Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 10 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.

Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.11 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

Answer 1

Solution :

Given that,

Population standard deviation = = 0.26

Margin of error = E = 0.11

At 80% confidence level

= 1-0.80% =1-0.80 =0.20

/2 =0.20/ 2= 0.10

Z/2 = Z_{0.10 = 1.960}

Z/2 = 1.960  

Z_/2 = 1.28

sample size = n = [Z_/2* / E] ²

n = [1.28 *0.26 /0.11 ]²

n = 9