Chapter 12.8 Logistic Regression

Chapter 12.8 Logistic Regression Logistic regression is an example of a large class of regression models called generalized linear models (GLM) An Observational Case Study: The Donner Party (Gayson, D.K., 1990, Donner Party deaths: A demographic assessment, Journal of Anthropological Research, 46, 223-42, and Ramsey, F.L. and Schafer, D.W., 2002, The Statistical Sleuth, 2nd Ed, Duxbury Press, p. 580. In 1846, the Donner and Reed families left Illinois for California by covered wagon (87 people, 20 wagons). They attempted a new and untried crossing of the region between Ft. Bridger, Wyoming and the Sacramento Valley. After numerous problems and delays in Utah, they reached the Eastern Serria Nevada in late October. They were stranded near Lake Tahoe by a series of snowstorms that left as much as 8 feet of snow by some accounts. By the time they were rescued in April of the following year, 40 members had died. Some (or perhaps all) of those that survived did so by resorting to cannibalism The researchers attempted to address questions such as whether females are better able to withstand harsh conditions than men, and whether the odds of survival varied with age. Grayson was able to reconstruct records on survival, age and gender for 45 individuals. Summary of Findings The odds of survival of females were estimated to be 4.9 times the odds of survival for men of the same age. An approximate 95% confidence interval for this odds ratio is 1.1 to 21.6 1

We call π/(1 π) the odds of the event of interest. Here, π is the probability of the event of interest (e.g., survival), so π = P (S). The complement of S is denoted by S. If π = 0.5, the odds of S (relative to S) is 1 If π = 0.75, the odds of S (relative to S) is 0.75/0.25 = 3, or 3 to 1 If π = 0.25, the odds of S (relative to S) is 0.25/0.75 = 0.33, or 1 in 3 The logistic regression model is of log(odds of S), though it is easy to recover the estimated odds of S, and the probability of S, from the fitted model Donner Party revisited Consider a plot of survivorship (= 1 implies survivorship and = 0 implies death) for each of participants, against age, by gender (Figure 1) Figure 1 illustrates two common failures when using ordinary least squares regression with binary response variables 1. First, the predicted values are completely inconsistent with the possible expected values for the response variable. Specifically, E(Y i ) = π i is the expected value of the ith (binary) response. Here π i is also the probability of surviving, so 0 π i 1, yet the lines are both below and above 0 and 1 even within the observed range of age 2. Secondly, last semester it was observed that the survivorship rate for males was consistent lower than that for females, but the regression lines indicate that older males have a greater survivorship rate than older females. The model is not consistent with a straightforward contingency table analysis. The two problems noted above 2

are absent from Figure 2 showing estimated probabilities from a logistic regression model. Survivorship 0.0 0.2 0.4 0.6 0.8 1.0 Males Females 20 30 40 50 60 Age Figure 1: Survivorship for the Donner Party participants. Fitted least squares lines (for each gender) are shown. n = 45 Figure 2 shows the data with the fitted logistic regression model super-imposed. The lines show the estimated probability of survival as a function of age and gender. Scope of Inference: Because the data are observational, the result cannot be used to infer that women are more apt to survive than men. There may be confounding variables that account for the apparent difference (e..g., behavior). Moreover, these 45 individuals are not a randomly sampled from any identifiable population to which inference may be made 3

Survivorship 0.0 0.2 0.4 0.6 0.8 1.0 Males Females 20 30 40 50 60 Age Figure 2: Survivorship for the Donner Party participants. Fitted logistic regression lines (for each gender) are shown. n = 45 A Retrospective Case Study (see Ramsey and Schafer, The Statistical Sleuth, 2nd ed.) Holst, P.A., Kromhout, D. and Brand, R. 1988. For debate: pet birds as an independent risk factor for lung cancer, British Medical Journal, 297, 13-21. A health survey in The Hague, Netherlands presented evidence of an association between keeping pet birds and increased risk of lung cancer. To investigate this link further, researchers conducted a case-control study of patients at four hospitals in The Hague. They identified 49 cases of lung cancer among patients that were younger than 65 and long-term residents of the city. The also selected 98 controls from the population of city residents having the same general age structure as 4

