z test for a single proportion overview

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$z$ test for a single proportion	Binomial test for a single proportion $z$ test for a single proportion $z$ test for the difference between two proportions Goodness of fit test Chi-squared test for the relationship between two categorical variables One sample $z$ test for the mean One sample $t$ test for the mean Paired sample $t$ test Two sample $z$ test Two sample $t$ test - equal variances not assumed Two sample $t$ test - equal variances assumed One way ANOVA Two way ANOVA Pearson correlation Regression (OLS)Logistic regression Mann-Whitney-Wilcoxon test Kruskal-Wallis test Sign test McNemar's test Cochran's Q test Marginal Homogeneity test / Stuart-Maxwell test Friedman test Wilcoxon signed-rank test One sample Wilcoxon signed-rank test Spearman's rho
Independent variable
None
Dependent variable
One categorical with 2 independent groups
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.
Alternative hypothesis
H₁ two sided: $\pi \neq \pi_0$ H₁ right sided: $\pi > \pi_0$ H₁ left sided: $\pi < \pi_0$
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
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$ 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.
Sampling distribution of $z$ if H₀ were true
Approximately the standard normal distribution
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 $\alpha$
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
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.
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.
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
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
Practice questions

z test for a single proportion - overview