Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 19 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 ? = 0.22 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. (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 error of estimate E = 0.11 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
________ hummingbirds

Answer 1

Solution:
Given in the question
N= 19 so degree of freedom = 18
Sample mean = 3.15
Population standard deviation = 0.22
alpha= 0.2 and alpha/2 = 0.1
so talpha/2 at 0.1 and df=18 is 1.330
So confidence interval is
3.15+/-1.330*(0.22/sqrt(19))
3.15+/-1.330*0.22/4.3588
Lower limit=3.15 - 1.330*0.22/4.3588 = 3.15-0.067128 = 3.0828
Upper Limit = 3.15 + 1.330*0.22/4.3588 = 3.15+0.067128 = 3.2117128
Margin of error = 1.330*0.22/sqrt(19) = 0.067128
Solution(2):
Margin of error = Zalpha/2*0.22/sqrt(n)
0.11 = 1.28*0.22/sqrt(n)
sqrt(n)=1.28*0.22/0.11 =
n= 2.56*2.56 = 6.5536 or 7