Question

Suppose you work in Ingle's seafood department. You are asked to find a 99% confidence interval of how many Black Tiger Shrimp a customer should expect to get in a pound. You sample 54 pounds of shrimp and obtain a mean of 50 shrimp per pound with σ = 15 per pound. What is this 99% confidence interval?

Answer 1

The 99% confidence interval for the average number of Black Tiger Shrimp per pound, based on a sample with a mean of 50 shrimp and a population standard deviation of 15, is approximately (44.75, 55.25).

Step-by-step explanation:

The student is asked to find a 99% confidence interval for the average number of Black Tiger Shrimp per pound based on a sample. Given a sample mean of 50 shrimp per pound and a population standard deviation (σ) of 15 per pound, along with a sample size (n) of 54, we can use the Z-distribution because the population standard deviation is known and the sample size is larger than 30.

To find the confidence interval, we use the formula:

1. Identify the appropriate Z-value for the 99% confidence level. The Z-value for a 99% confidence interval is typically 2.576 (you can find this value in Z-tables).
2. Compute the standard error of the mean (SEM) which is σ divided by the square root of n (σ/√n).
3. Calculate the margin of error by multiplying the Z-value with the SEM.
4. Find the lower and upper bounds of the confidence interval by subtracting and adding the margin of error from the sample mean, respectively.

Let's calculate:

• SEM = 15 / √54 ≈ 2.04
• Margin of error = 2.576 × 2.04 ≈ 5.25
• Lower bound = 50 - 5.25 ≈ 44.75
• Upper bound = 50 + 5.25 ≈ 55.25

Therefore, the confidence interval is approximately (44.75, 55.25). This means that we are 99% confident that the true mean number of Black Tiger Shrimp per pound lies between 44.75 and 55.25 shrimp.