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 = 0.32 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 two decimal places.)

lower limit ___

upper limit ____

margin of error___

(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

Solution :

Given that,

Z_/2 = 1.28

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

= 1.28 * (0.32 / 16)

= 0.10

Margin of error = 0.10

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

- E < < + E

3.15 - 0.10 < < 3.15 + 0.10

3.05 < < 3.25

Lower limit = 3.05

Upper limit = 3.25

(d)

Margin of error = E = 0.15

sample size = n = [Z_/2* / E] ²

n = [1.28 * 0.32 / 0.15]²

n = 7.5

Sample size = n = 8