SOLUTION: Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.� Suppose a small group of 15 Allen's hummingbirds has been under study in Arizona. The average we

Question 1110607: Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.� Suppose a small group of 15 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 = 0.36 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
(b) What conditions are necessary for your calculations? (Select all that apply.)
σ is unknown
uniform distribution of weights
σ is known
n is large
normal distribution of weights

(c) Interpret your results in the context of this problem.
The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.
There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.
The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.
There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.
The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.
(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.10 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

hummingbirds
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The width of the interval is z(0.90)*sigma/sqrt(n)
this is 1.28*0.36/sqrt(15)=0.1190 or 0.12
The interval is (3.03, 3.27) units gms.
I need sigma is known and normal distribution. They go together, and when I have them, n can be anything, so it is not required to be large.
The second choice that there is an 80% chance the interval is one of those... I like to say that if I construct 100 similar intervals for this sample size, 80 of them will contain the parameter. I don't know which 80, so the probability can be used. The important concept is not that there is an 80% chance the mean is in this interval. It either is or it isn't, which is why we use confidence and not probability.
The last part requires 1.28*0.36/sqrt(n) to be < 0.1
0.2123<.01 n, squaring everything
21.23 is the sample size required, or n=22
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