SOLUTION: You measure 34 watermelons weights, and find they have a mean weight of 59 ounces. Assume the population standard deviation is 3.3 ounces. Based on this, what is the maximal marg

Algebra ->  Probability-and-statistics -> SOLUTION: You measure 34 watermelons weights, and find they have a mean weight of 59 ounces. Assume the population standard deviation is 3.3 ounces. Based on this, what is the maximal marg      Log On


   



Question 1182877: You measure 34 watermelons weights, and find they have a mean weight of 59 ounces.
Assume the population standard deviation is 3.3 ounces.
Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight.

Answer as decimal, to 2 places

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
the margin of error is z*sigma/sqrt(n)
=2.576*3.3/sqrt(34)
=1.46 oz.