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 = 0.26 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 = 0.06 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) hummingbirds

Answer 1

Solution :

Given that,

= 3.15

= 0.26

n = 14

At 80% confidence level the z is ,

  = 1 - 80% = 1 - 0.80 = 0.2

/ 2 = 0.2 / 2 = 0.1

Z/2 = Z0.1 = 1.282

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

= 1.282 * (0.26 / 14)

= 0.09

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

- E < < + E

3.15 - 0.09 < < 3.15 + 0.09

3.06 < < 3.24

(3.06,3.24)

lower limit =3.06

upper limit =3.24

Margin of error = 0.09

d)

Population standard deviation = = 0.26

Margin of error = E = 0.06

Z_/2 = 1.282

sample size = n = [Z_/2* / E] ²

n = [1.282 * 0.26 / 0.06]²

n = 30.86

Sample size = n = 31