z test for a single proportion overview

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$z$ test for a single proportion	Cochran's Q test	Spearman's rho
Independent variable	Independent/grouping variable	Variable 1
None	One within subject factor ($\geq 2$ related groups)	One of ordinal level
Dependent variable	Dependent variable	Variable 2
One categorical with 2 independent groups	One categorical with 2 independent groups	One of ordinal level
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: $\pi_1 = \pi_2 = \ldots = \pi_I$ Here $\pi_1$ is the population proportion of 'successes' for group 1, $\pi_2$ is the population proportion of 'successes' for group 2, and $\pi_I$ is the population proportion of 'successes' for group $I.$	H₀: $\rho_s = 0$ Here $\rho_s$ is the Spearman correlation in the population. The Spearman correlation is a measure for the strength and direction of the monotonic relationship between two variables of at least ordinal measurement level. In words, the null hypothesis would be: H₀: there is no monotonic relationship between the two variables in the population.
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: not all population proportions are equal	H₁ two sided: $\rho_s \neq 0$ H₁ right sided: $\rho_s > 0$ H₁ left sided: $\rho_s < 0$
Assumptions	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 If the sample size is too small for $z$ to be approximately normally distributed, the binomial test for a single proportion should be used.	Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another	Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another Note: this assumption is only important for the significance test, not for the correlation coefficient itself. The correlation coefficient itself just measures the strength of the monotonic relationship between two variables.
Test statistic	Test statistic	Test statistic
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$ 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.	If a failure is scored as 0 and a success is scored as 1: $Q = k(k - 1) \dfrac{\sum_{groups} \Big (\mbox{group total} - \frac{\mbox{grand total}}{k} \Big)^2}{\sum_{blocks} \mbox{block total} \times (k - \mbox{block total})}$ Here $k$ is the number of related groups (usually the number of repeated measurements), a group total is the sum of the scores in a group, a block total is the sum of the scores in a block (usually a subject), and the grand total is the sum of all the scores. Before computing $Q$, first exclude blocks with equal scores in all $k$ groups.	$t = \dfrac{r_s \times \sqrt{N - 2}}{\sqrt{1 - r_s^2}} $ Here $r_s$ is the sample Spearman correlation and $N$ is the sample size. The sample Spearman correlation $r_s$ is equal to the Pearson correlation applied to the rank scores.
Sampling distribution of $z$ if H₀ were true	Sampling distribution of $Q$ if H₀ were true	Sampling distribution of $t$ if H₀ were true
Approximately the standard normal distribution	If the number of blocks (usually the number of subjects) is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom	Approximately the $t$ distribution with $N - 2$ degrees of freedom
Significant?	Significant?	Significant?
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 $\alpha$	If the number of blocks 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 $t$ observed in sample is at least as extreme as critical value $t^$ or Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$ Right sided: Check if $t$ observed in sample is equal to or larger than critical value $t^$ or Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$ Left sided: Check if $t$ observed in sample is equal to or smaller than critical value $t^*$ or Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$
Approximate $C\%$ confidence interval for $\pi$	n.a.	n.a.
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)	-	-
Equivalent to	Equivalent to	n.a.
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.	Friedman test, with a categorical dependent variable consisting of two independent groups.	-
Example context	Example context	Example context
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.	Subjects perform three different tasks, which they can either perform correctly or incorrectly. Is there a difference in task performance between the three different tasks?	Is there a monotonic relationship between physical health and mental health?
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 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	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 Cochran's Q test	Analyze > Correlate > Bivariate... Put your two variables in the box below Variables Under Correlation Coefficients, select Spearman
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 Jamovi will give you the exact $p$ value based on the binomial distribution, rather than the approximate $p$ value based on the normal distribution	Jamovi does not have a specific option for the Cochran's Q test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the $p$ value that would have resulted from the Cochran's Q test. Go to: ANOVA > Repeated Measures ANOVA - Friedman Put the $k$ variables containing the scores for the $k$ related groups in the box below Measures	Regression > Correlation Matrix Put your two variables in the white box at the right Under Correlation Coefficients, select Spearman Under Hypothesis, select your alternative hypothesis
Practice questions	Practice questions	Practice questions

z test for a single proportion - overview