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Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.
H_{0}: $\pi = \pi_0$
Here $\pi$ is the population proportion of 'successes', and $\pi_0$ is the population proportion of successes according to the null hypothesis.
H_{0}: $\rho = \rho_0$
Here $\rho$ is the Pearson correlation in the population, and $\rho_0$ is the Pearson correlation in the population according to the null hypothesis (usually 0). The Pearson correlation is a measure for the strength and direction of the linear relationship between two variables of at least interval measurement level.
Alternative hypothesis
Alternative hypothesis
Alternative hypothesis
H_{1} two sided: $m \neq m_0$
H_{1} right sided: $m > m_0$
H_{1} left sided: $m < m_0$
H_{1} two sided: $\pi \neq \pi_0$
H_{1} right sided: $\pi > \pi_0$
H_{1} left sided: $\pi < \pi_0$
H_{1} two sided: $\rho \neq \rho_0$
H_{1} right sided: $\rho > \rho_0$
H_{1} left sided: $\rho < \rho_0$
Assumptions
Assumptions
Assumptions of test for correlation
The population distribution of the scores is symmetric
Sample is a simple random sample from the population. That is, observations are independent of one another
Sample is a simple random sample from the population. That is, observations are independent of one another
In the population, the two variables are jointly normally distributed (this covers the normality, homoscedasticity, and linearity assumptions)
Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Note: these assumptions are only important for the significance test and confidence interval, not for the correlation coefficient itself. The correlation coefficient just measures the strength of the linear relationship between two variables.
Test statistic
Test statistic
Test statistic
Two 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:
For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score} - m_0)$. 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.
For each subject, compute the absolute value of the difference score $|\mbox{score} - m_0|$.
Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
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.
$X$ = number of successes in the sample
Test statistic for testing H0: $\rho = 0$:
$t = \dfrac{r \times \sqrt{N - 2}}{\sqrt{1 - r^2}} $
where $r$ is the sample correlation $r = \frac{1}{N - 1} \sum_{j}\Big(\frac{x_{j} - \bar{x}}{s_x} \Big) \Big(\frac{y_{j} - \bar{y}}{s_y} \Big)$ and $N$ is the sample size
Test statistic for testing values for $\rho$ other than $\rho = 0$:
$r_{Fisher} = \dfrac{1}{2} \times \log\Bigg(\dfrac{1 + r}{1 - r} \Bigg )$, where $r$ is the sample correlation
$\rho_{0_{Fisher}} = \dfrac{1}{2} \times \log\Bigg( \dfrac{1 + \rho_0}{1 - \rho_0} \Bigg )$, where $\rho_0$ is the population correlation according to H0
Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true
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.
where $r_{Fisher} = \frac{1}{2} \times \log\Bigg(\dfrac{1 + r}{1 - r} \Bigg )$ 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).
Then transform back to get the approximate $C$% confidence interval for $\rho$:
The Pearson correlation coefficient is a measure for the linear relationship between two quantitative variables.
The Pearson correlation coefficient squared reflects the proportion of variance explained in one variable by the other variable.
The Pearson correlation coefficient can take on values between -1 (perfect negative relationship) and 1 (perfect positive relationship). A value of 0 means no linear relationship.
The absolute size of the Pearson correlation coefficient is not affected by any linear transformation of the variables. However, the sign of the Pearson correlation will flip when the scores on one of the two variables are multiplied by a negative number (reversing the direction of measurement of that variable). For example:
the correlation between $x$ and $y$ is equivalent to the correlation between $3x + 5$ and $2y - 6$.
the absolute value of the correlation between $x$ and $y$ is equivalent to the absolute value of the correlation between $-3x + 5$ and $2y - 6$. However, the signs of the two correlation coefficients will be in opposite directions, due to the multiplication of $x$ by $-3$.
The Pearson correlation coefficient does not say anything about causality.
The Pearson correlation coefficient is sensitive to outliers.
Results significance test ($t$ and $p$ value) testing $H_0$: $\beta_1 = 0$ are equivalent to results significance test testing $H_0$: $\rho = 0$
Example context
Example context
Example context
Is the median mental health score of office workers different from $m_0 = 50$?
Is the proportion of smokers amongst office workers different from $\pi_0 = 0.2$?
Is there a linear relationship between physical health and mental health?
SPSS
SPSS
SPSS
Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to:
Analyze > Nonparametric Tests > One Sample...
On the Objective tab, choose Customize Analysis
On the Fields tab, specify the variable for which you want to compute the Wilcoxon signed-rank test
On the Settings tab, choose Customize tests and check the box for 'Compare median to hypothesized (Wilcoxon signed-rank test)'. Fill in your $m_0$ in the box next to Hypothesized median
Click Run
Double click on the output table to see the full results