Two sample t test  equal variances assumed  overview
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Two sample $t$ test  equal variances assumed  Sign test 


Independent/grouping variable  Independent variable  
One categorical with 2 independent groups  2 paired groups  
Dependent variable  Dependent variable  
One quantitative of interval or ratio level  One of ordinal level  
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. 
 
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: $\mu_1 < \mu_2$ 
 
Assumptions  Assumptions  

 
Test statistic  Test statistic  
$t = \dfrac{(\bar{y}_1  \bar{y}_2)  0}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}} = \dfrac{\bar{y}_1  \bar{y}_2}{s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}$
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 $\mu_2 > \mu_1$.  $W = $ number of difference scores that is larger than 0  
Pooled standard deviation  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 $t$ if H_{0} were true  Sampling distribution of $W$ if H_{0} were true  
$t$ distribution with $n_1 + n_2  2$ degrees of freedom  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 $\sqrt{nP(1P)} = \sqrt{n \times 0.5(1  0.5)}$. 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.  
Significant?  Significant?  
Two sided:
 If $n$ is small, the table for the binomial distribution should be used: Two sided:
If $n$ is large, the table for standard normal probabilities can be used: Two sided:
 
$C\%$ confidence interval for $\mu_1  \mu_2$  n.a.  
$(\bar{y}_1  \bar{y}_2) \pm t^* \times s_p\sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}$
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.    
Effect size  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.    
Visual representation  n.a.  
  
Equivalent to  Equivalent to  
One way ANOVA with an independent variable with 2 levels ($I$ = 2):

Two sided sign test is equivalent to
 
Example context  Example context  
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.  Do people tend to score higher on mental health after a mindfulness course?  
SPSS  SPSS  
Analyze > Compare Means > IndependentSamples T Test...
 Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples...
 
Jamovi  Jamovi  
TTests > Independent Samples TTest
 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
 
Practice questions  Practice questions  