A quality inspector must verify whether a machine that packages snack foods is working correctly. The inspector will randomly select a sample of packages and weigh the amount of snack food in each. Assume that the weights of food in packages filled by this machine have a standard deviation of 0.30 ounce. An estimate of the mean amount of snack food in each package must be reported with 99.6% confidence and a margin of error of no more than 0.12 ounces. What would be the minimum sample size for the number of packages the inspector must select?

Answer: 52Step-by-step explanation:Formula for sample size:-[tex]n= (\dfrac{z_{\alpha/2\cdot \sigma}}{E})^2[/tex], where [tex]\sigma[/tex] = population standard deviation.[tex]z_{\alpha/2}[/tex] = Two -tailed z-value for [tex]{\alpha[/tex] (significance level) E= margin of error.Given : tex]\sigma=\text{ 0.30 ounce}[/tex]⇒Significance level for 99.6% confidence level :[tex]\alpha=1-0.996=0.004[/tex]By using z-value table ,Two -tailed z-value for [tex]\alpha=0.01 [/tex]:[tex]z_{\alpha/2}=z_{0.002}=2.878[/tex]E= 0.12 ounces.Minimum sample size will be :-[tex]n= (\dfrac{2.878\cdot 0.30}{0.12})^2\\\\= (7.195)^2\\\\=51.768025\approx52[/tex]Hence, the minimum sample size  for the number of packages the inspector must select = 52