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You measure 39 backpacks' weights, and find they have a mean weight of 64 ounces. Assume the population standard deviation is 8.9 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean backpack weight. Give your answer as a decimal, to two places

User Mallioch
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Final answer:

The maximal margin of error associated with a 99% confidence interval can be calculated using the formula: ME = Z * (σ/√n), where Z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size. After calculation, the maximal margin of error comes to 1.62 ounces.

Step-by-step explanation:

The maximal margin of error (ME) associated with a 99% confidence interval can be calculated using the formula:

ME = Z * (σ/√n)

Where Z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size. In this case, the mean weight of the backpacks is 64 ounces, the population standard deviation is 8.9 ounces, and the sample size is 39.

First, we need to find the z-score for a 99% confidence level using a z-table or calculator. The z-score for a 99% confidence level is approximately 2.576.

Substituting the values into the formula:

ME = 2.576 * (8.9/√39) ≈ 1.61579

Therefore, the maximal margin of error associated with a 99% confidence interval for the true population mean backpack weight is approximately 1.62 ounces.

User Renan Lopes
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