Question

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7. The manufacturer wants to be certain that the mean battery life exceeds 25 hours in accelerated lifetime testing.
Construct a 90%, two sided confidence interval on mean life in the accelerated test.

Answer 1

Answer:

The confidence interval is  $25.16 < \mu < 26.85$

Step-by-step explanation:

From  the question we are given a data set

     25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is  

       $\= x = \frac{\sum x_i}{n}$

where is the sample size with values  n =  10

         $\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}$

          $\= x = 26$

The standard deviation is evaluated as

             $\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }$

substituting values

     $= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }$

                  $\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }$

     $\sigma = 1.625$

The  confidence level is given as 90% hence the level of significance is calculated as

    $\alpha = 100 -90$

    $\alpha =10$%

   $\alpha = 0.10$

Now the critical values of $\frac{\alpha }{2}$ is obtained from the normal distribution table as

      $Z_{\frac{\alpha }{2} } = 1.645$

The reason  we are obtaining  the critical values of $\frac{\alpha }{2}$ instead of  $\alpha$ is because  we are considering two tails of the area under the normal curve

  The margin of error is evaluated as

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

substituting values

           $MOE = 1.645 * \frac{1.625 }{\sqrt{10} }$

           $MOE = 0.845$

The  90%, two sided confidence interval is mathematically evaluated as

           $\= x - MOE < \mu < \= x + MOE$

           $26 - 0.845 < \mu < 26 + 0.845$

           $25.16 < \mu < 26.85$

Given that the lower and the upper limit is greater than  25 then we can assure the manufactures  that the battery life exceeds 25 hours