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 ($J \geqslant 2$)
Null hypothesis
Null hypothesis
H_{0}: $\pi = \pi_0$
$\pi$ is the population proportion of 'successes'; $\pi_0$ is the population proportion of successes according to the 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 $\pi_J$
Alternative hypothesis
Alternative hypothesis
H_{1} two sided: $\pi \neq \pi_0$
H_{1} right sided: $\pi > \pi_0$
H_{1} left sided: $\pi < \pi_0$
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
Assumptions
Assumptions
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
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
Test statistic
Test statistic
$z = \dfrac{p  \pi_0}{\sqrt{\dfrac{\pi_0(1  \pi_0)}{N}}}$
$p$ 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.
$X^2 = \sum{\frac{(\mbox{observed cell count}  \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
where 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
Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Approximate $C\%$ confidence interval for $\pi$
n.a.
Regular (large sample):
$p \pm z^* \times \sqrt{\dfrac{p(1  p)}{N}}$
where $z^*$ 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 $z^*$ 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)

Equivalent to
n.a.
When testing two sided: goodness of fit test, with 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
Is the proportion of smokers amongst office workers different from $\pi_0 = .2$? Use the normal approximation for the sampling distribution of the test statistic.
Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low}$ = .2, $\pi_{moderate}$ = .6, and $\pi_{high}$ = .2?
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
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)
Jamovi
Jamovi
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
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)