Chi Square for 2x2 contingency table, from the Unpaired Comparison of Two Proportions Program Page , tests the probability that the number of positive and negative cases in two groups are from a similar population, a goodness of fit test.   Until tests using confidence interval became popular in the 1990s, this was the principle test used to compare two proportions.

For Chi Square to be used, the total number in the study should exceed 30 cases, and each cell should have at least 5 cases.  Short of these numbers, the assumptions of Chi Squares distribution cannot be assured and there is a possibility of misinterpretation.

An example

We wish to study the difference in the preferences for specialty training between the male and female interns in a hospital. We found a group of 16 male interns, 10 of whom chose surgical specialties. We also found a group of 21 female interns, 5 of them chose surgical specialties. The Chi Squares is 4.15 and α=0.04. From this we can conclude that male and female interns differ significantly (at the p<0.05 decision level), in choosing surgical specialties for training.

Fisher's exact probability, from the Unpaired Comparison of Two Proportions Program Page , estimates the probability of observing the difference in proportions in two groups, departing from the null hypothesis that the two groups are the same. The calculation provided estimates the probability for the two tail test.

As the probability is estimated using permutation, no assumption of the underlying binomial distribution is made, and the power of the model is 100%. Because of this, the test remains valid even when the sample size is small. Fisher's test is therefore often used when the sample or cell size are insufficient for the chi square test.

The test uses Factorial numbers repeatedly, so that computing time increases exponentially with the number of cases. Therefore this test is only used when the conditions for a Chi Square Test cannot be satisfied.

An example

We repeat the study in the Chi Square Panel, but use Fisher's Exact Probability instead.

We wish to study the difference in the preferences for specialty training between the male and female interns in a hospital. We found a group of 16 male interns, 10 of whom chose surgical specialties. We also found a group of 21 female interns, 5 of them chose surgical specialties. Fisher's Exact Probability = 0.02, and we can conclude that male and female interns differ significantly (at the p<0.05 decision level), in choosing surgical specialties for training.

Risk difference, from the Unpaired Comparison of Two Proportions Program Page , is usually used to evaluate the results of a controlled trial. The risk is the proportion of the group with positive outcomes, so that risk in group 1 risk1=Pos1/(Pos1+Neg1), and risk in group 2 r2=Pos2/(Pos2+Neg2). Risk difference is then rd=r1-r2.

There are two methods to estimate the Standard Error (se) of the risk difference, from which the 95% confidence interval is calculated

• The traditional method is the most commonly used one, and this assumes the difference is a sample mean of a normally distributed measurement, with a Standard Error that can be estimated. This method is appropriate if the sample size in the study is at least 30 cases per group, and produces an increasingly misleading conclusion as sample size become smaller.  Should one of the group has 0 case in one of the cells then the program crashes with a division by zero error.
• The small sample or exact method makes no assumption about distribution, and can be used even if one of the cells contain zero case.   However it is not as powerful as the traditional method.

After a controlled trial, a piece of useful information is how many subjects need to change their treatment to affect a single case with a different outcome. This is known as the numbers needed to treat (NNT), which is the inverse of the risk difference, NNT = 1 / rd.

An example :

We wish to study whether preoperative antibiotics reduces postoperative infections in a randomised controlled trial. Group 1 are 11 patients that received antibiotics, and 4 of them had postoperative infections. Groups 2 are 15 patients that received no antibiotics and 12 of them had postoperative infections.

Risk of infection in the group receiving antibiotic is r1 = 4/11 = 0.31 (31%), and risk in the group that did not receive antibiotic is r2 = 12/15 = 0.8 (80%). Because the sample size is small, the small sample method is used. The risk difference is 0.31 - 0.8 = -0.49 (-49%), with the 95% confidence interval -0.71 to -0.12. In other words, those receiving antibiotics were 12% to 70% less likely to have postoperative infections than those not receiving antibiotics.

If the traditional method was used, the 95% confidence interval would be -0.81 to -0.17 (17% to 81%)

Number needed to treat is based on the risk difference, which is the same in both methods. NNT = 1 / RD = 1 / 0.49 = 2.03, rounded upwards to 3. In other words, for the reduction of 1 case of infection, three more patients will need to receive pre-operative antibiotics.

