Logistic regression - overview

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Logistic regression
Wilcoxon signed-rank test
Chi-squared test for the relationship between two categorical variables
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Independent variablesIndependent variableIndependent /column variable
One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables2 paired groupsOne categorical with $I$ independent groups ($I \geqslant 2$)
Dependent variableDependent variableDependent /row variable
One categorical with 2 independent groupsOne quantitative of interval or ratio levelOne categorical with $J$ independent groups ($J \geqslant 2$)
Null hypothesisNull hypothesisNull hypothesis
Model chi-squared test for the complete regression model:
  • H0: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
Wald test for individual regression coefficient $\beta_k$:
  • H0: $\beta_k = 0$
    or in terms of odds ratio:
  • H0: $e^{\beta_k} = 1$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
  • H0: $\beta_k = 0$
    or in terms of odds ratio:
  • H0: $e^{\beta_k} = 1$
in the regression equation $ \ln \big(\frac{\pi_{y = 1}}{1 - \pi_{y = 1}} \big) = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K $. Here $ x_i$ represents independent variable $ i$, $\beta_i$ is the regression weight for independent variable $ x_i$, and $\pi_{y = 1}$ represents the true probability that the dependent variable $ y = 1$ (or equivalently, the proportion of $ y = 1$ in the population) given the scores on the independent variables.
H0: $m = 0$

Here $m$ is the population median of the difference scores. A difference score is the difference between the first score of a pair and the second score of a pair.

Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
H0: there is no association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
  • H0: the distribution of the dependent variable is the same in each of the $I$ populations
If there is one random sample of size $N$ from the total population:
  • H0: the row and column variables are independent
Alternative hypothesisAlternative hypothesisAlternative hypothesis
Model chi-squared test for the complete regression model:
  • H1: not all population regression coefficients are 0
Wald test for individual regression coefficient $\beta_k$:
  • H1: $\beta_k \neq 0$
    or in terms of odds ratio:
  • H1: $e^{\beta_k} \neq 1$
    If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:
  • H1 right sided: $\beta_k > 0$
  • H1 left sided: $\beta_k < 0$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
  • H1: $\beta_k \neq 0$
    or in terms of odds ratio:
  • H1: $e^{\beta_k} \neq 1$
H1 two sided: $m \neq 0$
H1 right sided: $m > 0$
H1 left sided: $m < 0$
H1: there is an association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
  • H1: the distribution of the dependent variable is not the same in all of the $I$ populations
If there is one random sample of size $N$ from the total population:
  • H1: the row and column variables are dependent
AssumptionsAssumptionsAssumptions
  • In the population, the relationship between the independent variables and the log odds $\ln (\frac{\pi_{y=1}}{1 - \pi_{y=1}})$ is linear
  • The residuals are independent of one another
Often ignored additional assumption:
  • Variables are measured without error
Also pay attention to:
  • Multicollinearity
  • Outliers
  • The population distribution of the difference scores is symmetric
  • Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
Note: sometimes it considered sufficient for the data to be measured on an ordinal scale, rather than an interval or ratio scale. However, since the test statistic is based on ranked difference scores, we need to know whether a change in scores from, say, 6 to 7 is larger than/smaller than/equal to a change from 5 to 6. This is impossible to know for ordinal scales, since for these scales the size of the difference between values is meaningless.
  • Sample size is large enough for $X^2$ to be approximately chi-squared distributed under the null hypothesis. Rule of thumb:
    • 2 $\times$ 2 table: all four expected cell counts are 5 or more
    • Larger than 2 $\times$ 2 tables: average of the expected cell counts is 5 or more, smallest expected cell count is 1 or more
  • There are $I$ independent simple random samples from each of $I$ populations defined by the independent variable, or there is one simple random sample from the total population
Test statisticTest statisticTest statistic
Model chi-squared test for the complete regression model:
  • $X^2 = D_{null} - D_K = \mbox{null deviance} - \mbox{model deviance} $
    $D_{null}$, the null deviance, is conceptually similar to the total variance of the dependent variable in OLS regression analysis. $D_K$, the model deviance, is conceptually similar to the residual variance in OLS regression analysis.
Wald test for individual $\beta_k$:
The wald statistic can be defined in two ways:
  • Wald $ = \dfrac{b_k^2}{SE^2_{b_k}}$
  • Wald $ = \dfrac{b_k}{SE_{b_k}}$
SPSS uses the first definition.

