Question

You measure 44 watermelons' weights, and find they have an average weight of 54 grams. Assume the population standard deviation is 4.8 grams. Based on this, what is the margin of error associated with a 99% confidence interval for the true population mean watermelon weight. The sample size is sufficiently large that the normal distribution value can be used in deriving the margin of error. Give your answer as a decimal, to two places

Answer 1

Solution :

Given that,

Z_/2 = 2.576

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

= 2.576 * (4.8 / 44)

= 1.86

At 99% confidence interval estimate of the population mean is,

- E < < + E

54 - 1.86 < < 54 + 1.86

52.14 < < 55.86

(52.14 , 55.86)