Question

The cost of spare parts per machine are normally distributed with a population standard deviation of 158 dollars and an unknown population mean. If a random sample of 23 machines is taken and results in a sample mean of 1790 dollars, find the error bound (EBM) of the confidence interval with a 98% confidence level

Answer 1

Answer:

The 98% confidence interval is $1713.23 <   \mu  <  1866.76$

Step-by-step explanation:

From the question we are told that

   The population standard deviation is  $\sigma = 158$

    The sample size is  n  =  23

    The sample mean is  $\= x = 1790 \ dollars$

From the question we are told the confidence level is  98% , hence the level of significance is    

      $\alpha = (100 - 98 ) \%$

=>   $\alpha = 0.02$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } =  2.33$

Generally the margin of error is mathematically represented as  

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

      $E = 2.33 *  \frac{158 }{\sqrt{23} }$

      $E =76.76$

Generally 98% confidence interval is mathematically represented as  

      $\= x  -E <   \mu  <  \= x  +E$

       $1790  -76.76 <  \mu  <  1790  +76.76  $

        $1713.23 <   \mu  <  1866.76$