Question

You measure 39 turtles' weights, and find they have a mean weight of 70 ounces. Assume the population standard deviation is 2.5 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight.

Answer 1

Answer:Therefore, the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight is approximately 0.725 ounces.

Explanation:

The formula for the margin of error for a 90% confidence interval is:

margin of error = z*(sigma/sqrt(n))

where:

z is the z-score associated with the level of confidence (90% in this case)

sigma is the population standard deviation

n is the sample size

We can find the value of z using a z-score table or calculator. For a 90% confidence interval, the z-score is approximately 1.645.

Substituting the given values, we get:

margin of error = 1.645*(2.5/sqrt(39))

= 0.7246

Therefore, the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight is approximately 0.725 ounces.