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
Goodness of fit test
McNemar's test
$z$ test for the difference between two proportions
One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variables
Dependent variable
Dependent variable
Dependent variable
Dependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)
One categorical with 2 independent groups
One categorical with 2 independent groups
One categorical with 2 independent groups
Null hypothesis
Null hypothesis
Null hypothesis
Null hypothesis
H0: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
H0: the probability of drawing an observation from condition 1 is $\pi_1$, the probability of drawing an observation from condition 2 is $\pi_2$, $\ldots$,
the probability of drawing an observation from condition $J$ is $\pi_J$
Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options:
First score of pair is 0, second score of pair is 0
First score of pair is 0, second score of pair is 1 (switched)
First score of pair is 1, second score of pair is 0 (switched)
First score of pair is 1, second score of pair is 1
The null hypothesis H0 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) = P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is the same as the probability that a pair of scores switches from 1 to 0.
Other formulations of the null hypothesis are:
H0: $\pi_1 = \pi_2$, where $\pi_1$ is the population proportion of ones for the first paired group and $\pi_2$ is the population proportion of ones for the second paired group
H0: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)
H0: $\pi_1 = \pi_2$
Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2.
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.
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
H1: the population proportions are not all as specified under the null hypothesis
or equivalently
H1: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis
The alternative hypothesis H1 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0.
Other formulations of the alternative hypothesis are:
H1: $\pi_1 \neq \pi_2$
H1: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)
H1 two sided: $\pi_1 \neq \pi_2$
H1 right sided: $\pi_1 > \pi_2$
H1 left sided: $\pi_1 < \pi_2$
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: $e^{\beta_k} \neq 1$
Assumptions
Assumptions
Assumptions
Assumptions
Sample size is large enough for $X^2$ to be approximately chi-squared distributed. Rule of thumb: all $J$ expected cell counts are 5 or more
Sample is a simple random sample from the population. That is, observations are independent of one another
Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
Significance test: number of successes and number of failures are each 5 or more in both sample groups
Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures are each 10 or more in both sample groups
Plus four 90%, 95%, or 99% confidence interval: sample sizes of both groups are 5 or more
Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
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
Test statistic
Test statistic
Test statistic
Test statistic
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.
$X^2 = \dfrac{(b - c)^2}{b + c}$
Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0.
$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
Here $p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$,
$p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$,
$p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$,
$n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2.
Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1.$
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.
Sampling distribution of $X^2$ and of the Wald statistic if H0 were true
Approximately the chi-squared distribution with $J - 1$ degrees of freedom
If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom.
If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$.
Approximately the standard normal distribution
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 $\beta_k$:
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 $z$.
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Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$
Wald-type approximate $C\%$ confidence interval for $\beta_k$
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Regular (large sample):
$(p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}}$
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:
$(p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}}$
where $p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}$, $p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}$, 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)
$b_k \pm z^* \times SE_{b_k}$
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 $R^2_L$
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$R^2_L = \dfrac{D_{null} - D_K}{D_{null}}$
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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Equivalent to
Equivalent to
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Stuart-Maxwell test, with a categorical dependent variable consisting of two independent groups
Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$?
Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?
Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.
Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?
Put your categorical variable in the box below Test Variable List
Fill in the population proportions / probabilities according to $H_0$ in the box below Expected Values. If $H_0$ states that they are all equal, just pick 'All categories equal' (default)
Put the two paired variables in the boxes below Variable 1 and Variable 2
Under Test Type, select the McNemar test
SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
Put your independent (grouping) variable in the box below Row(s), and your dependent 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
Analyze > Regression > Binary Logistic...
Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
Jamovi
Jamovi
Jamovi
Jamovi
Frequencies > N Outcomes - $\chi^2$ Goodness of fit
Put your categorical variable in the box below Variable
Click on Expected Proportions and fill in the population proportions / probabilities according to $H_0$ in the boxes below Ratio. If $H_0$ states that they are all equal, you can leave the ratios equal to the default values (1)
Frequencies > Paired Samples - McNemar test
Put one of the two paired variables in the box below Rows and the other paired variable in the box below Columns
Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples - $\chi^2$ test of association
Put your independent (grouping) variable in the box below Rows, and your dependent variable in the box below Columns
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'