Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 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 = 0.20 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

(b)

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

normal distribution of weightsn is largeuniform distribution of weightsσ is unknownσ is known

(c)

Interpret your results in the context of this problem.

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.     The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval.

(d)

Find the sample size necessary for an 80% confidence level with a maximal margin of error  E = 0.06 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Answer 1

Solution :

Z/2 = Z0.025 = 1.282

Margin of error = E = Z/2 * ( /n)

E = 1.282 * (0.20/  14 )

E = 0.07

At 80% confidence interval estimate of the population mean is,

  ± E

3.15 ± 0.07

( 3.08, 3.22)  

lower limit = 3.08

upper limit = 3.22

margin of error = 0.07

b) normal distribution of weights

σ is known  

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

d) margin of error = E = 0.06

sample size = n = [Z/2* / E] 2

n = [1.282 * 0.20 / 0.06]2

n = 18.26

Sample size = n = 19 hummingbirds