Statistics Question

Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther. Suppose a small group of 16 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.38 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. (Round your answers to two decimal places.)

lower limit

upper limit 

margin of error

(b) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.12 for the mean weights of the hummingbirds.

Answer 1

Since the weights are normally distributed and the population standard deviation is known:

CI = sample mean +/- z-critical * sigma/(sample size)**0.5

For 80% confidence, z-critical = 1.28

The sample size is 16

The mean = 3.15

Sigma (the standard deviation) = 0.38

Plug in those numbers to get the upper and lower limits of the CI

Margin of error = z-critical * sigma/(sample size)**0.5

sample size >= (z-critical * sigma/E)**2

Plug in values for z-critical, sigma and E to get sample size. Round up to next higher integer.