Allen\'s hummingbird ( Selasphorus sasin ) has been studied by zoologist Bill Al
ID: 3218291 • Letter: A
Question
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 19 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 = 0.20 gram.
(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. (Round your answers to two decimal places.)
margin of error =
(b) What conditions are necessary for your calculations? (Select all that apply.)
normal distribution of weights
uniform distribution of weights
n is large
is unknown
is known
(c) Give a brief interpretation of 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.
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 that this interval contains the true average weight of Allen's hummingbirds is 0.80.
The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.
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.
(d) Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
______________hummingbirds
margin of error =
(b) What conditions are necessary for your calculations? (Select all that apply.)
normal distribution of weights
uniform distribution of weights
n is large
is unknown
is known
(c) Give a brief interpretation of 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.
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 that this interval contains the true average weight of Allen's hummingbirds is 0.80.
The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.
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.
(d) Find the sample size necessary for an 80% confidence level with a maximal error of estimate E = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
______________hummingbirds
Explanation / Answer
average weight = 3.15gms
sd = 0.2 gms
Thus the confidence interval (for 80%) is : (3.09 , 3.21)
using the formula
CI = x ± t/2 × (/n)
where x = Mean = Standard Deviation = 1 - (Confidence Level/100) t/2 = t-table value CI = Confidence Interval
b) condition necessary for calculation =
normal distribution of weights
n is large
is known
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