Risk Ratio (also called Relative Risks by some), from the Unpaired Comparison of Two Proportions Program Page , is used to evaluate risks in epidemiological studies, and to a less extent also in controlled trials. Relative Risk is the ratio of risks between the two groups rr = r1/r2 = (Pos1(Pos2+Neg2))/(Pos2(Pos1+Neg1)).

Risk Ratio addresses one of the problems associated with risk differences, that of comparing different effects or outcomes in a study. For example, in a study of fetal monitoring, there is an increase in the risk of Caesarean section with the use of monitoring, and possibly a reduction in fetal death. The problem is that a doubling of Caesarean section from 10% to 20% is 10% (rd=0.1), but a halving of fetal death is from 2 in a thousand to 1 in a thousand (rd=0.001). Risk Ratio is therefore a better comparison, as one is 2 and the other 0.5 (1/2).

Risk Ratio (rr) is a ratio, which has a log-normal distribution. All the calculations are therefore conducted using log(rr), and the final results are reverse transformed to the non-logarithm for clinical interpretation. The 95% interval of the non-log Risk Ratio is therefore assymmetrical around the mean rr value, being larger on the side of higher values. Graphical representation of Risk Ratio therefore uses a logarithm scale.

An example

We wish to study whether men are more likely to have car accidents than women. We collected a database of drivers, and found that out of 500 men 23 had an accident in the past year, while out of 530 women 10 had an accident during the same period.

Risk of car accident amongst men is r1 = 23/500 = 0.046 (4.6%), risk of accident in women is r2 = 10/530 = 0.019 (1.9%). Relative Risks is rr = 0.046/0.019 = 2.44, with 95% CI 1.17 to 5.07. In other words, male drivers have 1.17 to 5.07 times the risk of having a car accident compared with female drivers.

Odds Ratio, from the Unpaired Comparison of Two Proportions Program Page , is often used in situations where cause and effect are confused, and the concept of risk cannot be applied. An example is the relationship between social class and educational achievement, when it is uncertain whether the lack of education results in a lower social class or whether lower social class led to under-achievement in education.

Odds is the ratio of positives and negatives, so that odds in group 1 o1 = Pos1/Neg1, and in group 2 o2 = Pos2/Neg2. Odds ratio is or = o1/o2 = Pos1Neg2/Pos2Neg1.

Odds Ratio (or) is a ratio, which has a log-normal distribution. All the calculations are therefore conducted using log(or), and the final results are reverse transformed to the non-logarithm for clinical interpretation. The 95% interval of the non-log Odds Ratio is therefore assymmetrical around the mean or value, being larger on the side of higher values. Graphical representation of Odds Ratio therefore uses a logarithm scale.

An Example We wish to study the relationship between the deflexion of the fetal head and occipital posterior position in early labour. We do not know, and are unconcerned, which comes first, but merely want to know if there is an association.

We found 20 babies with occipital posterior position in early labour, 12 of whom has a deflexed head (Pos1=12) and 8 with a flexed head (Neg1=8) . We also found 35 babies with occipital anterior position, 10 of which had deflexed head (Pos2=10) and 25 with a flexed head (Neg2=25). The Odds ration is 3.75, the 95% confidence interval 1.4 to 9.9. We can therefore conclude that occipital posterior and deflexion of the fetal head in ealy labour are related.

Odds Ratio is also used extensively in multivariate assessment of risks such as in Logistic Regression and Bayesian Probability Models. A particular useful research model is the Retrospective Matched Paired Controlled Comparison of two groups, which allows a test of causal factors in groups with different outcomes. This is discussed in the Sample Size for Matched Paired Controlled Studies Explained and Tables Page and will not be repeated here.

The Marascuilo Procedure from the Comparison of Proportions in Multiple Groups Program Page is used to compare proportions when there are more than two groups with binary outcomes (yes/no). It firstly performs a test of overall homogeneity for a large contingency table. This is followed by multiple post hoc comparisons between pairs of groups in the data.

The Mascuilo's procedure is simpler and replaces using multiple comparisons of two groups with a Bonferroni's correction for multiple comparisons, so it is preferred when there are multiple groups to be compared at the same time.

