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Solnce55 [7]
2 years ago
11

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 10 Allen's hummingb

irds 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 = 0.32 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.) σ is known uniform distribution of weights σ is unknown normal distribution of weights n is large (c) Interpret your results in the context of this problem. We are 20% confident that the true average weight of Allen's hummingbirds falls within this interval. We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. The probability that this interval contains the true average weight 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 = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) hummingbirds
Mathematics
1 answer:
ruslelena [56]2 years ago
6 0

Answer: provided in the explanation segment

Step-by-step explanation:

(a). from the question, we can see that since that б is known, we can use standard normal, z.

we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?

⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;

x ± z * б / √m

which is

3.15 ± 1.28 * 0.32/√10

= 3.15 ± 0.1295 = 3.0205 or 3.2795

(b). normal distribution of weight (c) б is known

(c). option (a) and (e) are correct

(d).  from the question, let sample size be given as S

this gives';

1.28 * 0.32/√S = 0.15

√S = (1.28 * 0.32) / 0.15 = 2.73

S = 7.4529

cheers i hope this helps

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