Question

You measure 42 watermelons' weights, and find they have a mean weight of 59 ounces. Assume the population standard deviation is 12 ounces. Based on this, what is the margin of error associated with a 95% confidence interval for the true population mean watermelon weight. Round your answer to two decimal places. 0.92 Incorrect ounces

Answer 1

Solution :

Given that,

= 12

n = 42

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (12 / 42)

= 3.63

Margin of error = E = 3.63