Binomial test for a single proportion - overview
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Binomial test for a single proportion | Marginal Homogeneity test / Stuart-Maxwell test | $z$ test for the difference between two proportions |
You cannot compare more than 3 methods |
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Independent variable | Independent variable | Independent/grouping variable | |
None | 2 paired groups | One categorical with 2 independent groups | |
Dependent variable | Dependent variable | Dependent variable | |
One categorical with 2 independent groups | One categorical with $J$ independent groups ($J \geqslant 2$) | One categorical with 2 independent groups | |
Null hypothesis | Null hypothesis | Null hypothesis | |
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. | H0: for each category $j$ of the dependent variable, $\pi_j$ for the first paired group = $\pi_j$ for the second paired group.
Here $\pi_j$ is the population proportion in category $j.$ | 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. | |
Alternative hypothesis | Alternative hypothesis | Alternative hypothesis | |
H1 two sided: $\pi \neq \pi_0$ H1 right sided: $\pi > \pi_0$ H1 left sided: $\pi < \pi_0$ | H1: for some categories of the dependent variable, $\pi_j$ for the first paired group $\neq$ $\pi_j$ for the second paired group. | H1 two sided: $\pi_1 \neq \pi_2$ H1 right sided: $\pi_1 > \pi_2$ H1 left sided: $\pi_1 < \pi_2$ | |
Assumptions | Assumptions | Assumptions | |
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Test statistic | Test statistic | Test statistic | |
$X$ = number of successes in the sample | Computing the test statistic is a bit complicated and involves matrix algebra. Unless you are following a technical course, you probably won't need to calculate it by hand. | $z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
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 $\pi_2 > \pi_1.$ | |
Sampling distribution of $X$ if H0 were true | Sampling distribution of the test statistic if H0 were true | Sampling distribution of $z$ if H0 were true | |
Binomial($n$, $P$) distribution.
Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis). | Approximately the chi-squared distribution with $J - 1$ degrees of freedom | Approximately the standard normal distribution | |
Significant? | Significant? | Significant? | |
Two sided:
| If we denote the test statistic as $X^2$:
| Two sided:
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n.a. | n.a. | Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ | |
- | - | Regular (large sample):
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n.a. | n.a. | Equivalent to | |
- | - | When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels. | |
Example context | Example context | Example context | |
Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$? | Subjects are asked to taste three different types of mayonnaise, and to indicate which of the three types of mayonnaise they like best. They then have to drink a glass of beer, and taste and rate the three types of mayonnaise again. Does drinking a beer change which type of mayonnaise people like best? | Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic. | |
SPSS | SPSS | SPSS | |
Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| 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...
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Jamovi | n.a. | Jamovi | |
Frequencies > 2 Outcomes - Binomial test
| - | 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
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Practice questions | Practice questions | Practice questions | |