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Ton J. Cleophas and Aeilko H. ZwindermanStatistical Analysis of Clinical Data on a Pocket CalculatorStatistics on a Pocket Calculator10.1007/978-94-007-1211-9_12© Springer Science+Business Media B.V. 2011

12. Odds Ratios

Ton J. Cleophas^1,
2 and Aeilko H. Zwinderman^2,
3

(1)

Department of Medicine, Albert Schweitzer Hospital, Dordrecht, The Netherlands

(2)

European College of Pharmaceutical Medicine, Lyon, France

(3)

Department of Epidemiology and Biostatistics, Academic Medical Center, Amsterdam, The Netherlands

Ton J. Cleophas (Corresponding author)

Email: ajm.cleophas@wxs.nl

Aeilko H. Zwinderman

Email: a.h.zwinderman@amc.uva.nl

Abstract

The odds ratio test is just like the chi-square test applicable for testing cross-tabs. The advantage of the odds ratio test is that a odds ratio value can be calculated. The odds ratio value is just like the relative risk an estimate of the chance of having an event in group 1 compared to that of group 2. An odds ratio value of 1 indicates no difference between the two groups.

The odds ratio test is just like the chi-square test applicable for testing cross-tabs.

The advantage of the odds ratio test is that a odds ratio value can be calculated. The odds ratio value is just like the relative risk an estimate of the chance of having an event in group 1 compared to that of group 2. An odds ratio value of 1 indicates no difference between the two groups.

Example 1

	Events	No events
	Numbers of patients
Group 1	15 (a)	20 (b)	35 (a + b)
Group 2	15 (c )	5 (d)	20 (c + d)
	30 (a + c)	25 (b + d)	55 (a + b+ c+ d)

The odds of an event = the number of patients in a group with an event divided by the number without. In group 1 the odds of an event equals = a/b.

The odds ratio (OR) of group 1 compared to group 2

$\begin{array}{lll} = & \left( {{\text{a}}/{\text{b}}} \right)/\left( {{\text{c}}/{\text{d}}} \right) \\ = & \left( {{15}/{2}0} \right)/\left( {{15}/{5}} \right) \\ = & 0.{25} \\ {\text{lnOR}} = & { \ln }0.{25} = { } - {1}.{386 }\left( {{ \ln } = {\text{ natural logarithm}}} \right) \\ \end{array}$

The standard error (SE) of the above term

$\begin{array} {lll}= & \surd \left( {{1}/{\text{a}} + {1}/{\text{b}} + {1}/{\text{c}} + {1}/{\text{d}}} \right) \\ = & \surd \left( {{1}/{15} + {1}/{2}0 + {1}/{15} + {1}/{5}} \right) \\ = & \surd 0.{38333} \\ = & 0.{619} \\ \end{array}$

The odds ratio can be tested using the z-test (Chap. 10 ).

$\begin{array}{lll} {\text{The test}} - {\text{statistic}} = & {\text{z}} - {\text{value}} \\ = & \left( {\text{odds ratio}} \right)/{\text{SE}} \\ = & - {1}.{386}/0.{619} \\ = & - {2}.{239} \\ \end{array}$

If this value is smaller than −2 or larger than +2, then the odds ratio is significantly different from 1 with p < 0.05. An odds ratio of 1 means that there is no difference in events between group 1 and group 2. The bottom row of the t-table (page 21) gives the z-values matching Gaussian distributions. Look at a z-value of 1.96 right up at the upper row. We will find a p-value here of 0.05. And, so, a z-value larger than 1.96 indicates a p-value of <0.05. There is a significant difference in event between the two groups.

Example 2

	Events	No events
	Number of patients
Group 1	16 (a )	26 (b)	42 (a + b)
Group 2	5 (c )	30 (d)	35 (c + d)
	21 (a + c)	56 (b + d)	77 (a + b + c + d)

Test with OR whether there is a significant difference between group 1 and 2.

See for procedure also example 1.

$\begin{array}{lll} {\text{OR}} = & \left( {{16}/{26}} \right)/\left( {{5}/{3}0} \right) \\ = & {3}.{69} \\ \end{array}$

${\text{lnOR}} = {1}.{3}0{56 }({ \ln } = {\text{natural logarithm see the above example}})$

$\begin{array}{lll} {\text{SE}} = & \surd \left( {{1}/{16} + {1}/{26} + {1}/{5} + {1}/{3}0} \right) \\ = & \surd 0.{334333} \\ = & 0.{578} \\ \end{array}$

$\begin{array}{lll} {\text{z}} - {\text{value}} = & {1}.{3}0{56}/0.{578} \\ = & {2}.{259} \\ \end{array}$

Because this value is larger than 2, a p-value of <0.05 is observed, 0.024 to be precise (numerous “p-calculator for z-values” sites in Google will help you calculate an exact p-value if required.

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