Question

A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the cable can hold. The weight limit will be reported on cable packaging. The engineers take a random sample of 48 cables and apply weights to each of them until they break. The 48 cables have a mean breaking weight of 773 lb. The standard deviation of the breaking weight for the sample is 16.1 lb. Find the 95% confidence interval to estimate the mean breaking weight for this type cable.

Answer 1

Answer:

The 95%   confidence interval is   $768.44 < \mu$

Step-by-step explanation:

From the question we are told that  

    The  sample size is n = 48

    The sample mean is  $\= x = 773 \ lb$

     The standard deviation is  $\sigma = 16.1 \ lb$

   

Now given that the confidence level is  95% , then the level of significance is mathematically represented as

          $\alpha = 100 - 95$

          $\alpha = 5 \%$

          $\alpha = 0.05$

Next we obtain the  critical value of  $\frac{\alpha }{2}$ from the normal distribution table , the value is

              $Z_{\frac{\alpha }{2} } = 1.96$

The reason we are obtaining critical values of   $\frac{\alpha }{2}$ instead of    $\alpha$ is because  

$\alpha$ represents the area under the normal curve where the confidence level interval (   $1-\alpha$ ) did not cover which include both the left and right tail while

$\frac{\alpha }{2}$is just the area of one tail which what we required to calculate the margin of error

 The  margin of error is mathematically represented as

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

substituting values

     $MOE = 1.96 * \frac{ 16.1 }{ \sqrt{48} }$

     $MOE = 4.555$

The 95% confidence interval to estimate the mean breaking weight for this type cable is mathematically evaluated as

      $\= x - MOE < \mu < \= x - MOE$

substituting values

    $773 - 4.555 < \mu < 773 + 4.555$

     $768.44 < \mu$