the cancer cases. Data were gathered on the following variables: 1. Sex (1 =F, 0 =M) 2. Age in years 3. Socioeconomic status (1 =High, 0 =Low), determined by occupation of the household s principal wage earner 4. Years of smoking prior to diagnosis or examination 5. Average rate of smoking (cigarettes/day) 6. Indicator of birdkeeping. Birdkeeping was defined as keeping caged birds in the home for more than 6 consecutive months from 5 to 14 years before diagnosis (cases) or examination (controls) Scope of Inference: Inference extends to the population of lung cancer patients and unaffected individuals in The Hague in 1985 (the study year). Statistical analysis of these observational data cannot be used as evidence that birdkeeping causes lung cancer. However, there is medical rationale supporting this conclusion: people who keep birds inhale and expectorate excess allergens and dust particles which increases the likelihood of dysfunction of lung macrophages which in turn may lead to diminished immune system response. This is a retrospective study. A retrospective study is one in which two (sub)populations (cases and controls) are sampled at different rates. A comparison of the number of individuals in each sample provides no information regarding the probability that a randomly 5

sampled birdkeeper has lung cancer. However, a comparison of the cases and controls provides a relative measure of increased risk (the actual level of risk is not addressed) In essence, if the proportion of lung cancer cases among the cases (birdkeepers) is twice that among the controls (not birdkeepers), then the data indicate that the odds of lung cancer is twice as great for birdkeepers compared to nonbirdkeepers Generalized Linear Models (GLMs) A GLM is a probability model in which the mean, or expected value, of a response variable is related to a set of explanatory variables through a regression equation The usual multiple regression model is an example: µ = E(Y ) = β 0 + β 1 x 1 + + β p 1 x p 1 To handle distributions besides the normal distribution, we must model a nonlinear function of µ. For example, if the response variable is Poisson in distribution, then the GLM is log µ = β 0 + β 1 x 1 + + β p 1 x p 1 If the response variable is Binomial in distribution, then the GLM is where n is the number of trials ( ) µ log = β 0 + β 1 x 1 + + β p 1 x p 1 n µ If the response variable is Binary, (or Bernoulli) in distribution, then the GLM is ( ) π log = β 0 + β 1 x 1 + + β p 1 x p 1 1 π because µ = nπ = π when the distribution is Bernoulli 6

In general, there is some link function, say g(µ) that links the linear model to µ Provided that the link function is appropriate, nearly all of the ordinary multiple regression methods (residual analysis, hypothesis testing, confidence intervals) carry over to the GLM with only minor modifications Logistic regression is used for modeling the mean of a binary, or Bernoulli random variable. A Bernoulli random variable is a Binomial random variable for which the number of trials is n = 1. Only one of two outcomes (S or S) is possible, and the probability of S is denoted by P (S) = π. Consequently, P (S) = 1 P (S) = 1 π. The mean of a Binomial random variable is µ = nπ, and the variance is σ 2 = nπ(1 π) Thus, for a binary random variable Y defined as 1, if the outcome is S Y = 0, if the outcome is S, E(Y ) = µ = π and Var(Y ) = π(1 π) The logit function is the link between µ and the linear model. Let η denote the linear portion of the model, i.e., η = β 0 + β 1 x 1 + + β p 1 x p 1 7

estimates and specific values of x 1,..., x p 1 8 The logit function of π is ( ) π logit(π) = log 1 π Thus, for logistic regression logit(π) = η Recall that π/(1 π) is the the odds of S. If π = 0.5, the odds of S (relative to S) is 1 The logit function has the effect of stretching the possible values of π from 0 to 1, to to. This is very helpful with respect to the computational aspects of model fitting The inverse of the logit is important. We calculate it as follows ( ) π η =log 1 π e η = π 1 π 1 e =1 π = 1 η π π 1 1 + 1 e η = 1 π 1 + eη e η = 1 π eη 1 + e η =π Figure 3 show the logit and its inverse. When logistic regression is used to model the probability of an outcome, model fitting produces estimates of the parameters β 0,..., β p1 It is simple, then to obtain an estimate of the linear predictor η given the parameter

logit(p) -4-2 0 2 4 p 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.2 0.4 0.6 0.8 1.0 p -4-2 0 2 4 Linear predictor Figure 3: The logit function (left panel) and its inverse (right panel). It is necessary to be able to determine the estimated probability of the outcome of interest given specific values x 1,..., x p 1 This can be done in two steps: 1. Compute η = β 0 + β 1 x 1 +... + β p 1 x p 1 2. Compute π = exp( η) 1 + exp( η) (1) or more directly by computing π = exp( β 0 + β 1 x 1 +... + β p 1 x p 1 ) 1 + exp( β 0 + β 1 x 1 +... + β p 1 x p 1 ) 9 (2)

