaningful ranking, because it isn't. If we write the 'single' row first, we want to get the same P value. When we have ordered categories, we should use the information which the ordering gives us.

In the chi-squared test for trend, we not only use the order of the categories, but attach a numerical value to them. Hundley et al. (2002) have attached the values 1, 2, 3, and 4 to the categories '0-4', '5-9', '10-14', '15 or more'. This is reasonable, as these categories are derived from numbers of visits and each category except the last covers the same range of visits, 5. Often we do not have an underlying qualitative variable, such as when a subject's condition is classified as 'poor', 'good', fair', or 'excellent'. These are clearly in ascending order, but our decision as to what numerical value to attach is arbitrary, a matter of judgement. When we have no reason to decide otherwise we usually make the intervals equal, i.e. we score them 1, 2, 3, etc.

The chi-squared for trend statistic is always less than the chi-squared for association statistic. In this example, it is 9.33 compared to 11.36. However, it has fewer degrees of freedom and, if there really is a trend, will have a smaller P value. Hence it is a more powerful test when categories are ordered. The difference between the two chi-squared statistics also follows a Chi-squared distribution if the null hypothesis is true, with degrees of freedom equal to the difference between the two degrees of freedom. For the example we have chi-squared = 11.36 - 9.33 = 2.03, with degrees of freedom 3 - 1 = 2, P = 0.4. This is called the chi-squared about trend, and tests whether there is evidence for any association in the table which would not be explained by a linear trend. In this example, the data are clearly consistent with the null hypothesis that there is no non-linear association.

Many qualitative variables with more than two categories have ordered categories and should be analysed in some way which takes the ordering into account. There are some slight variations on the formulae used in this test and different programs might give slightly different answers. A different approach, which some programs use, is rank correlation, in particular Kendall's tau b. Stata does this and for Table 5 gives Kendall's tau-b = -0.1031, SE = 0.032. If we use this for a test of significance we get P = 0.001, slightly more significant than the chi-squared test for trend in this example. This may be because the category representing the lower extreme, '0-4' has a greater percentage in the 'traditional model' than the next category, '5-9', though after the percentages rise. Because there are only a few subjects in this category, this is insufficient to produce a significant association not explained by the trend.

There are two other points worth noting about Table 5. First, I have given the percentages for the columns, using the number of women in the visit category as the denominator. It would have been more logical to take them for the rows, with the number of women in the type of unit as the denominator, since it is the types of unit which we want to compare. This is what Hundley et al. (2002) did. My percentages were calculated to bring out the trend across the table. Second, there were 6 traditional units and 14 new model units. So the women were not really independent, as we might expect the number of visits to be related to the practice in the particular unit. An assumption of the test has been violated. It might have been better to estimate an average number for each unit and compare these with a two-sample method for continuous data. There are often many possible ways to analyse data, each of which may have advantages and disadvantages compared to the others.

For another example, Tappin et al. (2005) carried out a trial of motivational interviewing by midwives to help pregnant smokers to quit or cut down. Table 6 shows the change in self-reported smoking habits by treatment group.

Table 6. Change in self-reported smoking habit by pregnant women in a trial of motivational interviewing by midwives (Tappin et al., 2005)
Change in smoking habit Motivational interviewing Control
Quit 24 6.8% 31 7.5%
Cut down 31 8.8% 52 12.7%
Same 278 79.2% 286 69.6%
More 18 5.1% 42 10.2%
Total 341 100.0%           417       100.0%

The authors reported chi-squared for trend = 0.01, P=0.98, chi-squared non-linear = 11.26, P=0.004 (2 d.f.). There is a significant relationship overall, chi-squared = 11.26, 3 d.f., P = 0.01. All the chi-squared is in the part which is about trend. There is a relationship, but it is not a linear trend. The proportion of women in the motivation interviewing group goes down from quit to cut down, then goes up to same, then drops to more.

Figure 3 illustrates the two trend tests. Neither shows a completely smooth trend in one direction, but for the antenatal visits the three large groups clearly do show a rising trend. There is no discernable direction for the smoking study.

