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Allen\'s hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alth

ID: 3338048 • Letter: A

Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.t suppose a small group of 19 Allen's hummingbirds has been under study in Arizona. The average weight for these birds lsx-3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with -0.20 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 liit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) o is known uniform distribution of weights is unknown normal distribution of weights n is large (c) Tnterpret 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 The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80 There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hur nmingbirds in this region O The probablity to the true average welght of Allen's hummingblrds is equal to the sample mean, There is an 80% chance that the interval is one o the intervals containing the true average weight of Allen's hummingbirds in this region. d Find the sample size necessary for an 80% confidence level with a maximal margin of error F 10 or the mean weights at the hummingbird Round ip , the nearest hole ilmter. hummingblrds

Explanation / Answer

solution=

A) given

x bar = 3.15

n = 19

s = 0.20

% = 80 %

stanadard error SE =

CI: -SE < u < + SE

B) normal distribution of weights and is known

C) there is an 80% chance that the interval is one of the intervals containing the true average weight allens hummingbird in this region

D) 80 % OF Z Score is 1.2816

N = ( 1.2816 * 0.2 / 0.10 )^2 = 6.57 = 7 hummingbirds

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