Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 19 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 = 0.32 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 three decimal places.)

lower limit
upper limit
margin of error

(b) What conditions are necessary for your calculations? (Select all that apply.)

σ is known uniform distribution of weights normal distribution of weights n is large σ is unknown

(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.

The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.    

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.

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.

The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.

(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

Sample size = n = 19

Sample mean = = 3.15

Population standard deviation = = 0.32

a)

We have to construct 80% confidenc interval for the population mean.

Here population standard deviation is known so we have to use one sample z-confidence interval.

z confidence interval

Here E is a margin of error

Zc = 1.28    ( Using z table)

So confidence interval is ( 3.15 - 0.0940 , 3.15 + 0.0940) = > ( 3.056 , 3.244)

Lower limit: 3.056

Upper limit: 3.244

Margin of error: 0.094

b)

σ is known.

normal distribution of weights n is large.

c)

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.

d)

Confidence level = 0.80

Zc = 1.28    ( Using z table)

Margin of error = E = 0.15

Population standard deviation = = 0.32

We have to find sample size (n)