A useful interpretation of β 1 is obtained by considering the odds ratio. The odds ratio expresses how much more (or less) likely event A is to occur than event B. Suppose that A and B are two outcomes. To be concrete, suppose that A is the outcome that a heavy smoker has lung cancer, and B is the outcome that a nonsmoker has lung cancer. The increase in risk of lung cancer can be quantified by comparing the odds of having lung cancer for the smoker to the odds of having lung cancer to the nonsmoker. Suppose that the probability of A is π A = P (A). Specifically, given an individual is a heavy smoker, then the probability of having lung cancer is π A = P (A). Similarly, given that an individual is a nonsmoker, then the probability of having lung cancer is the probability of the outcome B, which is denoted by π B = P (B) The odds that A will occur is π A /(1 π A ) and the odds of B is π B /(1 π B ). The odds ratio is π A /(1 π A ) π B /(1 π B ) = π A(1 π B ) π B (1 π A ) For example, if P (A) = π A = 0.75, and P (B) = π B = 0.25, then the odds of A relative to B is π A (1 π B ) π B (1 π A ) = 0.75(1.25) 0.25(1.75) = 3 3 = 9 We say that the odds of the outcome A is 9 times the odds of the outcome B For example, if P (A) = π A = 0.75, and P (B) = π B = 0.6, then the odds of A relative to B is π A (1 π B ) π B (1 π A ) = 0.75(1.60) 0.60(1.75) = 5 4 8 5 = 2 Said another way, the odds of A is 3 : 1, and the odds of B is 0.6/0.4 = 1.5, or 3 : 2, hence, the odds of A is twice that of B 10

Interpretation of logistic regression coefficients Consider the logit model of π as a function of a single explanatory variable x ( ) π log = η = β 0 + β 1 x 1 π Let π A denote the probability when x = x A : π A 1 π A = e β 0+β 1 x A Let π B denote the probability when x = x B : π B 1 π B = e β 0+β 1 x B The ratio of the odds of success (i.e., π) when x = A relative to x = B is π A /(1 π A ) =eβ0+β1xa π B /(1 π B ) e β 0+β 1 x B =e β 0+β 1 x A (β 0 +β 1 x B ) =e β 1(x A x B ) This means that if x A differs from x B by one unit, i.e., x A x B = 1, then e β 1(x A x B ) = e β 1, and the odds of success will change by a multiplicative factor of e β 1 when x = x A compared to x = x B For example, suppose that β 1 = 0.5; then e β 1 = e 0.5 = 1.649. So, a one unit change in x will increase the odds of success by 1.65 times For example, suppose that β 1 = 0.5; then e β 1 = e 0.5 = 1/1.649 = 0.6065. 11

Thus, a one unit change in x will decrease the odds of success by a (multiplicative) factor of 0.6065. The odds of success are 100 60.6 = 39.4% less as a result of a 1 unit change in x The figure below plots β on the x-axis and e β on the y-axis, and shows how a one-unit change in x will change the odds of success exp(beta) 0 2 4 6-3 -2-1 0 1 2 beta Figure 4: The function e β plotted against β. Suppose that there are more than a single explanatory variable that is used to model the logit of π. A particular estimated coefficient β k is interpreted in much the same manner as in linear regression with multiple explanatory variables Suppose that all variables x 2,..., x p 1 are held constant, and we compare the effect of 12

