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

Question

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 12 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 = 0.28 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.) is unknown is known n is large normal distribution of weights uniform distribution of weights (c) Interpret your results in the context of this problem. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. 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 hummingbirds in this region. There is an 80% chance that the interval is 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 sample mean. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.13 for the mean weights of the hummingbirds. (Round up to the nearest whole

Explanation / Answer

a) std error of mean =std deviaiton/(n)1/2 =3.15/(12)1/2 =0.0808

for 80% CI; z=1.28

therefore 80% confidence interval =sample mean -/+ z*std error =3.05 ; 3.25

margin of error E =z*std errror =0.10

b)

is known

normal distribution of weights

c)

There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region

d)

here E =0.13

z=1.28

therefore sample size required =(z*std deviation/E)1/2 =~8

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.