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

Step-by-step explanation:

Formula for sample size:-

$n= (\dfrac{z_{\alpha/2\cdot \sigma}}{E})^2$

, 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 :-

$n= (\dfrac{2.878\cdot 0.30}{0.12})^2\\\\= (7.195)^2\\\\=51.768025\approx52$

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