Question

A sample of 29 boxes of cereal has a sample standard deviation of 0.81 ounces. Construct a 95% confidence interval to estimate the true standard deviation of the filling process for the boxes of cereal.
a. ( 0.427,1.144)
b. (0.515,1.105)
c. (0.726,0.726)

Answer 1

Answer:

The 95% confidence interval is   $0.6428 < \sigma < 1.0954$

Step-by-step explanation:

From the question we are told that

   The sample size is  n = 29

    The sample standard deviation is $s = 0.81$

  Generally the 95% confidence interval to estimate the true standard deviation  is mathematically represented as  

                      $\sqrt{\frac{(n- 1)s^2 }{ X^2_( \frac{\alpha }{2 } , n-1)} } < \sigma < \sqrt{\frac{(n- 1)s^2 }{ X^2_(1- \frac{\alpha }{2 } , n-1)} }$

Generally from the  chi -distribution table  the critical value of $\frac{\alpha }{2}$ at a degree of freedom of  $df = 29 - 1 = 28$  is  

            $X^2 _{\frac{\alpha }{2} , 28 } =44.46079184$

Generally from the  chi -distribution table  the critical value of $1 - \frac{\alpha }{2}$ at a degree of freedom of  $df = 29 - 1 = 28$  is  

             $X^2 _{(1 - \frac{\alpha }{2}) , 28 } = 15.30786055$

So

                 $\sqrt{\frac{(29- 1)0.81^2 }{ 44.46079184 } } < \sigma < \sqrt{\frac{(29- 1)0.81^2 }{15.315.307860551 }$

=>              $0.6428 < \sigma < 1.0954$