Figure 3. Two trend analyses (The area of the circle is proportional to the number of subjects in that smoking group.)     d

## Risk ratio

We now turn to methods for 2 by 2 tables only. Consider the venous leg ulcer bandaging trial shown in Table 4. We want an estimate of the size of the treatment effect. One we have already looked at is the difference between two proportions, for which we can find a standard error, a large sample confidence interval using the standard error, and a small sample confidence interval using exact probabilities. For Table 4 the two proportions are 0.538 and 0.284, or 53.8% and 28.4%, for elastic and inelastic bandaging respectively. The difference is 0.538 - 0.284 = 0.254 or 53.8% - 28.4% = 25.4 percentage points.

We can also carry out a test of the null hypothesis that the proportions healed are the same for the populations of ulcer patients treated by elastic and inelastic bandages. This can be done using the standard error for testing the null hypothesis. This test gives a test statistic which, follows a Standard Normal distribution if the null hypothesis is true. For Table 4 this is z = 2.98, which has two-tailed probability P = 0.0029. We would usually present this as P = 0.003. This is exactly equivalent to the chi-squared test for association, and always gives an identical P value. The test statistic here is the square root of the chi-squared.

Proportion who heal is called the risk of healing for that population. This sounds rather odd, since risk usually refers to something bad and healing is a good thing, but because we use exactly the same methods to analyse the proportions experiencing desirable events as for the proportions who experience undesirable events we use the same term for both. The difference is called the risk difference, absolute risk difference, or absolute risk reduction.

We can look at the difference in risk between the two treatment groups in a different way. The ratio of the risk of healing in the elastic bandage group to the risk in the inelastic bandage group is called the risk ratio. For Table 4, the risk ratio = 0.538/0.284 = 1.89. The risk ratio is also called the relative risk and the rate ratio, all of which can be conveniently abbreviated to RR.

Having calculated our estimate of effect, we would like a confidence interval for it. Ratios are rather difficult things to deal with statistically. Because risk ratio is a ratio, it has a very awkward distribution. Variations which makes the denominator smaller have a much bigger effect on the ratio than do those which make the denominator larger, and we get a skew distribution. We deal with this using logarithms (see lecture on Tranformations). Taking the log of a ratio gives us the difference between the logs of the two numbers. If we take the log of the rate ratio, we have something which is found by adding and subtracting log frequencies. The distribution becomes approximately Normal and, provided the frequencies are not small, it has a simple standard error.

For Table 4, the log risk ratio is loge(RR) = 0.6412. Its standard error is 0.2256 and so we can find a 95% confidence interval for loge(RR) by

0.6412 - 1.96×0.2256 to 0.6412 + 1.96×0.2256
= 0.1990 to 1.0834.

Most of us do not think easily in terms of logarithms, so we transform this back to the natural scale by the antilog or exponential: 95% CI for RR = exp(0.1990) to exp(1.0834) = 1.22 to 2.95. So we estimate that the proportion who will heal given elastic bandaging is between 1.22 and 2.95 times the proportion who heal given inelastic bandaging. Although we have a different confidence interval for the risk ratio, we use exactly the same P value as before.

The risk ratio is not in the middle of its confidence interval, unlike the risk difference. The confidence interval is symmetrical on the log scale, not the natural scale.

These confidence intervals are large sample approximations. There are several better ones which are used for small frequencies. A good guide is that if all four frequencies exceed five, the large sample method will be OK. We have problems when one of the frequencies is zero, as the relative risk may be zero or not able to be calculated at all. Estimation is very difficult, but we can find an upper or a lower limit even if we cannot estimate the RR.

## Odds ratios

The odds ratio is another method to estimate the relationship in a 2 by 2 table. First, we shall look at odds. This is a familiar concept from sports and gambling. In statistics the odds of an event is number of times it happens divided by the number of times it doesn't happen. For example, in Table 4 there were 65 patients who received elastic bandages, of whom 35 healed and 30 did not. The risk of healing = 35/65 = 0.538. The odds of healing = 35/30 = 1.17. So

• Risk = number experiencing event divided by number who could.
• Odds = number experiencing event divided by number who did not experience event.

Another way to look at this is that risk = 0.538 means that for every person treated, 0.538 people heal, or for every 100 people treated, 53.8 people heal. Odds = 1.17 means that for every person who do not heal, 1.17 people heal, or for every 100 people who do not heal, 117 people heal.

