$z$ test for a single proportion

This page offers all the basic information you need about the $z$ test for a single proportion. It is part of Statkat’s wiki module, containing similarly structured info pages for many different statistical methods. The info pages give information about null and alternative hypotheses, assumptions, test statistics and confidence intervals, how to find p values, SPSS how-to’s and more.

To compare the $z$ test for a single proportion with other statistical methods, go to Statkat's or practice with the $z$ test for a single proportion at Statkat's

When to use?

Deciding which statistical method to use to analyze your data can be a challenging task. Whether a statistical method is appropriate for your data is partly determined by the measurement level of your variables. The $z$ test for a single proportion requires one variable of the following type:

 One categorical with 2 independent groups

Note that theoretically, it is always possible to 'downgrade' the measurement level of a variable. For instance, a test that can be performed on a variable of ordinal measurement level can also be performed on a variable of interval measurement level, in which case the interval variable is downgraded to an ordinal variable. However, downgrading the measurement level of variables is generally a bad idea since it means you are throwing away important information in your data (an exception is the downgrade from ratio to interval level, which is generally irrelevant in data analysis).

If you are not sure which method you should use, you might like the assistance of our method selection tool or our method selection table.

Null hypothesis

The $z$ test for a single proportion tests the following null hypothesis (H0):

H0: $\pi = \pi_0$

$\pi$ is the population proportion of 'successes'; $\pi_0$ is the population proportion of successes according to the null hypothesis
Alternative hypothesis

The $z$ test for a single proportion tests the above null hypothesis against the following alternative hypothesis (H1 or Ha):

H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
Assumptions

Statistical tests always make assumptions about the sampling procedure that's been used to obtain the sample data. So called parametric tests also make assumptions about how data are distributed in the population. Non-parametric tests are more 'robust' and make no or less strict assumptions about population distributions, but are generally less powerful. Violation of assumptions may render the outcome of statistical tests useless, although violation of some assumptions (e.g. independence assumptions) are generally more problematic than violation of other assumptions (e.g. normality assumptions in combination with large samples).

The $z$ test for a single proportion makes the following 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.
Test statistic

The $z$ test for a single proportion is based on the following test statistic:

$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
$p$ 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.
Sampling distribution

Sampling distribution of $z$ if H0 were true:

Approximately the standard normal distribution
Significant?

This is how you find out if your test result is significant:

Two sided:
Right sided:
Left sided:
Approximate $C\%$ confidence interval for $\pi$

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

The $z$ test for a single proportion is equivalent to:

• When testing two sided: goodness of fit test, with 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.
Example context

The $z$ test for a single proportion could for instance be used to answer the question:

Is the proportion of smokers amongst office workers different from $\pi_0 = .2$? Use the normal approximation for the sampling distribution of the test statistic.
SPSS

How to perform the $z$ test for a single proportion in 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
Jamovi

How to perform the $z$ test for a single proportion in 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