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 previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with sigma = 0.28 gram. 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.) normal distribution of weights n is large uniform distribution of weights sigma is known sigma is unknown (c) Interpret your results in the context of this problem. 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. The probability to the true average weight of Allen's hummingbirds is equal to the sample mean. There is an 00% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.60. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.09 for the mean weights of the hummingbirds. (Round up to the with a nearest whole number)

Explanation / Answer

Sol:

alpha=1-80%-1-0.8=0.2

alpha/2=0.2/2=0.1

sample mean=x bar=3.15

sample standard deviation=0.28

smaple size=n=12

n-1=12-1=11

t critical at 11 df and 0.1 level of significance=1.363

lower limit=sample mean-margin of error

margin of error=t crit*sample sd/sqrt(n)

=1.363*0.28/sqrt(12)

=0.1102

lower limit=3.15-0.1102=3.039

upper limit=3.15+0.1102=3.2602

we are 89% confident that the true population mean lies in between 3.039 and 3.2602

Solutionb:

normal distribution of weights

sigma is unknown

Solutionc:

Answer C

SolutionD:

n=requd sample size

n=[1.363*0.28/0.09]2

n=17.98

n=18

erequd sample size=18

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