Question

9:38 1 LTE Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bll Alther.t Suppose a small group of 18 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-0.38 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.) (b) What conditions are necessary for your calculations? (Select all that apply.) O uniform distribution of weights oo is unknown o n is large OƠİS known o normal distribution of weights (c) Interpret your results in the context of this problem. O 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 O 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 ofthe intervals containing the true average weight of Allen's hummingbirds in this region There is a 20% chance that the interval s one of 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 E-0.16 for the mean weights of the hummingbirds. (Round up to the nearest whole number) Use the Student's t distribution to find te for a 0.95 confidence level when the sample is 22. (Round your answer to three decimall places.) Student's t distributions are symmetric about a value of t. What is that t value?

Answer 1

Solution:- Given that n = 18, x = 3.15 grams, σ = 0.38 grams

(a) 80% confidence interval for the average weights : X +/- σ*s/sqrt(n)
= 3.15 +/- 1.28*0.38/sqrt(18)
= 3.0354 , 3.2646
= 3.04, 3.26 (rounded)

(b) option b. σ is known ,

option e. normal distribution of weights

(c) option d.

(d) n = (Z*σ/E)^2 = (1.28*0.38/0.16)^2 = 9.2416 = 9

n = 9
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=> tc = 95% confidence interval n = 22 df = 21

t = 2.080