ANCOVA - overview

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ANCOVA
$z$ test for a single proportion
Two way MANOVA
You cannot compare more than 3 methods
Independent variablesIndependent variableIndependent/grouping variables
One or more categorical with independent groups, and one or more quantitative control variables of interval or ratio level (covariates)NoneTwo categorical, the first with $I$ independent groups and the second with $J$ independent groups ($I \geqslant 2$, $J \geqslant 2$)
Dependent variableDependent variableDependent variables
One quantitative of interval or ratio levelOne categorical with 2 independent groupsTwo or more quantitative of interval or ratio level
THIS TABLE IS YET TO BE COMPLETEDNull hypothesisTHIS TABLE IS YET TO BE COMPLETED
-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.
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n.a.Alternative hypothesisn.a.
-H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
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n.a.Assumptionsn.a.
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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
If the sample size is too small for $z$ to be approximately normally distributed, the binomial test for a single proportion should be used.
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n.a.Test statisticn.a.
-$z = \dfrac{p - \pi_0}{\sqrt{\dfrac{\pi_0(1 - \pi_0)}{N}}}$
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.
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n.a.Sampling distribution of $z$ if H0 were truen.a.
-Approximately the standard normal distribution-
n.a.Significant?n.a.
-Two sided: Right sided: Left sided: -
n.a.Approximate $C\%$ confidence interval for $\pi$n.a.
-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)
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n.a.Equivalent ton.a.
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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.
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n.a.Example contextn.a.
-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.-
n.a.SPSSn.a.
-Analyze > Nonparametric Tests > Legacy Dialogs > Binomial...
  • 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
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n.a.Jamovin.a.
-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
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