Answer:
[tex] s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
And replacing we got:
[tex] s= 0.286[/tex]
And then the estimator for the standard error is given by:
[tex] SE= \frac{0.286}{\sqrt{4}}= 0.143[/tex]
Step-by-step explanation:
For this case we have the following dataset given:
20.05, 20.56, 20.72, and 20.43
We can assume that the distribution for the sample mean is given by:
[tex] \bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And the standard error for this case would be:
[tex] SE= \frac{\sigma}{\sqrt{n}}[/tex]
And we can estimate the deviation with the sample deviation:
[tex] s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
And replacing we got:
[tex] s= 0.286[/tex]
And then the estimator for the standard error is given by:
[tex] SE= \frac{0.286}{\sqrt{4}}= 0.143[/tex]