a 1-unit change in x 1 on the odds of S. The change is best measured with respect to the odds ratio. Let π 1 denote the probability of S for some value, say x 1, and π 2 denote the probability of S at x 1 + 1. The odds ratio is ( ) π2 / ( π1 1 π 2 1 π 1 ) = eβ 0+β 1 (x 1 +1)+ +β p 1 x p 1 e β 0+β 1 x 1 + +β p 1 x p 1 = eβ 1(x 1 +1) e β 1x 1 = e β 1 We say that if x 1 changes by 1-unit, then the odds of S will change by a multiplicative factor of e β 1 provided that all other variables are held constant More generally, if x 1 changes from A to B, then the odds of S will change by a multiplicative factor of e β1(a B) provided that all other variables are held constant because ( ) πa / ( ) πb = eβ 0+β 1 A+ +β p 1 x p 1 1 π A 1 π B e β 0+β 1 B+ +β p 1 x p 1 = eβ 1A e β 1B = eβ 1(A B) Example: logistic regression produces a fitted model for the Donner party. The fitted model is of log-odds of π = Pr(Death), and it is logit( π) = 3.23 + 0.078x AGE 1.597x GENDER, where 1, if Female x GENDER = 0, if Male The odds of death increase by a multiplicative factor of e 0.078 = 1.081 for every year of life if gender is held constant (for both males and females). In other words, the risk increases by 8% for every year of life. An approximate 95% CI for this factor is 1.005 to 1.163 13

The odds of death of females are estimated to be e 1.597 = 0.2025 times the odds of death for males of the same age. An approximate 95% confidence interval for this odds ratio is 0.046 to 0.89 We also can say that the odds of death of males are estimated to be 1 e 1.597 = e1.597 = 4.9 times the odds of death for females of the same age. An approximate 95% confidence interval for this odds ratio is 1/0.89 = 1.124 to 1/0.046 = 21.71 To get a direct comparison of probabilities, we need to specify either the probability of death for females (at a specific age) to get the probability of death for males at the same age, or specify the probability of death for males to get the corresponding probability for females. For example, suppose that the probability of death for females is π F = 0.5. Then 4.9 π F 1 π F = π M 1 π M 4.9 = π M 1 π M Solving the righthand equation for π M yields π M = 0.83. More generally, π M = ( 1 + e β 1 π ) 1 F π F This formula yields π M = 0.710 when π F = 1/3 and π M = 0.907 when π F = 2/3 Estimation of Logistic Regression Coefficients In contrast to ordinary multiple regression, the regression parameters are estimated by maximizing the likelihood of the data (instead of minimizing the sum of the residuals) Logistic regression parameter estimates are maximum likelihood estimates 14

The likelihood of the data, in the case of discrete data 1, is the probability of obtaining the sample as a function of the parameters By maximizing the likelihood through the choice of the parameter estimates, we make the probability of obtaining the sample as large as possible. Any choice of parameter estimates results in a lesser probability An example involving 1 parameter is a random sample of n = 20 Bernoulli observations. Suppose the sample is y = (0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1), and that each observation is generated from the Bin(1, π) distribution. Probabilities from the Bin(1, π) distribution can be computed according to the formula P (Y = y 1 ) = π y 1 (1 π) 1 y 1, for y 1 = 0 or 1 Because the sample is random, the observations are independent, and the likelihood L(π; y) of observing this sample of 20 observations is the product of the individual probabilities, that is L(π; y) = P (Y 1 = 0, Y 2 = 0,..., Y 20 = 1) = P (Y 1 = 0) P (Y 2 = 0) P (Y 20 = 1) = π 0 (1 π) 1 π 0 (1 π) 1 π 1 (1 π) 0 = π 4 (1 π) 16 because there are four 1 s and sixteen 0 s in the sample y Figure 5 (left panel) is a graph of L(p; y) versus p shows how the likelihood of the sample varies with 0 < p < 1 1 In the case of continuous data, the definition is somewhat more complicated 15

