Sign test - overview

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Sign test
Binomial test for a single proportion
Wilcoxon signed-rank test
Cochran's Q test
Marginal Homogeneity test / Stuart-Maxwell test
One way ANOVA
Independent variableIndependent variableIndependent variableIndependent/grouping variableIndependent variableIndependent/grouping variable
2 paired groupsNone2 paired groupsOne within subject factor ($\geq 2$ related groups)2 paired groupsOne categorical with $I$ independent groups ($I \geqslant 2$)
Dependent variableDependent variableDependent variableDependent variableDependent variableDependent variable
One of ordinal levelOne categorical with 2 independent groupsOne quantitative of interval or ratio levelOne categorical with 2 independent groupsOne categorical with $J$ independent groups ($J \geqslant 2$)One quantitative of interval or ratio level
Null hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesisNull hypothesis
  • H0: P(first score of a pair exceeds second score of a pair) = P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
  • H0: the population median of the difference scores is equal to zero
A difference score is the difference between the first score of a pair and the second score of a pair.
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.
H0: $m = 0$

Here $m$ is the population median of the difference scores. A difference score is the difference between the first score of a pair and the second score of a pair.

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: $\pi_1 = \pi_2 = \ldots = \pi_I$

Here $\pi_1$ is the population proportion of 'successes' for group 1, $\pi_2$ is the population proportion of 'successes' for group 2, and $\pi_I$ is the population proportion of 'successes' for group $I.$
H0: for each category $j$ of the dependent variable, $\pi_j$ for the first paired group = $\pi_j$ for the second paired group.

Here $\pi_j$ is the population proportion in category $j.$
ANOVA $F$ test:
  • H0: $\mu_1 = \mu_2 = \ldots = \mu_I$
    $\mu_1$ is the population mean for group 1; $\mu_2$ is the population mean for group 2; $\mu_I$ is the population mean for group $I$
$t$ Test for contrast:
  • H0: $\Psi = 0$
    $\Psi$ is the population contrast, defined as $\Psi = \sum a_i\mu_i$. Here $\mu_i$ is the population mean for group $i$ and $a_i$ is the coefficient for $\mu_i$. The coefficients $a_i$ sum to 0.
$t$ Test multiple comparisons:
  • H0: $\mu_g = \mu_h$
    $\mu_g$ is the population mean for group $g$; $\mu_h$ is the population mean for group $h$
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
  • H1 two sided: P(first score of a pair exceeds second score of a pair) $\neq$ P(second score of a pair exceeds first score of a pair)
  • H1 right sided: P(first score of a pair exceeds second score of a pair) > P(second score of a pair exceeds first score of a pair)
  • H1 left sided: P(first score of a pair exceeds second score of a pair) < P(second score of a pair exceeds first score of a pair)
If the dependent variable is measured on a continuous scale, this can also be formulated as:
  • H1 two sided: the population median of the difference scores is different from zero
  • H1 right sided: the population median of the difference scores is larger than zero
  • H1 left sided: the population median of the difference scores is smaller than zero
H1 two sided: $\pi \neq \pi_0$
H1 right sided: $\pi > \pi_0$
H1 left sided: $\pi < \pi_0$
H1 two sided: $m \neq 0$
H1 right sided: $m > 0$
H1 left sided: $m < 0$
H1: not all population proportions are equalH1: for some categories of the dependent variable, $\pi_j$ for the first paired group $\neq$ $\pi_j$ for the second paired group.ANOVA $F$ test:
  • H1: not all population means are equal
$t$ Test for contrast:
  • H1 two sided: $\Psi \neq 0$
  • H1 right sided: $\Psi > 0$
  • H1 left sided: $\Psi < 0$
$t$ Test multiple comparisons:
  • H1 - usually two sided: $\mu_g \neq \mu_h$
AssumptionsAssumptionsAssumptionsAssumptionsAssumptionsAssumptions
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
  • Sample is a simple random sample from the population. That is, observations are independent of one another
  • The population distribution of the difference scores is symmetric
  • Sample of difference scores is a simple random sample from the population of difference scores. That is, difference scores are independent of one another
Note: sometimes it considered sufficient for the data to be measured on an ordinal scale, rather than an interval or ratio scale. However, since the test statistic is based on ranked difference scores, we need to know whether a change in scores from, say, 6 to 7 is larger than/smaller than/equal to a change from 5 to 6. This is impossible to know for ordinal scales, since for these scales the size of the difference between values is meaningless.
  • Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another
  • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
  • Within each population, the scores on the dependent variable are normally distributed
  • The standard deviation of the scores on the dependent variable is the same in each of the populations: $\sigma_1 = \sigma_2 = \ldots = \sigma_I$
  • 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
Test statisticTest statisticTest statisticTest statisticTest statisticTest statistic
$W = $ number of difference scores that is larger than 0$X$ = number of successes in the sampleTwo different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic. In order to compute each of the test statistics, follow the steps below:
  1. For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score}_2 - \mbox{score}_1)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
  2. For each subject, compute the absolute value of the difference score $|\mbox{score}_2 - \mbox{score}_1|$.
  3. Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
  4. Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:

