This page offers structured overviews of one or more selected methods. Add additional methods for comparisons (max. of 3) by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables
None
Dependent variable
Dependent variable
One categorical with 2 independent groups
One categorical with 2 independent groups
Null hypothesis
Null 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: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
Alternative hypothesis
Alternative 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: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
Assumptions
Assumptions
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
Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
Significance test: $N \times \pi_0$ and $N \times (1 - \pi_0)$ are each larger than 10
Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures in sample are each 15 or more
Plus four 90%, 95%, or 99% confidence interval: total sample size is 10 or more
Sample is a simple random sample from the population. That is, observations are independent of one another
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.
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
Here $p$ is the sample proportion of successes: $\dfrac{X}{N}$, $N$ is the sample size, and $\pi_0$ is the population proportion of successes according to the null hypothesis.
Sampling distribution of $X^2$ and of the Wald statistic if H0 were true
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 $z$.
Two sided:
Check if $z$ observed in sample is at least as extreme as critical value $z^*$ or
Find two sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
Right sided:
Check if $z$ observed in sample is equal to or larger than critical value $z^*$ or
Find right sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
Left sided:
Check if $z$ observed in sample is equal to or smaller than critical value $z^*$ or
Find left sided $p$ value corresponding to observed $z$ and check if it is equal to or smaller than $\alpha$
Wald-type approximate $C\%$ confidence interval for $\beta_k$
Approximate $C\%$ confidence interval for $\pi$
$b_k \pm z^* \times SE_{b_k}$
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).
Regular (large sample):
$p \pm z^* \times \sqrt{\dfrac{p(1 - p)}{N}}$
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)
With plus four method:
$p_{plus} \pm z^* \times \sqrt{\dfrac{p_{plus}(1 - p_{plus})}{N + 4}}$
where $p_{plus} = \dfrac{X + 2}{N + 4}$ and 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)
Goodness of fit measure $R^2_L$
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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n.a.
Equivalent to
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When testing two sided: goodness of fit test, with a categorical variable with 2 levels.
When $N$ is large, the $p$ value from the $z$ test for a single proportion approaches the $p$ value from the binomial test for a single proportion. The $z$ test for a single proportion is just a large sample approximation of the binomial test for a single proportion.
Example context
Example context
Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?
Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic.
SPSS
SPSS
Analyze > Regression > Binary Logistic...
Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
Put your dichotomous variable in the box below Test Variable List
Fill in the value for $\pi_0$ in the box next to Test Proportion
If computation time allows, SPSS will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution
Jamovi
Jamovi
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'
Frequencies > 2 Outcomes - Binomial test
Put your dichotomous variable in the white box at the right
Fill in the value for $\pi_0$ in the box next to Test value
Under Hypothesis, select your alternative hypothesis
Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution