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

Answer 1

Answer:

The margin of error is  $E = 1.96 * \frac{ 0.92}{\sqrt{28 } }$

Step-by-step explanation:

From the question we are told that

    The sample size is  $n = 28$

     The  sample  mean is  $\= x = 2.4 \ hr$

      The  standard deviation is  $\sigma = 0.92 \ hr$

     

Given that the confidence level is 95% the the level of significance can be evaluated as

             $\alpha = 100 -95$

            $\alpha = 5 \%$

             $\alpha = 0.05$

Next we obtain the critical value of  $\frac{\alpha }{2}$ from the normal distribution table,the value is  $Z_{\frac{\alpha }{2} } = Z_{\frac{0.05}{2} } = 1.96$

Generally the margin of error is mathematically represented as

           $E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

          $E = 1.96 * \frac{ 0.92}{\sqrt{28 } }$

         $E = 0.3408$