An Example : is as the default example data from the Comparison of Proportions in Multiple Groups Program Page .

We wish to study academic achievements in 3 social groups of students.

• Group 1 consists of 100 students from single parent family living in a deprived suburb. 60 of these students (0.6 or 60%) are judged to be under achievers.
• Group 2 consists of 80 students, also from deprived suburbs, but coming from stable two parent families. 20 of these students (0.25 or 25%) are judged to be under achievers.
• Group 3 consists of 60 students from intact families living in middle class suburbs. 10 of these students (0.17 or 17%) are judged to be under achievers.

An initial analysis using the Chi Squares Test indicates that the rates of under achievement are significantly different in the 3 groups (Chi Sq = 38.04, df=2, α<0.001).

Using the Marascuilo procedure to perform post hoc analysis, we found that group 1 is significantly different to group 2 (difference = 35%, α<0.001), and to group 3 (difference = 43.3%, α<0.001).

However, the difference between groups 2 and 3 (difference = 8.3%, α=0.47) is not statistically significant.

The Chi Square Test for contingency tables with multiple rows and columns tests that the distirbution between the columns in all the rows are similar. Like the Chi Squares Test for the 2x2 table, it is a goodness of fit test.

An Example

We wish to find out whether the shape of pelvis are similar between different ethnic groups. We reviewed all our pelvimetry, and found the following.

	Gynaecoid	Anthropoid	Android	Platyploid
Chinese	43	49	40	36
Vietnamese	49	48	42	47
Korean	33	36	31	41
Japanese	39	35	38	49

Chi Square = 5.31, df=9, p = 0.81, no statistical difference.

Please note : I used the random number generator to create the data to demonstrate the statistics. The data does not represent any reality. In any case, modern obstetrics generally considered these classifications of pelvis arbitrary and irrelevant anyway.

The large contingency table is also used for the calculation of the nonparametric Spearman's Correlation Coefficient. This is discussed in the Correlation and Regression Explained Page , and not repeated here

Standardization was originally developed for the analysis of death rates in epidemiological studies, to correct for age differences in different population groups. An example is the analysis of cancer deaths in two populations when the age distribution in these populations are dissimilar.

The calculation estimates the number of positive cases that would exist if the age distribution is standardized, making it possible to compare the rates (incidents) without the distortion of age distribution.

The method has been generalised so that comparison of rates between groups can be made after correction for one or more other variables.

Standardization requires the nomination of a standard population. The most common approach is to nominate one of the groups under comparison as the standard, and adjusts the number of positive cases in all other groups accordingly. Usually, the group that represents the control group is used.

When there is no clear choice which group should be the standard, the mean values across all the groups are used for standardization.

This section describes the use of the program in the Standardization of Proportions Program Page .

Input data This consists of 4 columns.

• Col 1 is the group designation. These are the groups being compared.
• Col 2 is the standardizing parameters designation, consisting of the subgroups. Please note that although these parameters are usually numerically designated, they are nominal and not ordinal in nature. In other words, the number assigned to them are merely names of subgroups (see example).
• Col 3 contains the number of subjects that are attribute negative (normal)
• Col 4 contains the number of subjects that are attribute positive (indexed outcome).

Results Output : These are in 2 sections.

Section 1 displays the input data, and can be referred to when the results are examined.

• The first table contains the total numbers in each row of data, and these numbers as percentage of the group total. This shows how the cases are distributed throughout the subgroups.
• The second table shows the number of positives, and these numbers as a percent of the row total. This shows the proportion of positives in all the subgroups.
• At the bottom of the second table are the total numbers and percent of positives for each group. These are the primary uncorrected proportions of positives in each group.

Section 2 displays the results of analysis. The program in the Standardization of Proportions Program Page calculates standardization multiple times, using in turn each group as the standard, as well as the means of all the groups. The format of these results are the same, and the user should choose only one of the results to present, basing the choice on the most appropriate standard.

The results are presented in 2 tables.

Table 1 examines each group as they are modified by standardization.

• The body of the table lists the standardized number of positives for each row for each group.
• At the right hand end of the table, the reference total number of cases for each row is presented. This is the number used to perform the standardization.
• Most of the table is useful only as a reference. The results are presented at the bottom of the table, and these are the standardized proportion of positives, its standard error, and 95% confidence interval, for each group.

