Question

A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is the 95% confidence interval for the mean score, , of all students taking the test

Answer 1

Answer:

The 95% confidence interval for the mean score, , of all students taking the test is

        $28.37< L\ 30.63$

Step-by-step explanation:

From the question we are told that

    The sample size is $n = 59$

    The mean score is  $\= x = 29.5$

     The standard deviation $\sigma = 5.2$

Generally the standard deviation of mean is mathematically represented as

                $\sigma _{\= x} = \frac{\sigma }{\sqrt{n} }$

substituting values

               $\sigma _{\= x} = \frac{5.2 }{\sqrt{59} }$

             $\sigma _{\= x} = 0.677$

The degree of freedom is mathematically represented as

          $df = n - 1$

substituting values

        $df = 59 -1$

        $df = 58$

Given that the confidence interval is 95%  then the level of significance is mathematically represented as

         $\alpha = 100 -95$

        $\alpha =$5%

        $\alpha = 0.05$

Now the critical value at  this significance level and degree of freedom is

       $t_{df , \alpha } = t_{58, 0.05 } = 1.672$

Obtained from the critical value table  

    So the the 95% confidence interval for the mean score, , of all students taking the test is mathematically represented as

      $\= x - t*(\sigma_{\= x}) < L\ \= x + t*(\sigma_{\= x})$

substituting value

      $(29.5 - 1.672* 0.677) < L\ (29.5 + 1.672* 0.677)$

       $28.37< L\ 30.63$