Note that the value of p that maximizes the likelihood is 0.2. Consequently, the maximum likelihood estimate of p is 0.2. likelihood 0.0 0.00001 0.00002 0.00003 0.00004 log-likelihood -18-16 -14-12 -10 0.0 0.1 0.2 0.3 0.4 0.5 0.6 p 0.0 0.1 0.2 0.3 0.4 0.5 0.6 p Figure 5: Left panel shows the likelihood L(p; y) = p 4 (1 p) 16 as a function of 0 < p < 1. The left panel shows log[l(p; y)] as a function of 0 < p < 1. In this example, the MLE is the same as our previous estimator of π, namely, the sample proportion p = # successes # trials = 4 20 It is easier to work with the log-likelihood, i.e., logl(π; y), instead of the likelihood because the logarithm converts products to sums, and the likelihood function is a product of individual terms arising from each observation Note that the value of p that maximizes logl(p; y) also maximizes L(p; y) (right panel 16

of Figure 5). Because log(x) is a strictly increasing function of x, the maximizing value of logl(p; y) is the same as the maximizing value of L(p; y) For the logistic regression model, the probability of survival for the ith individual is a function of the parameters β 0, β 1 and β 2. Two preliminary calculations help simplify the log-likelihood. They are 1. First π i = eβ 0+β 1 x AGE i +β 2 x GENDER i 1 + e β 0+β 1 x AGE i +β 1 x GENDER i = eη i 1 + e η i which leads to 1 π i = 1 1 + e β 0+β 1 x AGE i +β 2 x GENDER i = 1 1 + e η i 2. Tle log-probability for the ith observation can then be expressed as Then, log[p (Y i = y i )] = log[π y i i (1 π i) 1 y i ] = log π y i i + log(1 π i ) (1 y i) = y i log π i + (1 y i ) log(1 π i ) ( ) πi = y i log + log(1 π i ) 1 π i ( ) 1 = y i η i + log 1 + e η i = y i (β 0 + β 1 x AGE i 45 log[l(β 0, β 1, β 2 ; y)] = log[p (Y i = y i )] i=1 45 = i=1 y i (β 0 + β 1 x AGE i + β 2 x GENDER i + β 2 x GENDER i 17 ) log[1 + e β 0+β 1 x AGE i +β 2 x GENDER i ] ) log[1 + e β 0+β 1 x AGE i +β 2 x GENDER i ]

Numerical optimization methods are used to determine the values of β 0, β 1 and β 2 that maximize log[l(β 0, β 1, β 2 ; y)] For the Donner Party data, Table 1 2 shows log[l(β 0, β 1, β 2 ; y)] for a few of the possible values of β 0, β 1,and β 2 : Table 1: log[l(β 0, β 1, β 2 ; y)] as a function of β 0, β 1, β 2. β 0 β 1 β 2 log[l(β 0, β 1, β 2 ; y)] 1.50 0.50 1.25 27.7083 1.80 28.7931 0.80 1.25 26.1531 1.80 25.7272 1.70 0.50 1.25 29.0467 1.80 30.3692 0.80 1.25 25.7972 1.80 25.6904 1.63 0.078 1.60 25.6282 The last line in the table shows the value of log[l(β 0, β 1, β 2 ; y)] at the MLE s. These MLEs are different than those above because Ramsey and Schafer set π = P (Survival) whereas I set π = P (Death) Properties of MLEs If the model is correct and the sample size is large, then 1. MLE s are nearly unbiased 2. The standard errors of MLEs can be estimated, and the estimates are nearly unbiased 3. MLE s are more precise than nearly all other estimators 4. The distribution of an MLE is approximately normal (a consequence of the Central 2 From Ramsey, F.L. and Schafer, D.W. 2002. The Statistical Sleuth, 2nd Ed. p. 589. 18

Limit Theorem) Tests and Confidence Intervals Involving A Single Parameter The last property above implies that if β i is the MLE of β i, then β i N(βi, σ βi ) where σ βi is the standard error of the β i This gives the Wald test of H 0 : β i = 0 versus H 1 : β i 0 The test statistic is If H 0 is true, then Z N(0, 1). Z = β i σ βi. Suppose that the observed value of the test statistic is z; then, the p-value is 2P (Z z ). The normal approximation may be poor in some instances, and so the test must be treated with some caution. In essence, the p-value is may not be very accurate For example, in the Donner Party data analysis, the linear portion of a logistic regression model of P (Death) is η i = β 0 + β 1 x AGE i + β 2 x GENDER i + β 3 x AGE GENDER i, SPSS reports β 3 = 0.162, σ β3 = 0.092, and Sig. = 0.086 19

