Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 20 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 = 0.30 gram.

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

b.) 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

c.) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.14 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Answer 1

Solution :

a) Z/2 = Z_0.10 = 1.28

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

E = 1.28 * (0.30 /  20 )

E = 0.09

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

  ± E

.3.15 ± 0.09

( 3.06, 3.24 )  

lower limit = 3.06

upper limit = 3.24

margin of error = 0.09

c) margin of error = E = 0.14

sample size = n = [Z_/2* / E] ²

n = [1.28 * 0.30 / 0.14]²

n = 7.52

Sample size = n = 8  hummingbirds