  • $W_1 = \sum\, R_d^{+}$
    or
    $W_1 = \sum\, R_d^{-}$
    That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
    • tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one.
    • If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip').
  • $W_2 = \sum\, \mbox{sign}_d \times R_d$
    That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
If a failure is scored as 0 and a success is scored as 1:

$Q = k(k - 1) \dfrac{\sum_{groups} \Big (\mbox{group total} - \frac{\mbox{grand total}}{k} \Big)^2}{\sum_{blocks} \mbox{block total} \times (k - \mbox{block total})}$

Here $k$ is the number of related groups (usually the number of repeated measurements), a group total is the sum of the scores in a group, a block total is the sum of the scores in a block (usually a subject), and the grand total is the sum of all the scores.

Before computing $Q$, first exclude blocks with equal scores in all $k$ groups.
Computing the test statistic is a bit complicated and involves matrix algebra. Unless you are following a technical course, you probably won't need to calculate it by hand.ANOVA $F$ test:
  • $\begin{aligned}[t] F &= \dfrac{\sum\nolimits_{subjects} (\mbox{subject's group mean} - \mbox{overall mean})^2 / (I - 1)}{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2 / (N - I)}\\ &= \dfrac{\mbox{sum of squares between} / \mbox{degrees of freedom between}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square between}}{\mbox{mean square error}} \end{aligned} $
    where $N$ is the total sample size, and $I$ is the number of groups.
    Note: mean square between is also known as mean square model, and mean square error is also known as mean square residual or mean square within.
$t$ Test for contrast:
  • $t = \dfrac{c}{s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}}$
    Here $c$ is the sample estimate of the population contrast $\Psi$: $c = \sum a_i\bar{y}_i$, with $\bar{y}_i$ the sample mean in group $i$. $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $a_i$ is the contrast coefficient for group $i$, and $n_i$ is the sample size of group $i$.
    Note that if the contrast compares only two group means with each other, this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). In that case the only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$t$ Test multiple comparisons:
  • $t = \dfrac{\bar{y}_g - \bar{y}_h}{s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}}$
    $\bar{y}_g$ is the sample mean in group $g$, $\bar{y}_h$ is the sample mean in group $h$, $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $n_g$ is the sample size of group $g$, and $n_h$ is the sample size of group $h$.
    Note that this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). The only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
n.a.n.a.n.a.n.a.n.a.Pooled standard deviation
-----$ \begin{aligned} s_p &= \sqrt{\dfrac{(n_1 - 1) \times s^2_1 + (n_2 - 1) \times s^2_2 + \ldots + (n_I - 1) \times s^2_I}{N - I}}\\ &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - I}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} $

Here $s^2_i$ is the variance in group $i.$
Sampling distribution of $W$ if H0 were trueSampling distribution of $X$ if H0 were trueSampling distribution of $W_1$ and of $W_2$ if H0 were trueSampling distribution of $Q$ if H0 were trueSampling distribution of the test statistic if H0 were trueSampling distribution of $F$ and of $t$ 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.
Binomial($n$, $P$) distribution.

Here $n = N$ (total sample size), and $P = \pi_0$ (population proportion according to the null hypothesis).
Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true.

If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.

Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.
If the number of blocks (usually the number of subjects) is large, approximately the chi-squared distribution with $k - 1$ degrees of freedomApproximately the chi-squared distribution with $J - 1$ degrees of freedomSampling distribution of $F$:
  • $F$ distribution with $I - 1$ (df between, numerator) and $N - I$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
  • $t$ distribution with $N - I$ degrees of freedom
Significant?Significant?Significant?Significant?Significant?Significant?
If $n$ is small, the table for the binomial distribution should be used:
Two sided:
  • Check if $W$ observed in sample is in the rejection region or
  • Find two sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Right sided:
  • Check if $W$ observed in sample is in the rejection region or
  • Find right sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$
Left sided:
  • Check if $W$ observed in sample is in the rejection region or
  • Find left sided $p$ value corresponding to observed $W$ and check if it is equal to or smaller than $\alpha$

If $n$ is large, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
Two sided:
  • Check if $X$ observed in sample is in the rejection region or
  • Find two sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Right sided:
  • Check if $X$ observed in sample is in the rejection region or
  • Find right sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
Left sided:
  • Check if $X$ observed in sample is in the rejection region or
  • Find left sided $p$ value corresponding to observed $X$ and check if it is equal to or smaller than $\alpha$
For large samples, the table for standard normal probabilities can be used:
Two sided: Right sided: Left sided:
If the number of blocks is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
  • 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 $\alpha$
If we denote the test statistic as $X^2$:
  • 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 $\alpha$
$F$ test:
  • Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
  • Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$ (e.g. .01 < $p$ < .025 when $F$ = 3.91, df between = 4, and df error = 20)

