Standard deviation

Question

You measure 47 watermelons' weights, and find they have a mean weight of 71 ounces. Assume the population standard deviation is 11.2 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 3 decimal places.

±__ounces

Answer 1

given:

Xbar = 71 oz

n = 47

σ = 11.2 oz

α = (1 - .95) = .05

Z _α/2 = 1.95996

Calculate:

s_xbar (std error) ≅ 11.2/√47 = 1.32920

moe = ± Z _a/2 * s_{xbar or} ± 2.605 oz