Question

Apps P Gmad YouTube Maps p li 015 Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bll Alther.t 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 a0.34 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.) uniform distribution of weights B a is known normal distribution of weights n is large a is unknows (c) Interpret your results in the context of this problem. O The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20 O 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. O The probability to the true average weight of Allen's hummingbirds is equal tb the sample mean O 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. O The probability that this interval contains the true average veight of Allen's hummingbirds is 0.80. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E 0.06 for the mean weights of the hummingbirds. (Round up to the nearest vwhole number.) hummingbirds Wasch Talk to a Tuder Need Help? Read It

Answer 1

Solution: Given that x = 3.15, σ = 0.34, n = 17
80% Confidence interval for Z = 1.28

(a) 80% Confidence interval for the average weights of Allen's hummingbirds : X +/- Z*σ/sqrt(n)
= 3.15 +/- 1.28*0.34/sqrt(17)
= (3.04 , 3.26) rounded

Lower limit = 3.04

upper limit = 3.26

Margin of error = 0.11

(b) option B,C

(c) option D.

(d) E = 0.06, Z = 1.28, σ = 0.34

sample n = (Z*σ/E)^2 = (1.28*0.34/0.06)^2 = 52.61 = 53

n = 53