The p-value is approximately 0.086 because (ignoring rounding error) Z = β i σ βi = 0.162 0.092 = 1.73 and 2P (Z 1.73) = 0.086 Also, SPSS reports that e β 3 = e 0.162 = 1.175. This means that the difference in odds of death resulting from a 1 year increase in age is greater for females (GENDER = 1) than males by a factor of 1.175 given that age is held constant. A approximate 95% CI for this factor is 0.977 to 1.141 In my opinion, interaction between gender and age is not supported by the data, and I adopt the no-interaction model. Earlier, it was reported that the odds of death increase by a multiplicative factor of e 0.078 = 1.081 for every year of life if gender is held constant. So, if I compare a female at age= 20 versus a female at age= 30, the odds of death are 1.081 10 = 2.18 times greater for the older female. An approximate 95% CI for this increase obtained by computing the anit-logs of the upper and lower bound estimates on the parameter, specifically, 1.005 10 = 1.05 to 1.163 10 = 4.56 The Drop-in-Deviance Test More generally, testing the significance of a covariate or a factor in logistic regression is conducted using the same principles as in linear regression. The following test is preferred to the Wald statistic as the p-values associated with the test statistic are more accurate. Further, more than one variable can be tested for significance, and so this test 20

must be used if a factor with more than 2 levels is to be tested for significance Specifically, we compare the fit of two models: the full model which contains the covariate or the factor of interest, and a reduced model that is the same as the full except that the covariate or factor is not in the model If we are assessing the importance of a factor with k levels, so k 1 indicator variables are used to account for it, then all k 1 indicators are removed from the full model to get the reduced model Suppose that g variables are used to account for the term (if the term is a covariate g = 1), and the parameters that multiply the variables are β 1, β 2,..., β g. Then the null hypothesis of interest is H 0 : β 1 = β 2 = = β g = 0 and the research hypothesis is H 1 : at least one of β 1, β 2,..., β g is not 0. The drop-in-deviance test uses the the likelihood ratio statistic. This statistic is 2 times logarithm of the ratio of the two likelihoods, specifically, { L(full model; y) } LRT = 2log L(reduced model; y) = 2log[L(full model; y)] 2log[L(reduced model; y)]. If the H 0 is true, then the LRT is approximately chi-square in distribution and the degrees of freedom are g, i.e., LRT χ 2 g 21

The entire procedure is often called the drop-in-deviance test (analogous to the extrasums-of-squares test which uses an F -statistic). The deviance of a model, say model M, is 2 times the log-likelihood D(M) = 2log [L(M; y)] Recall that the likelihood was the probability of obtaining the sample as a function of a set of parameters. A large likelihood corresponds to a small deviance because of the sign change induced by the term 2 Hence, the larger the deviance, the worse the fit of the model. Similar to the error sums of squares in ordinary regression, we cannot interpret the deviance in isolation (without comparing it to some other model) The likelihood ratio statistic comparing two models can be expressed in terms of the deviance: LRT = D(reduced) D(full) Example: Donner Party data. Table 2 shows an analysis of deviance table Table 2: Likelihood Ratio Tests. -2 Log Likehood of Effect Chi-Square df Sig of Reduced Model Intercept 38.754 2.408 1.121 AGE 42.377 6.030 1.014 GENDER 41.381 5.034 1.025 The chi-square statistic is the difference in -2 log-likelihoods between the final model and a reduced model. The reduced model is formed by omitting an effect from the final model. The null hypothesis is that all parameters of that effect are equal to 0. 22

The test of significance for AGE compares the deviance of the full model (D = 36.347) to the deviance to the model without AGE but including GENDER (D = 42.377). The test statistic is LRT = 42.377 36.347 = 6.030, and P (χ 2 1 > 6.030) = 0.014. Hence, there is strong evidence (p-value 0.014) that there differences in the probability of survival at different ages A Case-Control Study - Birdkeeping and Lung Cancer The objective was to determine if there is an increased probability of lung cancer associated with birdkeeping, even after accounting for other factors (e.g., smoking) Factors (and covariates) are 1. Sex (1 =F, 0 =M) 2. Age in years 3. Socioeconomic status (1 =High, 0 =Low), determined by occupation of the household s principal wage earner 4. Years of smoking prior to diagnosis or examination 5. Average rate of smoking (cigarettes/day) 6. Indicator of birdkeeping. Birdkeeping was defined as keeping caged birds in the home for more than 6 consecutive months from 5 to 14 years before diagnosis (cases) or examination (controls) The first step is to examine the effect of smoking on lung cancer. 23 Figure 6 compares