Table 2 performs pair-wise comparison between the groups, with differences, Standard Errors and 95% confidence intervals. These are the final result output.

• The 95% confidence interval is calculated using difference ± 1.96 SE, and this is only correct if 1 comparison is made.
• For multiple comparisons (where there is more than 2 groups), the Bonferroni's correction is required. The 95% CI is then calculated as difference ±zSE, where z is obtained from the Probability of z Explained and Table Page using the Bonferroni's correction for a α of 0.05 (α_corrected = 0.05 / number of comparisons)

Example

		Employed(-)	Unemployed(+)
<secondray(1)	Male_unmarried(1)	80	20
<secondray(1)	Male_married(2)	165	35
<secondray(1)	Female_unmarried(3)	75	25
<secondray(1)	Female_married(4)	100	50
Secondary(2)	Male_unmarried(1)	450	50
Secondary(2)	Male_married(2)	650	50
Secondary(2)	Female_unmarried(3)	290	30
Secondary(2)	Female_married(4)	250	30
Tertiary(3)	Male_unmarried(1)	220	10
Tertiary(3)	Male_married(2)	890	12
Tertiary(3)	Female_unmarried(3)	430	25
Tertiary(3)	Female_married(4)	540	30

Standardization is usually performed on epidemiological data, involving large sample sizes. The following examples however uses computer generated data and contains only few subjects. This is done to demonstrate the use of the method

We wish to study the epidemiology of unemployment, in particular whether educational levels have an effect. The data is as shown to the right. We have 3 groups to compare.

• Group 1 are those who did not finish school
• Group 2 are those who finished school but had no tertiary or vocational education
• Group 3 are those who finished either tertiary education or vocational training.

However, we realise that the sex and marital status of the subject would also have an influence on employment, so we also subdivided our cases into 4 parameter groups

1	1	80	20
1	2	165	35
1	3	75	25
1	4	100	50
2	1	450	50
2	2	650	50
2	3	290	30
2	4	250	30
3	1	220	10
3	2	890	12
3	3	430	25
3	4	540	30

• Group A(1) are male unmarried
• Group B(2) are male married
• Group C(3) are female unmarried
• Group D(4) are female married.

We collected our data as shown above and to the right (we did not, I made up the data). The data input matrix is therefore as to the left

Please note that column 1 designate the groups that we primarily intend to address, that of educational level. Column 2 are the 4 qualifying groups of sex and marital status.

	Grp 1		Grp 2		Grp 3
Param	Row Tot	Row%	Row Tot	Row%	Row Tot	Row%
1	100	18.18	500	27.78	230	10.66
2	200	36.36	700	38.89	902	41.82
3	100	18.18	320	17.78	455	21.09
4	150	27.27	280	15.56	570	26.43
Grp Tot	550	100	1800	100	2157	100

The first output, table 1 to the left, shows the total numbers in each subgroup. Please note that the numbers in the subgroups are very much different amongst the three educational level groups, so that, without some form of correction, the effects of education on unemployment will be masked by its relationship with sex and marital status.

	Grp 1		Grp 2		Grp 3
Param	n+ve	%	n+ve	%	n+ve	%
1	20	20.00	50	10.00	10	4.35
2	35	17.50	50	7.14	12	1.33
3	25	25.00	30	9.38	25	5.49
4	50	33.33	30	10.71	30	5.26
Grp Tot	130	23.64	160	8.89	77	3.57

The second table (to the right) shows the percent unemployment (+ve) for each subgroup, and the uncorrected total unemployment rate in each group. Although it is pretty obvious that the 3 groups are different, the total unemployment rate between the groups cannot really be compared. The variation of unemployment between sex and marital status vary widely, and the proportion of different sex and marital status are quite different in the 3 groups.

	Grp1	Grp2	Grp3
Param	Std +	Std +	Std +	Std Total
1	56.7	28.4	12.3	283.6
2	102.6	41.9	7.8	586.3
3	71.4	26.8	15.7	285.7
4	115.6	37.2	18.3	346.8
Grp Tot	346.3	134.2	54.1	1502.3
Adj%	23.05	8.93	3.60
SE	7.54	2.97	1.99
95%CI	8.28	3.11	-0.31
95%CI	37.82	14.75	7.50

The first table of section 2 (left) shows the analysis. As none of the groups can be taken as standard, the standard is calculated from the means of the 3 groups (last column).

