Goodness of fit test  overview
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Goodness of fit test  Two sample $t$ test  equal variances assumed  One sample Wilcoxon signedrank test  $z$ test for the difference between two proportions 


Independent variable  Independent/grouping variable  Independent variable  Independent/grouping variable  
None  One categorical with 2 independent groups  None  One categorical with 2 independent groups  
Dependent variable  Dependent variable  Dependent variable  Dependent variable  
One categorical with $J$ independent groups ($J \geqslant 2$)  One quantitative of interval or ratio level  One of ordinal level  One categorical with 2 independent groups  
Null hypothesis  Null hypothesis  Null hypothesis  Null hypothesis  
 H_{0}: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.  H_{0}: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.  H_{0}: $\pi_1 = \pi_2$
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  Alternative hypothesis  
 H_{1} two sided: $\mu_1 \neq \mu_2$ H_{1} right sided: $\mu_1 > \mu_2$ H_{1} left sided: $\mu_1 < \mu_2$  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: $\pi_1 \neq \pi_2$ H_{1} right sided: $\pi_1 > \pi_2$ H_{1} left sided: $\pi_1 < \pi_2$  
Assumptions  Assumptions  Assumptions  Assumptions  



 
Test statistic  Test statistic  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.  $t = \dfrac{(\bar{y}_1  \bar{y}_2)  0}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}} = \dfrac{\bar{y}_1  \bar{y}_2}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s_p$ is the pooled standard deviation, $n_1$ 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 $s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}$ is the standard error of the sampling distribution of $\bar{y}_1  \bar{y}_2$. The $t$ value indicates how many standard errors $\bar{y}_1  \bar{y}_2$ 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 $\mu_2 > \mu_1$.  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:
 $z = \dfrac{p_1  p_2}{\sqrt{p(1  p)\Bigg(\dfrac{1}{n_1} + \dfrac{1}{n_2}\Bigg)}}$
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 $\pi_2 > \pi_1.$  
n.a.  Pooled standard deviation  n.a.  n.a.  
  $s_p = \sqrt{\dfrac{(n_1  1) \times s^2_1 + (n_2  1) \times s^2_2}{n_1 + n_2  2}}$      
Sampling distribution of $X^2$ if H_{0} were true  Sampling distribution of $t$ if H_{0} were true  Sampling distribution of $W_1$ and of $W_2$ if H_{0} were true  Sampling distribution of $z$ if H_{0} were true  
Approximately the chisquared distribution with $J  1$ degrees of freedom  $t$ distribution with $n_1 + n_2  2$ 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.  Approximately the standard normal distribution  
Significant?  Significant?  Significant?  Significant?  
 Two sided:
 For large samples, the table for standard normal probabilities can be used: Two sided:
 Two sided:
 
n.a.  $C\%$ confidence interval for $\mu_1  \mu_2$  n.a.  Approximate $C\%$ confidence interval for $\pi_1  \pi_2$  
  $(\bar{y}_1  \bar{y}_2) \pm t^* \times s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}$
where the critical value $t^*$ is the value under the $t_{n_1 + n_2  2}$ distribution with the area $C / 100$ between $t^*$ and $t^*$ (e.g. $t^*$ = 2.086 for a 95% confidence interval when df = 20). The confidence interval for $\mu_1  \mu_2$ can also be used as significance test.    Regular (large sample):
 
n.a.  Effect size  n.a.  n.a.  
  Cohen's $d$: Standardized difference between the mean in group $1$ and in group $2$: $$d = \frac{\bar{y}_1  \bar{y}_2}{s_p}$$ Cohen's $d$ indicates how many standard deviations $s_p$ the two sample means are removed from each other.      
n.a.  Visual representation  n.a.  n.a.  
      
n.a.  Equivalent to  n.a.  Equivalent to  
  One way ANOVA with an independent variable with 2 levels ($I$ = 2):
   When testing two sided: chisquared test for the relationship between two categorical variables, where both categorical variables have 2 levels.  
Example context  Example context  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 average mental health score different between men and women? Assume that in the population, the standard deviation of mental health scores is equal amongst men and women.  Is the median mental health score of office workers different from $m_0 = 50$?  Is the proportion of smokers different between men and women? Use the normal approximation for the sampling distribution of the test statistic.  
SPSS  SPSS  SPSS  SPSS  
Analyze > Nonparametric Tests > Legacy Dialogs > Chisquare...
 Analyze > Compare Means > IndependentSamples T Test...
 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...
 SPSS does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chisquared test instead. The $p$ value resulting from this chisquared test is equivalent to the two sided $p$ value that would have resulted from the $z$ test. Go to:
Analyze > Descriptive Statistics > Crosstabs...
 
Jamovi  Jamovi  Jamovi  Jamovi  
Frequencies > N Outcomes  $\chi^2$ Goodness of fit
 TTests > Independent Samples TTest
 TTests > One Sample TTest
 Jamovi does not have a specific option for the $z$ test for the difference between two proportions. However, you can do the chisquared test instead. The $p$ value resulting from this chisquared 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
 
Practice questions  Practice questions  Practice questions  Practice questions  