Likelihood ratio chi-squared test for individual $\beta_k$:
  • $X^2 = D_{K-1} - D_K$
    $D_{K-1}$ is the model deviance, where independent variable $k$ is excluded from the model. $D_{K}$ is the model deviance, where independent variable $k$ is included in the model.
Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic. In order to compute each of the test statistics, follow the steps below:
  1. For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score}_2 - \mbox{score}_1)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
  2. For each subject, compute the absolute value of the difference score $|\mbox{score}_2 - \mbox{score}_1|$.
  3. Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
  4. Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:

  • $W_1 = \sum\, R_d^{+}$
    or
    $W_1 = \sum\, R_d^{-}$
    That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
    • tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one.
    • If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip').
  • $W_2 = \sum\, \mbox{sign}_d \times R_d$
    That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here for each cell, the expected cell count = $\dfrac{\mbox{row total} \times \mbox{column total}}{\mbox{total sample size}}$, the observed cell count is the observed sample count in that same cell, and the sum is over all $I \times J$ cells.
Sampling distribution of $X^2$ and of the Wald statistic if H0 were trueSampling distribution of $W_1$ and of $W_2$ if H0 were trueSampling distribution of $X^2$ if H0 were true
Sampling distribution of $X^2$, as computed in the model chi-squared test for the complete model:
  • chi-squared distribution with $K$ (number of independent variables) degrees of freedom
Sampling distribution of the Wald statistic:
  • If defined as Wald $ = \dfrac{b_k^2}{SE^2_{b_k}}$: approximately the chi-squared distribution with 1 degree of freedom
  • If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$: approximately the standard normal distribution
Sampling distribution of $X^2$, as computed in the likelihood ratio chi-squared test for individual $\beta_k$:
  • chi-squared distribution with 1 degree of freedom
Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.

Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.
Approximately the chi-squared distribution with $(I - 1) \times (J - 1)$ degrees of freedom
Significant?Significant?Significant?
For the model chi-squared test for the complete regression model and likelihood ratio chi-squared test for individual $\beta_k$:
  • Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
  • Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
For the Wald test:
  • If defined as Wald $ = \dfrac{b_k^2}{SE^2_{b_k}}$: same procedure as for the chi-squared tests. Wald can be interpret as $X^2$
  • If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$: same procedure as for any $z$ test. Wald can be interpreted as $z$.
For large samples, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
  • Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
  • Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Wald-type approximate $C\%$ confidence interval for $\beta_k$n.a.n.a.
$b_k \pm z^* \times SE_{b_k}$
where the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
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Goodness of fit measure $R^2_L$n.a.n.a.
$R^2_L = \dfrac{D_{null} - D_K}{D_{null}}$
There are several other goodness of fit measures in logistic regression. In logistic regression, there is no single agreed upon measure of goodness of fit.
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Example contextExample contextExample context
Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?Is the median of the differences between the mental health scores before and after an intervention different from 0?Is there an association between economic class and gender? Is the distribution of economic class different between men and women?
SPSSSPSSSPSS
Analyze > Regression > Binary Logistic...
  • Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the Wilcoxon test
Analyze > Descriptive Statistics > Crosstabs...
  • Put one of your two categorical variables in the box below Row(s), and the other categorical variable in the box below Column(s)
  • Click the Statistics... button, and click on the square in front of Chi-square
  • Continue and click OK
JamoviJamoviJamovi
Regression > 2 Outcomes - Binomial
  • Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
  • If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
  • Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
T-Tests > Paired Samples T-Test
  • Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line
  • Under Tests, select Wilcoxon rank
  • Under Hypothesis, select your alternative hypothesis
Frequencies > Independent Samples - $\chi^2$ test of association
  • Put one of your two categorical variables in the box below Rows, and the other categorical variable in the box below Columns
Practice questionsPractice questionsPractice questions