Question

A researcher surveys middle-school students on their study habits. She finds that in a random sample of 28 middle-school students, the mean amount of time that they spend working on the computer each night is 2.4 hours with a standard deviation of 0.92 hours. She uses the sample statistics to compute a 95% confidence interval for the population mean - the the mean amount of time that all middle-school students spend working on the computer each night.

What is the margin of error for this confidence interval? Round to two decimal places.

Answer 1

We have given that,              
              
Sample mean =2.4      
Sample standard deviation =0.92  
Sample size =28      
Level of significance=0.05      
Degree of freedom =27      
              
t critical value is (by using t table)=   2.052      

$E= t*\frac{S}{\sqrt{n}}= 2.052 *(\frac { 0.92 } {\sqrt{ 28 }})=0.36$