Question

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The 90% confidence interval for the true average age of all students in the university is

Answer 1

Answer:

The 90%  confidence level is  $19.15< L < 20.85$

Step-by-step explanation:

From the question we are told that

     The sample size is  $n = 64$

     The mean age is  $\= x = 20 \ years$

      The standard deviation  is   $\sigma = 4 \ years$

 

Generally  the degree of freedom for this data set is mathematically represented as

        $df = n - 1$

substituting values

        $df = 64 - 1$

        $df = 63$

Given that the level of confidence is  90%  the significance level is mathematically evaluated as

          $\alpha = 100 - 90$

         $\alpha =$10 %  

         $\alpha = 0.10$

Now   $\frac{\alpha }{2} = \frac{0.10}{2} = 0.05$

Since we are considering a on tail experiment

The  critical value for half of  this significance level at the calculated  degree of freedom is obtained from the critical value table as

           $t_{df, \frac{ \alpha}{2} } = t_{63, 0.05 } = 1.669$

   The margin for error is mathematically represented as

          $MOE = t_{df , \frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values  

          $MOE = 1.699 * \frac{4 }{\sqrt{64} }$

         $MOE = 0.85$

he 90% confidence interval for the true average age of all students in the university is evaluated as follows

           $\= x - MOE < L < \= x + E$

substituting values  

         $20 - 0. 85 < L < 20 + 0.85$

         $19.15< L < 20.85$