Chi-squared test for the relationship between two categorical variables - overview
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Chi-squared test for the relationship between two categorical variables | $z$ test for a single proportion | Sign test |
You cannot compare more than 3 methods |
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Independent /column variable | Independent variable | Independent variable | |
One categorical with $I$ independent groups ($I \geqslant 2$) | None | 2 paired groups | |
Dependent /row variable | Dependent variable | Dependent variable | |
One categorical with $J$ independent groups ($J \geqslant 2$) | One categorical with 2 independent groups | One of ordinal level | |
Null hypothesis | Null hypothesis | Null hypothesis | |
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: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis. |
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Alternative hypothesis | Alternative hypothesis | Alternative 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 two sided: $\pi \neq \pi_0$ H1 right sided: $\pi > \pi_0$ H1 left sided: $\pi < \pi_0$ |
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Assumptions | Assumptions | Assumptions | |
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Test statistic | Test statistic | Test statistic | |
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
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. | $W = $ number of difference scores that is larger than 0 | |
Sampling distribution of $X^2$ if H0 were true | Sampling distribution of $z$ if H0 were true | Sampling distribution of $W$ if H0 were true | |
Approximately the chi-squared distribution with $(I - 1) \times (J - 1)$ degrees of freedom | Approximately the standard normal distribution | The exact distribution of $W$ under the null hypothesis is the Binomial($n$, $P$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $P = 0.5$.
If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $nP = n \times 0.5$ and standard deviation $\sqrt{nP(1-P)} = \sqrt{n \times 0.5(1 - 0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W - n \times 0.5}{\sqrt{n \times 0.5(1 - 0.5)}}$$ follows approximately the standard normal distribution if the null hypothesis were true. | |
Significant? | Significant? | Significant? | |
| Two sided:
| If $n$ is small, the table for the binomial distribution should be used: Two sided:
If $n$ is large, the table for standard normal probabilities can be used: Two sided:
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n.a. | Approximate $C\%$ confidence interval for $\pi$ | n.a. | |
- | Regular (large sample):
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n.a. | Equivalent to | Equivalent to | |
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Two sided sign test is equivalent to
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Example context | Example context | Example context | |
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. | Do people tend to score higher on mental health after a mindfulness course? | |
SPSS | SPSS | SPSS | |
Analyze > Descriptive Statistics > Crosstabs...
| Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
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Jamovi | Jamovi | Jamovi | |
Frequencies > Independent Samples - $\chi^2$ test of association
| Frequencies > 2 Outcomes - Binomial test
| Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the two sided $p$ value that would have resulted from the sign test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
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Practice questions | Practice questions | Practice questions | |