Question

The Ball Corporation's beverage can manufacturing plant in Fort Atkinson, Wisconsin, uses a metal supplier that provides metal with a known thickness standard deviation o= .000965 mm. Assume a random sample of 58 sheets of metal resulted in an c = 3193 mm. Calculate the 98 percent confidence interval for the true mean metal thickness. (Round your answers to 4 decimal places.) The 98% confidence interval is from

Answer 1

Random Sample n = 58

Standard deviation $\sigma$ = 0.000965

Mean $\bar{x}$ = 0.3193

Confidence interval when $\sigma$ is known given by = $\bar{x}$ $\pm$$z_{\alpha /2}$ * $\sigma$ / $\sqrt{n}$

Where $z_{\alpha /2}$ is normal table value at $\alpha$ level of significance .

For $\alpha$ = 1 - 0.98 = 0.02

      $\alpha$ / 2 = 0.01

Value of $z_{\alpha /2}$ is 2.3263              ( can be found from Normal table or any other software like R)

by using R

{

> qnorm(1-0.02/2)         # values of $z_{\alpha /2}$
[1] 2.326348

}

To calculate 100(1-$\alpha$)% or 98 % percent confidence interval for true mean metal thickness is given by -

                            CI     = $\bar{x}$ $\pm$$z_{\alpha /2}$ * $\sigma$ / $\sqrt{n}$ = 0.3193 $\pm$ 2.3263 * 0.000965 / $\sqrt{58}$

                                    = 0.3193 $\pm$ 0.0002947671

Thus 98 % percent confidence interval for true mean metal thickness is - ( 0.3190 , 0.3196 ) { round to four decimals }

98 % percent confidence interval is from 0.3190 to 0.3196