This page offers structured overviews of one or more selected methods. Add additional methods for comparisons 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
Chi-squared test for the relationship between two categorical variables
One categorical with $I$ independent groups ($I \geqslant 2$)
None
Dependent variable
Dependent /row variable
Dependent variable
One categorical with $J$ independent groups ($J \geqslant 2$)
One categorical with $J$ independent groups ($J \geqslant 2$)
One categorical with 2 independent groups
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 $\pi_J$
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
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
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
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
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
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 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
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.
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
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.
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
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.
Find $p$ value corresponding to observed $X^2$ 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 $\pi$
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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.
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
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 $\pi_{high} = 0.2$?
Is there an association between economic class and gender? Is the distribution of economic class different between men and women?
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)
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
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 > 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
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