ANCOVA - overview
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 | Spearman's rho | $z$ test for the difference between two proportions |
You cannot compare more than 3 methods |
---|---|---|---|
Independent variables | Variable 1 | Independent/grouping variable | |
One or more categorical with independent groups, and one or more quantitative control variables of interval or ratio level (covariates) | One of ordinal level | One categorical with 2 independent groups | |
Dependent variable | Variable 2 | Dependent variable | |
One quantitative of interval or ratio level | One of ordinal level | One categorical with 2 independent groups | |
THIS TABLE IS YET TO BE COMPLETED | Null hypothesis | 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. | H0: $\pi_1 = \pi_2$
Here $\pi_1$ is the population proportion of 'successes' for group 1, and $\pi_2$ is the population proportion of 'successes' for group 2. | |
n.a. | Alternative hypothesis | Alternative hypothesis | |
- | H1 two sided: $\rho_s \neq 0$ H1 right sided: $\rho_s > 0$ H1 left sided: $\rho_s < 0$ | H1 two sided: $\pi_1 \neq \pi_2$ H1 right sided: $\pi_1 > \pi_2$ H1 left sided: $\pi_1 < \pi_2$ | |
n.a. | Assumptions | Assumptions | |
- |
|
| |
n.a. | Test statistic | Test statistic | |
- | $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. | $z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
Here $p_1$ is the sample proportion of successes in group 1: $\dfrac{X_1}{n_1}$, $p_2$ is the sample proportion of successes in group 2: $\dfrac{X_2}{n_2}$, $p$ is the total proportion of successes in the sample: $\dfrac{X_1 + X_2}{n_1 + n_2}$, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. Note: we could just as well compute $p_2 - p_1$ in the numerator, but then the left sided alternative becomes $\pi_2 < \pi_1$, and the right sided alternative becomes $\pi_2 > \pi_1.$ | |
n.a. | Sampling distribution of $t$ if H0 were true | Sampling distribution of $z$ if H0 were true | |
- | Approximately the $t$ distribution with $N - 2$ degrees of freedom | Approximately the standard normal distribution | |
n.a. | Significant? | Significant? | |
- | Two sided:
| Two sided:
| |
n.a. | n.a. | Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ | |
- | - | Regular (large sample):
| |
n.a. | n.a. | Equivalent to | |
- | - | When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels. | |
n.a. | Example context | Example context | |
- | Is there a monotonic relationship between physical health and mental health? | Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic. | |
n.a. | SPSS | SPSS | |
- | Analyze > Correlate > Bivariate...
| SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
| |
n.a. | Jamovi | Jamovi | |
- | Regression > Correlation Matrix
| Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chi-squared test instead. The $p$ value resulting from this chi-squared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Frequencies > Independent Samples - $\chi^2$ test of association
| |
Practice questions | Practice questions | Practice questions | |