Standard deviation

Question

You measure 48 watermelons' weights, and find they have a mean weight of 76 ounces. Assume the population standard deviation is 7.1 ounces. Based on this, what is the margin of error associated with a 99% confidence interval for the true population mean watermelon weight.

Round your answer to 3 decimal places.

±___ ounces

Answer 1

Since we know population standard deviation we estimate the standard deviation of the mean (standard error) as :

se = 7.1/√48 or 1.02480. ounces

Using he formula for a confidence interval (and for generating a margin of error) assuming a normally distributed variable (Since population s.d. is known we use normal distribution instead of students t distribution.) we get the following.

moe = Z_crit * se

Z_crit is the Z value such that P(Z<x) = .995 (note: this is because alpha is always split for confidence intervals). The Z_crit is 2.5758.

Thus moe = ± 2.640 ounces. (to 3 significant digits after the decimal place)