This page offers structured overviews of one or more selected methods. Add additional methods for comparisons (max. of 3) by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
ANCOVA
$z$ test for a single proportion
$z$ test for a single proportion
You cannot compare more than 3 methods
Independent variables
Independent variable
Independent variable
One or more categorical with independent groups, and one or more quantitative control variables of interval or ratio level (covariates)
None
None
Dependent variable
Dependent variable
Dependent variable
One quantitative of interval or ratio level
One categorical with 2 independent groups
One categorical with 2 independent groups
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Null hypothesis
Null hypothesis
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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: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
n.a.
Alternative hypothesis
Alternative hypothesis
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H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
n.a.
Assumptions
Assumptions
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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
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
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.
$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
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.
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 $\alpha$
n.a.
Approximate $C\%$ confidence interval for $\pi$
Approximate $C\%$ confidence interval for $\pi$
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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)
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)
n.a.
Equivalent to
Equivalent to
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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.
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.
n.a.
Example context
Example context
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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 proportion of smokers amongst office workers different from $\pi_0 = 0.2$? Use the normal approximation for the sampling distribution of the test statistic.
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
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
n.a.
Jamovi
Jamovi
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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 > 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