z test for a single proportion - overview

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$z$ test for a single proportion
Logistic regression
Chi-squared test for the relationship between two categorical variables
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Independent variableIndependent variablesIndependent /column variable
NoneOne or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variablesOne categorical with $I$ independent groups ($I \geqslant 2$)
Dependent variableDependent variableDependent /row variable
One categorical with 2 independent groupsOne categorical with 2 independent groupsOne categorical with $J$ independent groups ($J \geqslant 2$)
Null hypothesisNull 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.
Model chi-squared test for the complete regression model:
  • H0: $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
Wald test for individual regression coefficient $\beta_k$:
  • H0: $\beta_k = 0$
    or in terms of odds ratio:
  • H0: $e^{\beta_k} = 1$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
  • H0: $\beta_k = 0$
    or in terms of odds ratio:
  • H0: $e^{\beta_k} = 1$
in the regression equation $ \ln \big(\frac{\pi_{y = 1}}{1 - \pi_{y = 1}} \big) = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K $. Here $ x_i$ represents independent variable $ i$, $\beta_i$ is the regression weight for independent variable $ x_i$, and $\pi_{y = 1}$ represents the true probability that the dependent variable $ y = 1$ (or equivalently, the proportion of $ y = 1$ in the population) given the scores on the independent variables.
H0: there is no association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
  • H0: the distribution of the dependent variable is the same in each of the $I$ populations
If there is one random sample of size $N$ from the total population:
  • H0: the row and column variables are independent
Alternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
Model chi-squared test for the complete regression model:
  • H1: not all population regression coefficients are 0
Wald test for individual regression coefficient $\beta_k$:
  • H1: $\beta_k \neq 0$
    or in terms of odds ratio:
  • H1: $e^{\beta_k} \neq 1$
    If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$ (see 'Test statistic'), also one sided alternatives can be tested:
  • H1 right sided: $\beta_k > 0$
  • H1 left sided: $\beta_k < 0$
Likelihood ratio chi-squared test for individual regression coefficient $\beta_k$:
  • H1: $\beta_k \neq 0$
    or in terms of odds ratio:
  • H1: $e^{\beta_k} \neq 1$
H1: there is an association between the row and column variable

More precisely, if there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
  • H1: the distribution of the dependent variable is not the same in all of the $I$ populations
If there is one random sample of size $N$ from the total population:
  • H1: the row and column variables are dependent
AssumptionsAssumptionsAssumptions
  • 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.
  • In the population, the relationship between the independent variables and the log odds $\ln (\frac{\pi_{y=1}}{1 - \pi_{y=1}})$ is linear
  • The residuals are independent of one another
Often ignored additional assumption:
  • Variables are measured without error
Also pay attention to:
  • Multicollinearity
  • Outliers
  • Sample size is large enough for $X^2$ to be approximately chi-squared distributed under the null hypothesis. Rule of thumb:
    • 2 $\times$ 2 table: all four expected cell counts are 5 or more
    • Larger than 2 $\times$ 2 tables: average of the expected cell counts is 5 or more, smallest expected cell count is 1 or more
  • There are $I$ independent simple random samples from each of $I$ populations defined by the independent variable, or there is one simple random sample from the total population
Test statisticTest 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.
Model chi-squared test for the complete regression model:
  • $X^2 = D_{null} - D_K = \mbox{null deviance} - \mbox{model deviance} $
    $D_{null}$, the null deviance, is conceptually similar to the total variance of the dependent variable in OLS regression analysis. $D_K$, the model deviance, is conceptually similar to the residual variance in OLS regression analysis.
Wald test for individual $\beta_k$:
The wald statistic can be defined in two ways:
  • Wald $ = \dfrac{b_k^2}{SE^2_{b_k}}$
  • Wald $ = \dfrac{b_k}{SE_{b_k}}$
SPSS uses the first definition.

Likelihood ratio chi-squared test for individual $\beta_k$:
  • $X^2 = D_{K-1} - D_K$
    $D_{K-1}$ is the model deviance, where independent variable $k$ is excluded from the model. $D_{K}$ is the model deviance, where independent variable $k$ is included in the model.
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here for each cell, the expected cell count = $\dfrac{\mbox{row total} \times \mbox{column total}}{\mbox{total sample size}}$, the observed cell count is the observed sample count in that same cell, and the sum is over all $I \times J$ cells.
Sampling distribution of $z$ if H0 were trueSampling distribution of $X^2$ and of the Wald statistic if H0 were trueSampling distribution of $X^2$ if H0 were true
Approximately the standard normal distributionSampling distribution of $X^2$, as computed in the model chi-squared test for the complete model:
  • chi-squared distribution with $K$ (number of independent variables) degrees of freedom
Sampling distribution of the Wald statistic:
  • If defined as Wald $ = \dfrac{b_k^2}{SE^2_{b_k}}$: approximately the chi-squared distribution with 1 degree of freedom
  • If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$: approximately the standard normal distribution
Sampling distribution of $X^2$, as computed in the likelihood ratio chi-squared test for individual $\beta_k$:
  • chi-squared distribution with 1 degree of freedom
Approximately the chi-squared distribution with $(I - 1) \times (J - 1)$ degrees of freedom
Significant?Significant?Significant?
Two sided: Right sided: Left sided: For the model chi-squared test for the complete regression model and likelihood ratio chi-squared test for individual $\beta_k$:
  • 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 $\alpha$
For the Wald test:
  • If defined as Wald $ = \dfrac{b_k^2}{SE^2_{b_k}}$: same procedure as for the chi-squared tests. Wald can be interpret as $X^2$
  • If defined as Wald $ = \dfrac{b_k}{SE_{b_k}}$: same procedure as for any $z$ test. Wald can be interpreted as $z$.
  • 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 $\alpha$
Approximate $C\%$ confidence interval for $\pi$Wald-type approximate $C\%$ confidence interval for $\beta_k$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)
$b_k \pm z^* \times SE_{b_k}$
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).
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n.a.Goodness of fit measure $R^2_L$n.a.
-$R^2_L = \dfrac{D_{null} - D_K}{D_{null}}$
There are several other goodness of fit measures in logistic regression. In logistic regression, there is no single agreed upon measure of goodness of fit.
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Equivalent ton.a.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.
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Example contextExample 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.Can body mass index, stress level, and gender predict whether people get diagnosed with diabetes?Is there an association between economic class and gender? Is the distribution of economic class different between men and women?
SPSSSPSSSPSS
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 > Regression > Binary Logistic...
  • Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Covariate(s)
Analyze > Descriptive Statistics > Crosstabs...
  • Put one of your two categorical variables in the box below Row(s), and the other categorical 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
JamoviJamoviJamovi
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 > 2 Outcomes - Binomial
  • Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
  • If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
  • Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
Frequencies > Independent Samples - $\chi^2$ test of association
  • Put one of your two categorical variables in the box below Rows, and the other categorical variable in the box below Columns
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