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.
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).
| 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.
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).
| 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
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.
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.
Figure 2. Upper 5% point of the Chi-squared distribution with 4 degrees of freedom,
as shown in Table 3
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.
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.
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.
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.
Table 5 shows data from a study of maternity care in Scotland.
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.
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
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
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.
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
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.
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
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.
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.
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.
Fletcher A, Cullum N, Sheldon TA. (1997)
A systematic review of compression treatment for venous leg ulcers.
British Medical Journal 315: 576-580 .
Hundley V, Penney G, Fitzmaurice A, van Teijlinen E, Graham E. (2002)
A comparison of data obtained from service providers and service users
to assess the quality of maternity care.
Midwifery 18, p 126-135.
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.
To Introduction to Statistics for Clinical Trials index.
To Martin Bland's M.Sc. index.
This page maintained by Martin Bland.
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.
Fisher's exact test
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
Yates' correction
The chi-squared test for trend
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
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 area of the circle is proportional to the number of subjects in that smoking group.)
d
Risk ratio
= 0.1990 to 1.0834.
Odds ratios
SE loge(OR) = 0.3679.
= 0.3598 to 1.8020.
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
References
Last updated: 10 September, 2009.