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 = 0.34 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.)

lower limit
upper limit
margin of error

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

σ is unknownn is largeuniform distribution of weightsnormal distribution of weightsσ is known

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

We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval.

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

We are 20% confident that the true average weight of Allen's hummingbirds falls within this interval.

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 = 0.11 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
how many  hummingbirds

Answer 1

Average, = 3.15 grams

Standard deviation, = 0.34 grams

n = 16

The sampling distribution of the sample mean of weight's of Allen's hummingbirds follows Normal distribution with Mean = 3.15 grams and Standard deviation = 0.34/√16 = 0.085 grams

(a) Corresponding to 80% confidence interval, the critical z score = 1.282

Margin of error = z*SE = 1.282*0.085 = 0.109

Lower limit = 3.15 - 0.109 ≈ 3.04 grams

upper limit = 3.15 + 0.109 ≈ 3.26 grams

Margin of error = 0.11 grams

(b) The necessary conditions are:

Normal distribution of weights

is known

(c) We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval

(d) E = 0.11

-> 1.282*0.34/√n = 0.11

-> n = 15.7 ≈ 16