z test for a single proportion overview

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$z$ test for a single proportion	Two sample $z$ test	McNemar's test
Independent variable	Independent/grouping variable	Independent variable
None	One categorical with 2 independent groups	2 paired groups
Dependent variable	Dependent variable	Dependent variable
One categorical with 2 independent groups	One quantitative of interval or ratio 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: $\mu_1 = \mu_2$ Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.	Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options: First score of pair is 0, second score of pair is 0 First score of pair is 0, second score of pair is 1 (switched) First score of pair is 1, second score of pair is 0 (switched) First score of pair is 1, second score of pair is 1 The null hypothesis H₀ is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) = P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is the same as the probability that a pair of scores switches from 1 to 0. Other formulations of the null hypothesis are: H₀: $\pi_1 = \pi_2$, where $\pi_1$ is the population proportion of ones for the first paired group and $\pi_2$ is the population proportion of ones for the second paired group H₀: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1)
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 two sided: $\mu_1 \neq \mu_2$ H₁ right sided: $\mu_1 > \mu_2$ H₁ left sided: $\mu_1 < \mu_2$	The alternative hypothesis H₁ is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0. Other formulations of the alternative hypothesis are: H₁: $\pi_1 \neq \pi_2$ H₁: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1)
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.	Within each population, the scores on the dependent variable are normally distributed Population standard deviations $\sigma_1$ and $\sigma_2$ are known 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	Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
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.	$z = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$ Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $\sigma^2_1$ is the population variance in population 1, $\sigma^2_2$ is the population variance in population 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis. The denominator $\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}$ is the standard deviation of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $z$ value indicates how many of these standard deviations $\bar{y}_1 - \bar{y}_2$ is removed from 0. Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$.	$X^2 = \dfrac{(b - c)^2}{b + c}$ Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0.
Sampling distribution of $z$ if H₀ were true	Sampling distribution of $z$ if H₀ were true	Sampling distribution of $X^2$ if H₀ were true
Approximately the standard normal distribution	Standard normal distribution	If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom. If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$.
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 $\alpha$	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 $\alpha$	For test statistic $X^2$: 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 $\alpha$ If $b + c$ is small, the table for the binomial distribution should be used, with as test statistic $b$: Check if $b$ observed in sample is in the rejection region or Find two sided $p$ value corresponding to observed $b$ and check if it is equal to or smaller than $\alpha$
Approximate $C\%$ confidence interval for $\pi$	$C\%$ confidence interval for $\mu_1 - \mu_2$	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)	$(\bar{y}_1 - \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$ 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). The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test.	-
n.a.	Visual representation	n.a.
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Equivalent to	n.a.	Equivalent to
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.	-	Stuart-Maxwell test, with a categorical dependent variable consisting of two independent groups Cochran's Q test, with two related groups Two sided sign test: $b = W$, and $X^2 = z^2$
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.	Is the average mental health score different between men and women? Assume that in the population, the standard devation of the mental health scores is $\sigma_1 = 2$ amongst men and $\sigma_2 = 2.5$ amongst women.	Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?
SPSS	n.a.	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 > 2 Related Samples... Put the two paired variables in the boxes below Variable 1 and Variable 2 Under Test Type, select the McNemar test
Jamovi	n.a.	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	-	Frequencies > Paired Samples - McNemar test Put one of the two paired variables in the box below Rows and the other paired variable in the box below Columns
Practice questions	Practice questions	Practice questions

z test for a single proportion - overview