Another way to define the odds is that it the probability of the event divided by one minus the probability of the event. Hence the odds of healing is 0.538/(1 - 0.538) = 1.17.

The odds ratio is the odds of healing given elastic bandages divided by the odds of healing given inelastic bandages. As we have seen, the odds of healing given elastic bandages = 35/30 = 1.17. Similarly, the odds of healing given inelastic bandages = 19/48 = 0.40.

Odds ratio = (35/30) / (19/48) = 1.17 / 0.40 = 2.95.

For every person who does not heal, 2.95 times as many will heal with elastic bandages as will heal with inelastic bandages.

'Odds ratio' is often abbreviated to 'OR'. Like RR, OR has an awkward distribution and we estimate the confidence interval in the same way. We use the log odds ratio. When we do this the distribution becomes approximately Normal and, provided the frequencies are not small, it has a simple standard error.

loge(OR) = loge(2.95) = 1.0809.
SE loge(OR) = 0.3679.

As usual, we will not go into the details of how this was calculated, as we would expect anyone who wanted to do statistical analysis would use a computer, but Bland (2000) gives details. We find the 95% CI for loge(OR) using (1.96 standard errors by the usual large sample method:

1.0809 - 1.96 × 0.3679 to 1.0809 + 1.96 × 0.3679
= 0.3598 to 1.8020.

To find the 95% CI for the odds ratio itself we antilog these limits:

95% CI for OR = exp(0.3598) to exp(1.8020) = 1.43 to 6.06.

Hence our estimated odds ratio is OR = 2.95, 95% CI = 1.43 to 6.06.

As for RR, OR is not in the middle of its confidence interval. The interval is symmetrical on the log scale, not the natural scale.

One of the great things about the odds ratio is that it doesn't matter which way round we do it. We can find the odds ratio for treatment by elastic bandage given healing. 35 patients healed and had elastic bandaging compared to 19 to healed with inelastic bandaging, so the odds of elastic bandaging for those who healed was 35/19. Similarly, the odds of elastic bandaging for those who did not heal was 30/48. Hence

Odds ratio for treatment: OR (35/19) / (30/48) = 0.95 as before.

Both these versions of the odds ratio are the same:

(35/30) / (19/48) = (35/19) / (30/48) = (35×48) / (30×19) = 0.95

So both versions of the odds ratio = (35×48)/(30×19). We also call this the ratio of cross products, because the numerator is the top left cell frequency multiplied by the bottom right and the denominator is the top right cell frequency multiplied by the bottom left.

Switching the order of the rows or columns inverts the odds ratio. Table 7 shows the data of Table 4 with the order of the columns reversed.

Table 7. Leg ulcer wound healing by type of bandage in a randomised trial, with order of columns reversed (Callam et al., 1992)
Bandage Did not heal Healed Total
n % n % n %
Elastic 30 46.2 35 53.8 65 100.0
Inelastic 48 71.6 19 28.4 67 100.0
Total 78 59.1 54 40.9 132 100.0

We can calculate the odds ratio for not healing given elastic bandage from the ratio of cross-products in this table: OR = (30/35)/(48/19) = 0.339 = 1/2.95. So this is one over the odds ratio for healing. In fact, there are only two possible odds ratios for a 2 by 2 table, reflecting the directions in which we might look at the relationship. On the log scale, these are equal and opposite: loge(2.95) = 1.082 and loge(0.339) = -1.082.

Odds ratios have the same problems when frequencies are small as do relative risks.

## References

Bland M. (2000) An Introduction to Medical Statistics, 3rd edition. Oxford University Press.

Callam MJ, Harper DR, Dale JJ, Brown D, Gibson B, Prescott RJ, Ruckley CV. (1992) Lothian Forth Valley leg ulcer healing trial-part 1: elastic versus non-elastic bandaging in the treatment of chronic leg ulceration. Phlebology 7: 136-41.

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Meadows J, Jenkinson S, Catalan J. (1994) Who chooses to have the HIV antibody test in the antinatal clinic? Midwifery 10, 44-48.

Northeast ADR, Layer GT, Wilson NM, Browse NL, Burnand KG. (1990) Increased compression expedites venous ulcer healing. Royal Society of Medicine Venous Forum. London: Royal Society of Medicine.