Question

A newspaper article estimated that the mean cost for preventive dental care was $165 per year.
They indicated that the estimate was based on a sample of 100 people and that the margin of
error was 2.8. Assuming a 95% confidence level was used, calculate the value of the standard
error.

Answer 1

Let $M$ be the margin of error and $SE$ be the standard error, then m is given by,

$M=Z_{\alpha/2}*SE$ , Where $SE=\sigma/\sqrt{n}$

Given $M=2.8$ and $Z_{\alpha/2}=Z_{0.05/2}=Z_{0.025}=1.96$

Therefore, $SE=M/Z_{0.025}=2.8/1.96=1.4286(4\displaystyle\space decimal\displaystyle\space places).$

Standard error is 1.4286.