Sign test - overview
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Sign test | Two sample $z$ test | $z$ test for the difference between two proportions |
You cannot compare more than 3 methods |
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Independent variable | Independent/grouping variable | Independent/grouping variable | |
2 paired groups | One categorical with 2 independent groups | One categorical with 2 independent groups | |
Dependent variable | Dependent variable | Dependent variable | |
One of ordinal level | One quantitative of interval or ratio level | One categorical with 2 independent groups | |
Null hypothesis | Null hypothesis | Null hypothesis | |
| H0: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2. | 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. | |
Alternative hypothesis | Alternative hypothesis | Alternative hypothesis | |
| H1 two sided: $\mu_1 \neq \mu_2$ H1 right sided: $\mu_1 > \mu_2$ H1 left sided: $\mu_1 < \mu_2$ | H1 two sided: $\pi_1 \neq \pi_2$ H1 right sided: $\pi_1 > \pi_2$ H1 left sided: $\pi_1 < \pi_2$ | |
Assumptions | Assumptions | Assumptions | |
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Test statistic | Test statistic | Test statistic | |
$W = $ number of difference scores that is larger than 0 | $z = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $\sigma^2_1$ is the population variance in population 1, $\sigma^2_2$ is the population variance in population 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis. The denominator $\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2}}$ is the standard deviation of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $z$ value indicates how many of these standard deviations $\bar{y}_1 - \bar{y}_2$ is removed from 0. Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$. | $z = \dfrac{p_1 - p_2}{\sqrt{p(1 - p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
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 $\pi_2 > \pi_1.$ | |
Sampling distribution of $W$ if H0 were true | Sampling distribution of $z$ if H0 were true | Sampling distribution of $z$ if H0 were true | |
The exact distribution of $W$ under the null hypothesis is the Binomial($n$, $P$) distribution, with $n =$ number of positive differences $+$ number of negative differences, and $P = 0.5$.
If $n$ is large, $W$ is approximately normally distributed under the null hypothesis, with mean $nP = n \times 0.5$ and standard deviation $\sqrt{nP(1-P)} = \sqrt{n \times 0.5(1 - 0.5)}$. Hence, if $n$ is large, the standardized test statistic $$z = \frac{W - n \times 0.5}{\sqrt{n \times 0.5(1 - 0.5)}}$$ follows approximately the standard normal distribution if the null hypothesis were true. | Standard normal distribution | Approximately the standard normal distribution | |
Significant? | Significant? | Significant? | |
If $n$ is small, the table for the binomial distribution should be used: Two sided:
If $n$ is large, the table for standard normal probabilities can be used: Two sided:
| Two sided:
| Two sided:
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n.a. | $C\%$ confidence interval for $\mu_1 - \mu_2$ | Approximate $C\%$ confidence interval for $\pi_1 - \pi_2$ | |
- | $(\bar{y}_1 - \bar{y}_2) \pm z^* \times \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}$
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). The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test. | Regular (large sample):
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n.a. | Visual representation | n.a. | |
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Equivalent to | n.a. | Equivalent to | |
Two sided sign test is equivalent to
| - | When testing two sided: chi-squared test for the relationship between two categorical variables, where both categorical variables have 2 levels. | |
Example context | Example context | Example context | |
Do people tend to score higher on mental health after a mindfulness course? | Is the average mental health score different between men and women? Assume that in the population, the standard devation of the mental health scores is $\sigma_1 = 2$ amongst men and $\sigma_2 = 2.5$ amongst women. | Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic. | |
SPSS | n.a. | SPSS | |
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
| - | 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...
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Jamovi | n.a. | Jamovi | |
Jamovi does not have a specific option for the sign test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the two sided $p$ value that would have resulted from the sign test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
| - | 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
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Practice questions | Practice questions | Practice questions | |