$t$ Test for contrast two sided: $t$ Test for contrast right sided: $t$ Test for contrast left sided:
$t$ Test multiple comparisons two sided:
  • Check if $t$ observed in sample is at least as extreme as critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
  • Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons right sided
  • Check if $t$ observed in sample is equal to or larger than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
  • Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons left sided
  • Check if $t$ observed in sample is equal to or smaller than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
  • Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
n.a.n.a.n.a.n.a.n.a.$C\%$ confidence interval for $\Psi$, for $\mu_g - \mu_h$, and for $\mu_i$
-----Confidence interval for $\Psi$ (contrast):
  • $c \pm t^* \times s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}$
    where the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for $\mu_g - \mu_h$ (multiple comparisons):
  • $(\bar{y}_g - \bar{y}_h) \pm t^{**} \times s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}$
    where $t^{**}$ depends upon $C$, degrees of freedom ($N - I$), and the multiple comparison procedure. If you do not want to apply a multiple comparison procedure, $t^{**} = t^* = $ the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$. Note that $n_g$ is the sample size of group $g$, $n_h$ is the sample size of group $h$, and $N$ is the total sample size, based on all the $I$ groups.
Confidence interval for single population mean $\mu_i$:
  • $\bar{y}_i \pm t^* \times \dfrac{s_p}{\sqrt{n_i}}$
    where $\bar{y}_i$ is the sample mean in group $i$, $n_i$ is the sample size of group $i$, and the critical value $t^*$ is the value under the $t_{N - I}$ distribution with the area $C / 100$ between $-t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). Note that $n_i$ is the sample size of group $i$, and $N$ is the total sample size, based on all the $I$ groups.
n.a.n.a.n.a.n.a.n.a.Effect size
-----
  • Proportion variance explained $\eta^2$ and $R^2$:
    Proportion variance of the dependent variable $y$ explained by the independent variable: $$ \begin{align} \eta^2 = R^2 &= \dfrac{\mbox{sum of squares between}}{\mbox{sum of squares total}} \end{align} $$ Only in one way ANOVA $\eta^2 = R^2.$ $\eta^2$ (and $R^2$) is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population.

  • Proportion variance explained $\omega^2$:
    Corrects for the positive bias in $\eta^2$ and is equal to: $$\omega^2 = \frac{\mbox{sum of squares between} - \mbox{df between} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}$$ $\omega^2$ is a better estimate of the explained variance in the population than $\eta^2.$

  • Cohen's $d$:
    Standardized difference between the mean in group $g$ and in group $h$: $$d_{g,h} = \frac{\bar{y}_g - \bar{y}_h}{s_p}$$ Cohen's $d$ indicates how many standard deviations $s_p$ two sample means are removed from each other.
n.a.n.a.n.a.n.a.n.a.ANOVA table
-----
ANOVA table

Click the link for a step by step explanation of how to compute the sum of squares.
Equivalent ton.a.n.a.Equivalent ton.a.Equivalent to
Two sided sign test is equivalent to --Friedman test, with a categorical dependent variable consisting of two independent groups.-OLS regression with one categorical independent variable transformed into $I - 1$ code variables:
  • $F$ test ANOVA is equivalent to $F$ test regression model
  • $t$ test for contrast $i$ is equivalent to $t$ test for regression coefficient $\beta_i$ (specific contrast tested depends on how the code variables are defined)
Example contextExample contextExample contextExample contextExample contextExample context
Do people tend to score higher on mental health after a mindfulness course?Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$?Is the median of the differences between the mental health scores before and after an intervention different from 0?Subjects perform three different tasks, which they can either perform correctly or incorrectly. Is there a difference in task performance between the three different tasks?Subjects are asked to taste three different types of mayonnaise, and to indicate which of the three types of mayonnaise they like best. They then have to drink a glass of beer, and taste and rate the three types of mayonnaise again. Does drinking a beer change which type of mayonnaise people like best?Is the average mental health score different between people from a low, moderate, and high economic class?
SPSSSPSSSPSSSPSSSPSSSPSS
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the Sign test
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
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the Wilcoxon test
Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
  • Put the $k$ variables containing the scores for the $k$ related groups in the white box below Test Variables
  • Under Test Type, select Cochran's Q test
Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
  • Put the two paired variables in the boxes below Variable 1 and Variable 2
  • Under Test Type, select the Marginal Homogeneity test
Analyze > Compare Means > One-Way ANOVA...
  • Put your dependent (quantitative) variable in the box below Dependent List and your independent (grouping) variable in the box below Factor
or
Analyze > General Linear Model > Univariate...
  • Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factor(s)
JamoviJamoviJamoviJamovin.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
  • Put the two paired variables in the box below Measures
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
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 Tests, select Wilcoxon rank
  • Under Hypothesis, select your alternative hypothesis
Jamovi does not have a specific option for the Cochran's Q test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the $p$ value that would have resulted from the Cochran's Q test. Go to:

ANOVA > Repeated Measures ANOVA - Friedman
  • Put the $k$ variables containing the scores for the $k$ related groups in the box below Measures
-ANOVA > ANOVA
  • Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factors
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