z test for a single proportion - overview

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$z$ test for a single proportion
Spearman's rho
Independent variableVariable 1
NoneOne of ordinal level
Dependent variableVariable 2
One categorical with 2 independent groupsOne of ordinal level
Null hypothesisNull 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: $\rho_s = 0$

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 hypothesisAlternative hypothesis
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
H1 two sided: $\rho_s \neq 0$
H1 right sided: $\rho_s > 0$
H1 left sided: $\rho_s < 0$
AssumptionsAssumptions
  • 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 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 statisticTest 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.
$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 trueSampling distribution of $t$ if H0 were true
Approximately the standard normal distributionApproximately the $t$ distribution with $N - 2$ degrees of freedom
Significant?Significant?
Two sided: Right sided: Left sided: Two sided: Right sided: Left sided:
Approximate $C\%$ confidence interval for $\pi$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)
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Equivalent ton.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.
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Example contextExample 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 there a monotonic relationship between physical health and mental health?
SPSSSPSS
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 > Correlate > Bivariate...
  • Put your two variables in the box below Variables
  • Under Correlation Coefficients, select Spearman
JamoviJamovi
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
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 questionsPractice questions