One sample z test for the mean  overview
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One sample $z$ test for the mean  One sample Wilcoxon signedrank test 


Independent variable  Independent variable  
None  None  
Dependent variable  Dependent variable  
One quantitative of interval or ratio level  One of ordinal level  
Null hypothesis  Null hypothesis  
$\mu = \mu_0$
$\mu$ is the unknown population mean; $\mu_0$ is the population mean according to the null hypothesis  $m = m_0$
$m$ is the unknown population median; $m_0$ is the population median according to the null hypothesis  
Alternative hypothesis  Alternative hypothesis  
Two sided: $\mu \neq \mu_0$ Right sided: $\mu > \mu_0$ Left sided: $\mu < \mu_0$ 
 
Assumptions  Assumptions  

 
Test statistic  Test statistic  
$z = \dfrac{\bar{y}  \mu_0}{\sigma / \sqrt{N}}$
$\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to H0, $\sigma$ is the population standard deviation, $N$ is the sample size. The denominator $\sigma / \sqrt{N}$ is the standard deviation of the sampling distribution of $\bar{y}$. The $z$ value indicates how many of these standard deviations $\bar{y}$ is removed from $\mu_0$  Two different types of test statistics can be used; 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 $z$ if H0 were true  Sampling distribution of $W_1$ and of $W_2$ if H0 were true  
Standard normal  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 a 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 a 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: the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated if ties are present in the data.  
Significant?  Significant?  
Two sided:
 For large samples, the table for standard normal probabilities can be used: Two sided:
 
$C\%$ confidence interval for $\mu$  n.a.  
$\bar{y} \pm z^* \times \dfrac{\sigma}{\sqrt{N}}$
where $z^*$ 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) The confidence interval for $\mu$ can also be used as significance test.    
Effect size  n.a.  
Cohen's $d$: Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y}  \mu_0}{\sigma}$$ Indicates how many standard deviations $\sigma$ the sample mean $\bar{y}$ is removed from $\mu_0$    
Visual representation  n.a.  
  
Example context  Example context  
Is the average mental health score of office workers different from $\mu_0$ = 50? Assume that the standard deviation of the mental health scores in the population is $\sigma$ = 3.  Is the median mental health score different from 50?  
n.a.  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...
 
n.a.  Jamovi  
  TTests > One Sample TTest
 
Practice questions  Practice questions  