Preface to First Edition - lib

120 LOGISTIC REGRESSION AND GENERALISED LINEAR MODELSTable 7.4backpain data. Number of drivers (D) and non-drivers (¯D), suburban(S) and city inhabitants (¯S) either suffering from a herniated disc (cases)or not (controls).Controls¯D D¯S S ¯S S Total¯D ¯S 9 0 10 7 26Cases S 2 2 1 1 6D ¯S 14 1 20 29 64S 22 4 32 63 121Total 47 7 63 100 217The last of the data sets to be considered in this chapter is shown in Table7.4. These data arise from a study reported in Kelsey and Hardy (1975)which was designed to investigate whether driving a car is a risk factor for lowback pain resulting from acute herniated lumbar intervertebral discs (AHLID).A case-control study was used with cases selected from people who had recentlyhad X-rays taken of the lower back and had been diagnosed as having AHLID.The controls were taken from patients admitted to the same hospital as a casewith a condition unrelated to the spine. Further matching was made on ageand gender and a total of 217 matched pairs were recruited, consisting of 89female pairs and 128 male pairs. As a further potential risk factor, the variablesuburban indicates whether each member of the pair lives in the suburbs orin the city.7.2 Logistic Regression and Generalised Linear Models7.2.1 Logistic RegressionOne way of writing the multiple regression model described in the previouschapter is as y ∼ N (µ,σ 2 ) where µ = β 0 + β 1 x 1 + · · · + β q x q . This makesit clear that this model is suitable for continuous response variables with,conditional on the values of the explanatory variables, a normal distributionwith constant variance. So clearly the model would not be suitable for applyingto the erythrocyte sedimentation rate in Table 7.1, since the response variableis binary. If we were to model the expected value of this type of response, i.e.,the probability of it taking the value one, say π, directly as a linear function ofexplanatory variables, it could lead to fitted values of the response probabilityoutside the range [0, 1], which would clearly not be sensible. And if we writethe value of the binary response as y = π(x 1 , x 2 , ...,x q ) + ε it soon becomesclear that the assumption of normality for ε is also wrong. In fact here ε mayassume only one of two possible values. If y = 1, then ε = 1−π(x 1 , x 2 , ...,x q )© 2010 by Taylor and Francis Group, LLC