This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the righthand column. To practice with a specific method click the button at the bottom row of the table
One categorical with $J$ independent groups ($J \geqslant 2$)
One categorical with 2 independent groups
One categorical with 2 independent groups
Null hypothesis
Null hypothesis
Null hypothesis
H_{0}: the population proportions in each of the $J$ conditions are $\pi_1$, $\pi_2$, $\ldots$, $\pi_J$
or equivalently
H_{0}: 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 H_{0} 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:
H_{0}: $\pi_1 = \pi_2$, 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
H_{0}: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)
H_{0}: $\pi = \pi_0$
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
Alternative hypothesis
H_{1}: the population proportions are not all as specified under the null hypothesis
or equivalently
H_{1}: the probabilities of drawing an observation from each of the conditions are not all as specified under the null hypothesis
The alternative hypothesis H_{1} 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:
H_{1}: $\pi_1 \neq \pi_2$
H_{1}: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)
H_{1} two sided: $\pi \neq \pi_0$
H_{1} right sided: $\pi > \pi_0$
H_{1} left sided: $\pi < \pi_0$
Assumptions
Assumptions
Assumptions
Sample size is large enough for $X^2$ to be approximately chisquared 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: $N \times \pi_0$ 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
$X^2 = \sum{\frac{(\mbox{observed cell count}  \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
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.
$z = \dfrac{p  \pi_0}{\sqrt{\dfrac{\pi_0(1  \pi_0)}{N}}}$
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.
Approximately the chisquared distribution with $J  1$ degrees of freedom
If $b + c$ is large enough (say, > 20), approximately the chisquared 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 $b$.
Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
If $b + c$ is small, the table for the binomial distribution should be used, with as test statistic $b$:
Check if $b$ observed in sample is in the rejection region or
Find two sided $p$ value corresponding to observed $b$ and check if it is equal to or smaller than $\alpha$
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 $\alpha$
n.a.
n.a.
Approximate $C\%$ confidence interval for $\pi$


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)
n.a.
Equivalent to
Equivalent to

StuartMaxwell test, with a categorical dependent variable consisting of two independent groups
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
Example context
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 amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic.
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 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
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
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