Im stuck on this problem.

Question

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.)

normal distribution of weightsn is largeuniform distribution of weightsσ is knownσ is unknown

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

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 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 to the true average weight of Allen's hummingbirds is equal to the sample mean.

The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) 

 hummingbirds

Answer 1

SE = 0.841621* 0.36/sqrt(15) = 0.078707934425954152...

lower = 3.15 - SE = 3.22807....

upper = 3.15 + SE= 3.071292...

(b) Normal Distribution

(c) 80% of the intervals contain the actual mean; So 80% certain the mean is in the interval

(d) 0.15 = 0.841621 * 0.36/sqrt(n)

sqrt(n) = 0.841621(0.36 / 0.15

n = 4.08

so 5