Question

a gigantic warehouse located in tampa, florida, stores approximately 60 million empty aluminum beer and soda cans. recently, a big fire occurred at the warehouse. the smoke from the fire contaminated many of the cans with blackspot, rendering them unusable. a local university statistician was hired by the insurance company to estimate p, the true proportion of cans in the warehouse that were contaminated by the fire. how many aluminum cans should be randomly sampled to estimate p to within .02 with 90% confidence?

Answer 1

To estimate the proportion of contaminated cans within a margin of error of 0.02 and 90% confidence, a sample size of at least 1693 cans is needed.

Step-by-step explanation:

To estimate p, the true proportion of cans in the warehouse that were contaminated by the fire, to within .02 with 90% confidence, we need to determine the appropriate sample size. The sample size formula for a proportion is:

n = (Z^2 * p * (1-p)) / E^2

Where:

• Z is the z-value corresponding to the desired confidence level
• p is the estimated proportion of the population that displays the attribute of interest
• E is the desired margin of error

Since p is not given, we use the most conservative estimate, which is 0.5. The Z-value for a 90% confidence level is 1.645. Plugging these values into the formula, we get:

n = (1.645^2 * 0.5 * (1-0.5)) / 0.02^2

n = (2.708025 * 0.25) / 0.0004

n = (0.67700625) / 0.0004

n = 1692.51

So, at least 1693 cans should be sampled to estimate the proportion p with a margin of error of 0.02 and 90% confidence.