Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 17 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 = 0.20 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.)

(b) What conditions are necessary for your calculations?

(c) Interpret your results in the context of this problem.

(d) 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

Given

n=17

xbar=3.5

= 0.2

A) 80% confidence interval for population average weight

(xbar +/- z_{alpha /2}*​​​​​​)

(3.5 +/- z_0.1 *​​​​​​)

(3.5 +/- 1.2815*0.0485)

(3.5 +/- 0.0622)

(3.4378, 3.5622)

B) The necessary condition is sample is random and independent.

C)

Margine of error = E =0.16

We know that

E=Zalpha/2 *​​​​​​

n=(z_0.1)^2 (*()

n=(1.2815*1.2815)*(0.2/0.16)^2

n= 1.6422*1.5625

n= 2.56

n=3