Question

Suppose a certain species bird has an average weight of mu = 3.55 grams. based on previous studies, we can assume that the weights of these birds have a normal distribution with sigma = 0.28 grams. for a small group of 17 birds, find a 80% confidence interval for the average weights of these birds.

Answer 1

Given:
μ = 3.55 g, population mean
σ = 0.28 g, population standard deviation
n = 17, sample size
80% confidence level

The interval for the mean at the 80% confidence level is

`\mu \pm z^(*) (\sigma)/( √(n) )`

From standard tables, obtain z* = 1.28.
Therefore the confidence interval is
3.55 +/- 1.28*(0.28/√17) = 3.55 +/- 0.0869 = (3.463, 3.637)

Answer:
The 80% confidence interval for the average weights of the birds is (3.463, 3.637) g.
That is, 3.463 g < AverageWeight < 3.637 g