Goodness of fit test  overview
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Goodness of fit test  Wilcoxon signedrank test 


Independent variable  Independent variable  
None  2 paired groups  
Dependent variable  Dependent variable  
One categorical with $J$ independent groups ($J \geqslant 2$)  One quantitative of interval or ratio level  
Null hypothesis  Null hypothesis  
 H_{0}: $m = 0$
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.  
Alternative hypothesis  Alternative hypothesis  
 H_{1} two sided: $m \neq 0$ H_{1} right sided: $m > 0$ H_{1} left sided: $m < 0$  
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 of the differences between the mental health scores before and after an intervention different from 0?  
SPSS  SPSS  
Analyze > Nonparametric Tests > Legacy Dialogs > Chisquare...
 Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 
Jamovi  Jamovi  
Frequencies > N Outcomes  $\chi^2$ Goodness of fit
 TTests > Paired Samples TTest
 
Practice questions  Practice questions  