> Secure l https://www.webassign.net/web/Student/Assignment-R., a | Allen\'s hum

Question

> Secure l https://www.webassign.net/web/Student/Assignment-R., a | Allen's hummingbird (Selasphorus sasin) has been studied by zoologist eill Alther 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 -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 two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your cakculations? (Select all that apply.) is unknown n is large normal distribution of weights uniform distribution of weights u 2 is known c) Interpret your results in the context of this problem There is an 80% chance that the interval a 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 There is a 20% chance that the interval a ona of the intervals containing the tee average weight of Allen's hummingbirds in this region The probablity that this interval centains te true avenge weight of Alan, hummingrats o.tc. The probability that this interval contains the true average weight of Alien's hummingbirds is 0.8o. sample mean. (d) Find the sample sae necessary for an B0% cenficence ere tha memal ragn s, ener E . om er the mean weights of the hummingbirds. (Round up to the nearest whole number) hummingbirds

Explanation / Answer

Solution:-

1) n = 17, X = 3.15, = 0.26

a) Z = 1.28

=> 80% confidence interval = X +/- Z*/sqrt(n)
= 3.15 +/- 1.28*0.26/sqrt(17)
= 3.15 +/- 0.08
= (3.07 , 3.23)

lower limit = 3.07
upper limit = 3.23
margin of error = 0.08

d) E = 0.13

=> n = (Z*s/E)^2 = (1.28*0.26/0.13)^2 = 6.5536 = 7

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2)

Solution:- n = 42, X = 37.5, = 8

a) Z = 2.58

=> 80% confidence interval = X +/- Z*/sqrt(n)
= 37.5 +/- 2.58*8/sqrt(42)
= 37.5 +/- 3.1848
= (34.32 , 40.68)

lower limit = 34.32
upper limit = 40.68
margin of error = 3.18

d) E = 2.7

=> n = (Z*s/E)^2 = (2.58*8/2.7)^2 = 58.4375 = 58

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