Question

A veterinary tech weighs a sample of 205
adult male rats. The sample mean weight is 320.62
grams and the sample standard deviation is 64.75
grams. What is the margin of error with a 99.9%
confidence level (z=3.29
)?

Answer 1

To calculate the margin of error at a 99.9% confidence level, we use the formula:

Margin of error = z * (standard deviation / sqrt(sample size))

Given:
z = 3.29
sample standard deviation = 64.75 grams
sample size = 205

Margin of error = 3.29 * (64.75 / sqrt(205))
Margin of error = 3.29 * (64.75 / 14.32)
Margin of error = 3.29 * 4.526 = 14.90

Therefore, the margin of error at a 99.9% confidence level is 14.90 grams.