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.

Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.16 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Answer 1

Solution :

Given that,

standard deviation = =0.22

Margin of error = E = 0.16

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 =0.10

Z/2 = Z0.10= 1.28 Using z table    )

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

n = ( 1.28*0.22 /0.16 )2

n =3.0976

Sample size = n =3