The results are in the last 4 rows. Adjusted for sex and marital status, the standardized unemployment rate (Adj%) are 23%(95% CI = 8% to 38%) for those with less than secondary education (group 1), 9%(95% CI = 3% to 15%) for those that finished secondary education (group 2), and 4%(0 to 8%) for those with post-secondary education (group 3).

Grp	Grp	Diff%	SE	95%CI
1	2	14.12	1.63	10.94	17.31
1	3	19.45	1.27	16.97	21.94
2	3	5.33	0.79	3.78	6.88

The last table (to the right) compares the 3 groups. The differences and the standard error of the differences are as shown in the table, but the 95% confidence intervals requires recalculation. As there are 3 groups and therefore 3 comparisons, the 95% CI (p=0.025, z=1.96) will need to be recalibrated. For 95% confidence interval when there are 3 comparisons, Bonferroni's correction means p = 0.025 / 3 = 0.0083, and z for p=0.0083, as looked up in our z table in the Probability of z Explained and Table Page is 2.4. The 95% CI are therefore difference ±2.4SE instead of ±1.96SE. This difference is trivial in this example when the percent values are rounded to integers, but the difference would be more noticeable if there are many groups to compare.

The differences are therefore very similar to that without Bonferroni's correction, and are as follows:

• Between those with and without secondary education (grps 1 and 2), 14% (95%CI = 10% to 18%)
• Between those without secondary education and those with tertiary education (grps 1 and 3), 19% (95%CI = 16% to 22%)
• Between those with secondary education and those with tertiary education (grps 2 and 3), 5% (95%CI = 3% to 7%)

The program in the Regression Analysis Using Proportions Program Page is a special case of the contingency table analysis.

The entry data consists of 3 columns.

The first column is not a count at all, but is an interval measurement to be used as the x variable in the regression.

The second and the third column are the number of cases that have positive (N_pos) and negative (N_neg)attributes

Algorithm

1. The program calculates the positive proportion for each row p = N_pos / (N_pos + N_neg), which is the y value.
2. The data is tabulated, showing the numbers and the probability for each row.
3. Statistical calculation consists of the partition of the chi square, from the total to that attributable to a trend, and that is residual after the trend is removed.
   • The chi square for the trend evaluates whether there is a significant trend throughout the rows
   • The residual chi square evaluates whether there remains significant heterogeneity other than the trend.
4. The proportion is used as the y variable, and a regression analysis performed against the x variable (value in column 1). The result is the change in proportion per unit value of x (not per row but per unit value).

An Example

1990	10	100
1992	8	50
1995	60	300

The example uses the default data from the Regression Analysis Using Proportions Program Page , as shown in the table to the left.

We want to study changing business environment by looking at the proportion of business failures over the years. In 1990 there were 110 business start ups, 10 of which failed (100 successful). In 1992 there were 58 start ups, and 8 failed (50 successful). We ran out of research money in 1993 and 1994 so did not collect any data, but in 1995 we found 361 start ups and 61 of which failed (300 successful). The data matrix for input is therefore as shown on the left.

Row	Row Val (x=year)	Npos	Nneg	Ntotal	ppos
1	1990	10	100	110	0.0909
2	1992	8	50	58	0.1379
3	1995	61	300	361	0.169
Total		79	450	529	0.1493

The data is firstly displayed by the program, as shown to the right.

We can see that failure rates were 9.1% for 1990, 13.8% for 1992, and 16.9% for 1995, and the overall failure rate is 14.9%

	Chi Sq.	df	p
Total	4.11	2	0.13
Regression	4.02	1	0.045
Residual	0.09	1	0.76

The program now partitions the chi squares, as shown in the table to the left.

The analysis shows that there is a regression that is significant at p<0.05 level. Once this is partitioned, the residual variation is not statistically significant.

Finally, the regression coefficient is calculated. Change in proportion per unit row value = 0.015, indicates that, between 1990 and 1995, business failures increases by 1.5% per year.