ANCOVA - overview
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ANCOVA | $z$ test for the difference between two proportions | McNemar's test |
You cannot compare more than 3 methods |
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Independent variables | Independent/grouping variable | Independent variable | |
One or more categorical with independent groups, and one or more quantitative control variables of interval or ratio level (covariates) | One categorical with 2 independent groups | 2 paired groups | |
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 | |
THIS TABLE IS YET TO BE COMPLETED | Null hypothesis | Null hypothesis | |
- | 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. | Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options:
Other formulations of the null hypothesis are:
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n.a. | Alternative hypothesis | Alternative hypothesis | |
- | H1 two sided: $\pi_1 \neq \pi_2$ H1 right sided: $\pi_1 > \pi_2$ H1 left sided: $\pi_1 < \pi_2$ | The alternative hypothesis H1 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0. Other formulations of the alternative hypothesis are:
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n.a. | Assumptions | Assumptions | |
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n.a. | Test statistic | Test statistic | |
- | $z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
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 $\pi_2 > \pi_1.$ | $X^2 = \dfrac{(b - c)^2}{b + c}$
Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0. | |
n.a. | Sampling distribution of $z$ if H0 were true | Sampling distribution of $X^2$ if H0 were true | |
- | Approximately the standard normal distribution | If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom. If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$. | |
n.a. | Significant? | Significant? | |
- | Two sided:
| For test statistic $X^2$:
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n.a. | Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ | n.a. | |
- | Regular (large sample):
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n.a. | Equivalent to | Equivalent to | |
- | When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels. |
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n.a. | Example context | Example context | |
- | Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic. | Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders? | |
n.a. | SPSS | SPSS | |
- | 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...
| Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
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n.a. | Jamovi | Jamovi | |
- | 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
| Frequencies > Paired Samples - McNemar test
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Practice questions | Practice questions | Practice questions | |