Question

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 1

Answer: 52

Explanation:

Formula for sample size:-

`n= ((z_(\alpha/2\cdot \sigma))/(E))^2`

, where
`\sigma` = population standard deviation.

`z_(\alpha/2)` = Two -tailed z-value for
`{\alpha` (significance level)

E= margin of error.

Given : tex]\sigma=\text{ 0.30 ounce}[/tex]

⇒Significance level for 99.6% confidence level :
`\alpha=1-0.996=0.004`

By using z-value table ,Two -tailed z-value for
`\alpha=0.01`:

`z_(\alpha/2)=z_(0.002)=2.878`

E= 0.12 ounces.

Minimum sample size will be :-

`n= ((2.878\cdot 0.30)/(0.12))^2\\\\= (7.195)^2\\\\=51.768025\approx52`

Hence, the minimum sample size for the number of packages the inspector must select = 52