Question

You measure 20 turtles' weights, and find they have a mean weight of 45 ounces. Assume the population standard deviation is 11.5 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight. Give your answer as a decimal, to two places

Answer 1

Answer: the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight = 4.23

Step-by-step explanation:

Formula for margin of error : $E=\dfrac{z^*\times (Population\ standard\ deviation)}{\sqrt{Sample\ size}}$

, where z* = Critical z-value.

Given: population standard deviation = 11.5 ounces

Sample size = 20

Z value for 90% confidence level = 1.645

margin of error (E) = $\dfrac{1.645\times11.5}{\sqrt{20}}$

$\approx4.23$

Hence, the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight = 4.23