Goodness of fit test  overview
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Goodness of fit test  One sample Wilcoxon signedrank test 


Independent variable  Independent variable  
None  None  
Dependent variable  Dependent variable  
One categorical with $J$ independent groups ($J \geqslant 2$)  One of ordinal level  
Null hypothesis  Null hypothesis  
 H_{0}: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.  
Alternative hypothesis  Alternative hypothesis  
 H_{1} two sided: $m \neq m_0$ H_{1} right sided: $m > m_0$ H_{1} left sided: $m < m_0$  
Assumptions  Assumptions  

 
Test statistic  Test statistic  
$X^2 = \sum{\frac{(\mbox{observed cell count}  \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
Here the expected cell count for one cell = $N \times \pi_j$, the observed cell count is the observed sample count in that same cell, and the sum is over all $J$ cells.  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:
 
Sampling distribution of $X^2$ if H_{0} were true  Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true  
Approximately the chisquared distribution with $J  1$ degrees of freedom  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.  
Significant?  Significant?  
 For large samples, the table for standard normal probabilities can be used: Two sided:
 
Example context  Example context  
Is the proportion of people with a low, moderate, and high social economic status in the population different from $\pi_{low} = 0.2,$ $\pi_{moderate} = 0.6,$ and $\pi_{high} = 0.2$?  Is the median mental health score of office workers different from $m_0 = 50$?  
SPSS  SPSS  
Analyze > Nonparametric Tests > Legacy Dialogs > Chisquare...
 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...
 
Jamovi  Jamovi  
Frequencies > N Outcomes  $\chi^2$ Goodness of fit
 TTests > One Sample TTest
 
Practice questions  Practice questions  