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

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 :

, 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) =

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

hi,

6x times 4 is 24x and 5 times 4 is 20, so the answer is 24x+20.

hope this and for a little humor i am groot.