Question

A random sample of 16 ATM transactions at the Last National Bank of Flat Rock revealed a mean transaction time of 2.8 minutes with a standard deviation of 1.2 minutes. The width (in minutes) of the 95 percent confidence interval for the true mean transaction time is plus/minus (to three decimal places):

Answer 1

Answer:

The  width is  $W = 1.2786$

Step-by-step explanation:

From the question we are told that

    The  sample size is  n = 16    

      The mean time is  $\= x = 2.8 \ minute$    

     The standard deviation is  $s = 1.2 \ minutes$

Generally the degree of freedom is mathematically represented as  

         $df = n - 1$

=>      $df = 16 - 1$

=>     $df = 15$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the t distribution table the critical value  of  $\frac{\alpha }{2}$ at a degree of freedom of  df = 15 is  

   $t_{\frac{\alpha }{2} , 15 } =  2.131$

Generally the margin of error is mathematically represented as  

      $E = t_{\frac{\alpha }{2}, 15 } *  \frac{\sigma }{\sqrt{n} }$

=>    $E = 2.131  *  \frac{ 1.2 }{\sqrt{16} }$

=>    $E = 0.6393$

Generally the width the confidence 95% confidence interval  

       $W = 2 * E$

=>    $W = 2 * 0.6393$

=>    $W = 1.2786$