z test for a single proportion - overview

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$z$ test for a single proportion
Kruskal-Wallis test
Paired sample $t$ test
You cannot compare more than 3 methods
Independent variableIndependent/grouping variableIndependent variable
NoneOne categorical with $I$ independent groups ($I \geqslant 2$)2 paired groups
Dependent variableDependent variableDependent variable
One categorical with 2 independent groupsOne of ordinal levelOne quantitative of interval or ratio level
Null hypothesisNull hypothesisNull 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.
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
  • H0: the population medians for the $I$ groups are equal
Else:
Formulation 1:
  • H0: the population scores in any of the $I$ groups are not systematically higher or lower than the population scores in any of the other groups
Formulation 2:
  • H0: P(an observation from population $g$ exceeds an observation from population $h$) = P(an observation from population $h$ exceeds an observation from population $g$), for each pair of groups.
Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
H0: $\mu = \mu_0$

Here $\mu$ is the population mean of the difference scores, and $\mu_0$ is the population mean of the difference scores according to the null hypothesis, which is usually 0. A difference score is the difference between the first score of a pair and the second score of a pair.
Alternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
If the dependent variable is measured on a continuous scale and the shape of the distribution of the dependent variable is the same in all $I$ populations:
  • H1: not all of the population medians for the $I$ groups are equal
Else:
Formulation 1:
  • H1: the poplation scores in some groups are systematically higher or lower than the population scores in other groups
Formulation 2:
  • H1: for at least one pair of groups:
    P(an observation from population $g$ exceeds an observation from population $h$) $\neq$ P(an observation from population $h$ exceeds an observation from population $g$)
H1 two sided: $\mu \neq \mu_0$
H1 right sided: $\mu > \mu_0$
H1 left sided: $\mu < \mu_0$
AssumptionsAssumptionsAssumptions
  • 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.
  • Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
  • Difference scores are normally distributed in the population
  • Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
Test statisticTest statisticTest 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.

$H = \dfrac{12}{N (N + 1)} \sum \dfrac{R^2_i}{n_i} - 3(N + 1)$

Here $N$ is the total sample size, $R_i$ is the sum of ranks in group $i$, and $n_i$ is the sample size of group $i$. Remember that multiplication precedes addition, so first compute $\frac{12}{N (N + 1)} \times \sum \frac{R^2_i}{n_i}$ and then subtract $3(N + 1)$.

Note: if ties are present in the data, the formula for $H$ is more complicated.
$t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$
Here $\bar{y}$ is the sample mean of the difference scores, $\mu_0$ is the population mean of the difference scores according to the null hypothesis, $s$ is the sample standard deviation of the difference scores, and $N$ is the sample size (number of difference scores).

The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$.
Sampling distribution of $z$ if H0 were trueSampling distribution of $H$ if H0 were trueSampling distribution of $t$ if H0 were true
Approximately the standard normal distribution

For large samples, approximately the chi-squared distribution with $I - 1$ degrees of freedom.

For small samples, the exact distribution of $H$ should be used.

$t$ distribution with $N - 1$ degrees of freedom
Significant?Significant?Significant?
Two sided: Right sided: Left sided: For large samples, the table with critical $X^2$ values can be used. If we denote $X^2 = H$:
  • Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
  • Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Two sided: Right sided: Left sided:
Approximate $C\%$ confidence interval for $\pi$n.a.$C\%$ confidence interval for $\mu$
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)
-$\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$
where the critical value $t^*$ is the value under the $t_{N-1}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20).

The confidence interval for $\mu$ can also be used as significance test.
n.a.n.a.Effect size
--Cohen's $d$:
Standardized difference between the sample mean of the difference scores and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Cohen's $d$ indicates how many standard deviations $s$ the sample mean of the difference scores $\bar{y}$ is removed from $\mu_0.$
n.a.n.a.Visual representation
--
Paired sample t test
Equivalent ton.a.Equivalent to
  • 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.
-
  • One sample $t$ test on the difference scores.
  • Repeated measures ANOVA with one dichotomous within subjects factor.
Example contextExample contextExample 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.Do people from different religions tend to score differently on social economic status? Is the average difference between the mental health scores before and after an intervention different from $\mu_0 = 0$?
SPSSSPSSSPSS
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
Analyze > Nonparametric Tests > Legacy Dialogs > K Independent Samples...
  • Put your dependent variable in the box below Test Variable List and your independent (grouping) variable in the box below Grouping Variable
  • Click on the Define Range... button. If you can't click on it, first click on the grouping variable so its background turns yellow
  • Fill in the smallest value you have used to indicate your groups in the box next to Minimum, and the largest value you have used to indicate your groups in the box next to Maximum
  • Continue and click OK
Analyze > Compare Means > Paired-Samples T Test...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
JamoviJamoviJamovi
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
ANOVA > One Way ANOVA - Kruskal-Wallis
  • Put your dependent variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable
T-Tests > Paired Samples T-Test
  • Put the two paired variables in the box below Paired Variables, one on the left side of the vertical line and one on the right side of the vertical line
  • Under Hypothesis, select your alternative hypothesis
Practice questionsPractice questionsPractice questions