the relationship between age and smoking, and whether the individuals kept birds, for the cases and the controls Birdkeeping No Birdkeeping 50 LUNGCANCER NOCANCER 40 Years of smoking 30 20 10 0 40 45 50 55 60 65 40 45 50 55 60 65 Age Figure 6: Age versus years smoking, by case (lung cancer) and control (no lung cancer). n = 147 Figure 6 shows that smoking is strongly associated with lung cancer because of the cases, all but one has been smoking for more than 10 years, and all but 4 have been smoking for more than 20 years Comparing cases (LUNGCANCER) to controls (NOCANCER) reveals that relatively more individuals were birdkeepers among the cases than the controls Birdkeeping does not appear to be associated with years of smoking or age. The researchers (main) objective was to determine whether there is an association 24

between birdkeeping and incidence of lung cancer. This statement implies the model fitting strategy should be to fit a rich model with all available explanatory variables besides birdkeeping, and determine whether the indicator of birdkeeping significantly improves the fit of the rich model. If birdkeeping is significant, then the differences in the odds of contracting lung cancer effect between birdkeepers and others should be estimated The drop-in-deviance statistic obtained from adding birdkeeping to the main-effectsonly model is LRT = 11.7 and the associated p-value is 0.0006 The odds associated of contracting cancer is estimated to be 3.77 times greater when comparing two individuals that differ according to whether they keep birds or not, given that all other variables are held constant. An approximate 95% confidence interval for this increase is 1.7 to 8.5 We may be interested in the significance of the other variables, though it must be admitted that there may be other variables that are associated with the incidence of lung cancer that were not measured. I would not try to describe, in general, what variables are associated with lung cancer because of these unobserved, or latent variables However, looking among those variables that are available, it is of interest which show and association with lung cancer. A good-fitting model (found by fitting all main effects and sequentially eliminating non-significant variables) is shown in Table 2 Table 3: Parameter estimates from the final model (Birdkeeping data). Variable Parameter estimate Estimated standard error Wald statistic Approx. p-value Intercept 3.1802 0.636 5.00 < 0.0005 Years smoking 0.058 0.0168 3.46 < 0.001 Bird keeping 1.476 0.3959 3.73 < 0.001 25

The odds associated of contracting cancer is estimated to be e 10 0.058 = 1.7 times greater when comparing two individuals that differ according to 10-years in the length of time that they have smoked, given that all other factors are the same The odds associated of contracting cancer is estimated to be e 10 1.476 = 4.38 times greater when comparing two individuals that differ according to whether they keep birds or not, given that all other factors are the same AIC and BIC Two measures of model fit (and variants on each) are sometimes used in model fitting when the researcher has a large collection of explanatory variables and little scientific understanding of the problem to guide model fitting. These are Akaike s information criterion (AIC) and Bayes information criterion (BIC) These measures use a penalty term that depends on the number of variables in the model. The difference between the two is the size of the penalty. The BIC penalty is greater than that using AIC when the sample size is large The AIC penalty is 2p, where p is the number of model parameters. The AIC measure of the fit of model M 1 is AIC(M 1 ) = D(M 1 )+2p, where D(M 1 ) is the deviance associated with model M 1 The BIC penalty is p ln(n), where n is the number of observations. The BIC measure of the fit of model M 1 is BIC(M 1 ) = D(M 1 ) + p ln(n) Figure 7 shows the penalty functions as a they vary with n and p = 10. AIC does not depend on n, and so the penalty function is constant. A larger penalty is imposed by BIC to compensate for an increase in test power with 26

