Binomial test for a single proportion overview

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Binomial test for a single proportion	Friedman test	$z$ test for the difference between two proportions
Independent variable	Independent/grouping variable	Independent/grouping variable
None	One within subject factor ($\geq 2$ related groups)	One categorical with 2 independent groups
Dependent variable	Dependent variable	Dependent variable
One categorical with 2 independent groups	One of ordinal level	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: the population scores in any of the related groups are not systematically higher or lower than the population scores in any of the other related groups Usually the related groups are the different measurement points. Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.	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: the population scores in some of the related groups are systematically higher or lower than the population scores in other related groups	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
Sample is a simple random sample from the population. That is, observations are independent of one another	Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another	Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb: Significance test: number of successes and number of failures are each 5 or more in both sample groups Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures are each 10 or more in both sample groups Plus four 90%, 95%, or 99% confidence interval: sample sizes of both groups are 5 or more Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
Test statistic	Test statistic	Test statistic
$X$ = number of successes in the sample	$Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i - 3 \times N(k + 1)$ Here $N$ is the number of 'blocks' (usually the subjects - so if you have 4 repeated measurements for 60 subjects, $N$ equals 60), $k$ is the number of related groups (usually the number of repeated measurements), and $R_i$ is the sum of ranks in group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N \times k(k + 1)} \times \sum R^2_i$ and then subtract $3 \times N(k + 1)$. Note: if ties are present in the data, the formula for $Q$ is more complicated.	$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$ 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 $Q$ 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).	If the number of blocks $N$ is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom. For small samples, the exact distribution of $Q$ should be used.	Approximately the standard normal distribution
Significant?	Significant?	Significant?
Two sided: Check if $X$ observed in sample is in the rejection region or Find two sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$ Right sided: Check if $X$ observed in sample is in the rejection region or Find right sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$ Left sided: Check if $X$ observed in sample is in the rejection region or Find left sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$	If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$: Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or 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 $\pi_1 - \pi_2$
-	-	Regular (large sample): $(p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}}$ 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: $(p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}}$ where $p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}$, $p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}$, 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
-	-	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 $\pi_0 = 0.2$?	Is there a difference in depression level between measurement point 1 (pre-intervention), measurement point 2 (1 week post-intervention), and measurement point 3 (6 weeks post-intervention)?	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... 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	Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples... Put the $k$ variables containing the scores for the $k$ related groups in the white box below Test Variables Under Test Type, select the Friedman test	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... Put your independent (grouping) variable in the box below Row(s), and your dependent variable in the box below Column(s) Click the Statistics... button, and click on the square in front of Chi-square Continue and click OK
Jamovi	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	ANOVA > Repeated Measures ANOVA - Friedman Put the $k$ variables containing the scores for the $k$ related groups in the box below Measures	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 Put your independent (grouping) variable in the box below Rows, and your dependent variable in the box below Columns
Practice questions	Practice questions	Practice questions

Binomial test for a single proportion - overview