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 = 0.40 gram.

When finding an 80% confidence interval, what is the critical value for confidence level?

z_c =

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

lower limit upper limit margin of error

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

uniform distribution of weights

σ is unknown

n is large

σ is known

normal distribution of weights

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.

Answer 1

Solution :

Z/2 = Z0.10 = 1.28

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

E = 1.28 * (0.40 /  16 )

E = 0.13

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

  ± E

3.15 ± 0.13   

( 3.02, 3.28 )  

lower limit = 3.02

upper limit = 3.28

margin of error = 0.13

b) σ is known

normal distribution of weights  

c) margin of error = E = 0.06

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

n = [1.28 * 0.40 / 0.06]2

n = 72.81

Sample size = n = 73 hummingbirds.