Penalty 0 10 20 30 40 50 50 100 150 200 Sample size Figure 7: Penalty functions for AIC (horizontal line) and BIC (curve) when p = 5. larger sample size. If the researcher believes that some, but not all explanatory variables exert an influence on response variable, they may choose the model with the smallest BIC (versus AIC). On the other hand, if the researcher is inclined to believe that all explanatory variables exert some influence on the response and wants to find a parsimonious model, then the researcher will choose the model with the smallest AIC. Cautionary Remarks These criteria (AIC and BIC) are sometimes useful, and sometimes not. It is incorrect to assume that they relieve the scientist from the obligation of careful and considered model fitting. Many statisticians use these methods sparingly, if at all, in part because there has been abundant criticism of AIC for being too liberal (that is, it tends to include variables that are not associated with the response in certain situations). 27

In general, it is best to keep in mind that finding a best model is an unrealistic goal (given the many variables that have may some influence on the response variable, but have not been measured). A realistic objective of model fitting is to learn which variables, among those that have been measured, are associated with the response variable and to obtain a description of how the explanatory variables relate to the response. For example, in the birdkeeping analysis, from among the available variables, it was found that only years of smoking and birdkeeping had an significant effect on the odds of contracting lung cancer, and that gender, age, socioeconomic status, and smoking rate were not significant. The odds of contracting lung cancer where much greater (about 4.4 times greater) for birdkeepers versus those that did not keep birds. This is association is larger than the change in the odds associated with 10 years of smoking, as the odds of contracting lung cancer are estimated to be 1.7 times greater when comparing individuals that differ by 10 years in the length of time that they smoked A goodness of fit test when the number of trials is greater than one It is sometimes possible to test whether the assumed model fits the data Generally, these are situations in which a separate variance parameter is not used in the model. In addition, there must be multiple observations observed at each combination of factor levels and covariate values These conditions are satisfied if the assumed model is binomial, and the number of trials is greater than one For example, we can carry out a goodness of fit test for the Challenger data because there were 6 trials associated with each launch and the number of failed o-rings (out of 28

the 6) is the observed response variable Let m denote the number of trials (m = 6) for the Challenger data. The idea is that a model-free, good estimate of the probability of failure associated with each launch (and hence, observation on the number of failures) is the sample proportion of the m that failed. For example, if y = 2 failed when temperature = 53, then the estimated probability of failure is 2/6 when temperature is 53 There are two ways to get a set of estimates that correspond to this best model 1. Compute each the sample proportions by hand 2. Fit a saturated model that has the same number of parameters as observations (say, with an intercept and n 1 indicator variables that identify each observation. Then compute the fitted probabilities according to the saturated model The term saturated is used because model cannot contain any other parameters. For this model p = n Then, compare the fit of the saturated model to the simpler model. If the fits are nearly the same, then we conclude that model fit is good Formally, we set H 0 : the model fits the data versus H a : the model does not fit the data The test statistic is the likelihood ratio statistic 29

{ L(saturated model; y) } LRT = 2log L(simple model; y) = 2log[L(saturated model; y)] 2log[L(simple model; y)]. If H 0 is true, then the LRT is approximately chi-square in distribution and the degrees of freedom are n p, i.e., LRT χ 2 n p We reject H 0 if Pr(χ 2 n p LRT ) is smaller than some α (e.g., 0.05) Example: The Challenger data using the logit model and temperature. SPSS reports the LRT goodness of fit statistic as Deviance= 18.086 with 21 degrees of freedom. The associated p-value (not reported) is 0.644 Specifically, SPSS definition and calculation of the deviance different than I described above. The SPSS calculation is { L(saturated model; y) } Deviance = 2log L(simple model; y) = 2log[L(saturated model; y)] 2log[L(simple model; y)]. While the SPSS definition is somewhat unconventional, it does lend itself to the goodness of fit test. Moreover, their definition can be used for significance testing of variables without any additional work When testing the significance of a variable, we use the likelihood ratio statistic, and I advocate using the difference in deviances between the models with and without the variable. Using the SPSS form, we obtain 30

LRT = D(M 1 ) D(M 2 ) =2log[L(saturated model; y)] 2log[L(M 1 ; y)] {2log[L(saturated model; y)] 2log[L(M 2 ; y)]} =2log[L(M 2 ; y)]] 2log[L(M 1 ; y)] This is the correct test statistic A cautionary remark: most of the time deviance statistic that appears in the SPSS output cannot be used in a goodness of fit test. The conditions stated above must hold. Most notably, the number of trials must be greater than one for the majority of the observations 31