Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 20 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 = 0.22 gram. (a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.) (b) What conditions are necessary for your calculations? (Select all that apply.) -n is large -σ is known -σ is unknown -normal distribution of weights -uniform distribution of weights (c) Interpret your results in the context of this problem. -The probability to the true average weight of Allen's hummingbirds is equal to the sample mean. -There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. -There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. -The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. -The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) ______ hummingbirds

Answer 1

Solution:

Sample size = n = 20

The average weight for these birds is x = 3.15 grams.That is: Sample mean =

Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution,

with σ = 0.22 gram

Part a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?

Formula:

where

For c = 80% , find area = ( 1 + c) / 2 = ( 1+0.80) / 2 = 1.80 /2 = 0.90

Look in z table for area = 0.9000 or its closest area and find z value:

Area 0.8997 is closest to 0.9000 and it corresponds to 1.2 and 0.08

thus z = 1.28

Thus

The margin of error =

Thus

Part b) What conditions are necessary for your calculations?

σ is known

normal distribution of weights

Part c) Interpret your results in the context of this problem.

There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.

Part d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.15 for the mean weights of the hummingbirds.

hummingbirds