Question

DETAILS BBUNDERSTAT12 7.1.015. ummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther. Suppose a small group of 11 Allen's hummingbirds has been under study in Artrona studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with = 0.26 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 wer limit pper limit margin of error What conditions are necessary for your calculations? (Select all that apply.) On is large is known normal distribution of weights uniform distribution of weights O is unknown 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. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval. We are 20% confident that the true average weight of Allen's hummingbirds falls within this interval. ind the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.12 for the mean weights of the hummingbirds. (Round hummingbirds

Answer 1

solutions - aj so l. contidence interval = X I 2012 GILE 3.15 + 1.2828 0.26 lower limit 3.05 upper limit 3.25 0.10 Margim of error by is known B a normal distribution of weight. average с we are soy confident true weight falls within this range. ME = 2012 0.12 = 1.282 x 0.26 m 7.72 so 8 sample size =