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

Given $\alpha=0.05, df=n-1=100-1=99$ degrees of freedom.

The critical t-value is $t_c=1.984217.$

The margin of error is $ME=t_c\times\dfrac{s}{\sqrt{n}}=2.8.$

Then the standard error is

$$SE=\dfrac{s}{\sqrt{n}}=\dfrac{ME}{t_c}=\dfrac{2.8}{1.984217}=1.411$$