ANCOVA - overview
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ANCOVA | $z$ test for a single proportion | One sample Wilcoxon signed-rank test |
You cannot compare more than 3 methods |
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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 of ordinal level | |
THIS TABLE IS YET TO BE COMPLETED | Null hypothesis | 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. | H0: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis. | |
n.a. | Alternative hypothesis | Alternative hypothesis | |
- | H1 two sided: $\pi \neq \pi_0$ H1 right sided: $\pi > \pi_0$ H1 left sided: $\pi < \pi_0$ | H1 two sided: $m \neq m_0$ H1 right sided: $m > m_0$ H1 left sided: $m < m_0$ | |
n.a. | Assumptions | Assumptions | |
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n.a. | Test statistic | 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. | Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic.
In order to compute each of the test statistics, follow the steps below:
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n.a. | Sampling distribution of $z$ if H0 were true | Sampling distribution of $W_1$ and of $W_2$ if H0 were true | |
- | Approximately the standard normal distribution | Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true. Sampling distribution of $W_2$: If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true. If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used. Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated. | |
n.a. | Significant? | Significant? | |
- | Two sided:
| For large samples, the table for standard normal probabilities can be used: Two sided:
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n.a. | Approximate $C\%$ confidence interval for $\pi$ | n.a. | |
- | Regular (large sample):
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n.a. | Equivalent to | n.a. | |
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n.a. | Example context | 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. | Is the median mental health score of office workers different from $m_0 = 50$? | |
n.a. | SPSS | SPSS | |
- | Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
| Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to:
Analyze > Nonparametric Tests > One Sample...
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n.a. | Jamovi | Jamovi | |
- | Frequencies > 2 Outcomes - Binomial test
| T-Tests > One Sample T-Test
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Practice questions | Practice questions | Practice questions | |