Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 12 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.

When finding an 80% confidence interval, what is the critical value for confidence level? (Give your answer to two decimal places.)

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

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

uniform distribution of weights

normal distribution of weights

n is large

σ is unknown

σ is known

Interpret your results in the context of this problem.

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.     

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.

Which equation is used to find the sample size n for estimating μ when σ is known?

n =

zσ σ

E

	2

n =

zσ E

σ

	2

     

n =

zσ σ E

n =

zσ E σ

Find the sample size necessary for an 80% confidence level with a maximal margin of error  E = 0.09 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Answer 1

Solution :

Z/2 = Z0.10 = 1.28

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

E = 1.28 * (0.22 /  12 )

E = 0.08

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

  ± E

3.15 ± 0.08   

( 3.07, 3.23)  

lower limit = 3.07

upper limit = 3.23

margin of error = 0.08

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 = 0.09

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

n = [1.28 * 0.22 / 0.09]2

n = 9.78

Sample size = n = 10 hummingbirds