# ANCOVA: overview

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ANCOVA
$z$ test for the difference between two proportions
Independent variablesIndependent 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
Dependent variableDependent variable
One quantitative of interval or ratio levelOne categorical with 2 independent groups
THIS TABLE IS YET TO BE COMPLETEDNull hypothesis
-$\pi_1 = \pi_2$
$\pi_1$ is the unknown proportion of "successes" in population 1; $\pi_2$ is the unknown proportion of "successes" in population 2
n.a.Alternative hypothesis
-Two sided: $\pi_1 \neq \pi_2$
Right sided: $\pi_1 > \pi_2$
Left sided: $\pi_1 < \pi_2$
n.a.Assumptions
-
• Sample size is large enough for $z$ to be approximately normally distributed. Rule of thumb:
• Significance test: number of successes and number of failures are each 5 or more in both sample groups
• Regular (large sample) 90%, 95%, or 99% confidence interval: number of successes and number of failures are each 10 or more in both sample groups
• Plus four 90%, 95%, or 99% confidence interval: sample sizes of both groups are 5 or more
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
n.a.Test statistic
-$z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
$p_1$ 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, $n_2$ 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 $z$ if H0 were true
-Approximately standard normal
n.a.Significant?
-Two sided:
Right sided:
Left sided:
n.a.Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$
-Regular (large sample):
• $(p_1 - p_2) \pm z^* \times \sqrt{\dfrac{p_1(1 - p_1)}{n_1} + \dfrac{p_2(1 - p_2)}{n_2}}$
where $z^*$ 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:
• $(p_{1.plus} - p_{2.plus}) \pm z^* \times \sqrt{\dfrac{p_{1.plus}(1 - p_{1.plus})}{n_1 + 2} + \dfrac{p_{2.plus}(1 - p_{2.plus})}{n_2 + 2}}$
where $p_{1.plus} = \dfrac{X_1 + 1}{n_1 + 2}$, $p_{2.plus} = \dfrac{X_2 + 1}{n_2 + 2}$, and $z^*$ 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
-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
-Is the proportion smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.
n.a.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...
• Put your independent (grouping) variable in the box below Row(s), and your dependent variable in the box below Column(s)
• Click the Statistics... button, and click on the square in front of Chi-square
• Continue and click OK
n.a.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
• Put your independent (grouping) variable in the box below Rows, and your dependent variable in the box below Columns
Practice questionsPractice questions