Question

A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 55 specimens and counts the number of seeds in each. Use her sample results (mean = 15.1, standard deviation = 20.7) to find the 80% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Give a step by step breakdown.

80% C.I. = __(11.523,18.677)________________

Answer 1

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

From given data, we have

Xbar = 15.1

σ = 20.7

n = 55

Confidence level = 80%

Critical Z value = 1.28155

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 15.1 ± 1.28155*20.7/sqrt(55)

Confidence interval = 15.1 ± 1.28155*2.791187431

Confidence interval = 15.1 ± 3.5770

Lower limit = 15.1 - 3.5770 = 11.523

Upper limit = 15.1 + 3.5770 = 18.677

Confidence interval = (11.523, 18.677)